John M's ActivityTypepadTypepadtag:typepad.com,2003:profile.typepad.com/services/activity/atom/tag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://activitystrea.ms/schema/1.0/personhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6e00e00995bbf6883301bb095b3bb3970d John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T21:47:13Ztag:api.typepad.com,2009:6a00d83451b33869e201bb095b3bb2970dhttp://activitystrea.ms/schema/1.0/comment2016-12-03T21:47:13Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8b831ae970b John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T20:20:11Ztag:api.typepad.com,2009:6a00d83451bd4869e201b7c8b831ad970bhttp://activitystrea.ms/schema/1.0/comment2016-12-03T20:20:11Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307"Improvements in US Air Pollution"tag:api.typepad.com,2009:6a00d83451bd4869e200d8341c5c0553efEnvironmental Economicshttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d241eb6b970c John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T18:07:10Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d241eb6a970chttp://activitystrea.ms/schema/1.0/comment2016-12-03T18:07:10Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Minimum Wage Increases and Earnings in Low-Wage Jobstag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8b823be970b John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T17:49:14Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8b823bd970bhttp://activitystrea.ms/schema/1.0/comment2016-12-03T17:49:14Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Links for 12-03-16tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8b81a45970b John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T16:31:06Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8b81a44970bhttp://activitystrea.ms/schema/1.0/comment2016-12-03T16:31:05Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d241b754970c John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T08:37:32Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d241b753970chttp://activitystrea.ms/schema/1.0/comment2016-12-03T08:37:32Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Paul Krugman: Seduced and Betrayed by Donald Trumptag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8b7ebc0970b John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T07:34:41Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8b7ebbf970bhttp://activitystrea.ms/schema/1.0/comment2016-12-03T07:34:41Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8b7eb54970b John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T07:30:41Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8b7eb53970bhttp://activitystrea.ms/schema/1.0/comment2016-12-03T07:30:41Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb095af010970d John M posted something http://activitystrea.ms/schema/1.0/post2016-12-03T07:11:30Ztag:api.typepad.com,2009:6a00d83451b33869e201bb095af00f970dhttp://activitystrea.ms/schema/1.0/comment2016-12-03T07:11:30Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb0957b977970d John M posted something http://activitystrea.ms/schema/1.0/post2016-11-28T04:10:42Ztag:api.typepad.com,2009:6a00d83451b33869e201bb0957b976970dhttp://activitystrea.ms/schema/1.0/comment2016-11-28T04:10:42Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Should We Worry About the Top 1%, or Praise Them?tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8b4af09970b John M posted something http://activitystrea.ms/schema/1.0/post2016-11-28T03:56:46Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8b4af08970bhttp://activitystrea.ms/schema/1.0/comment2016-11-28T03:56:46Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307A Secular Religion That Lasted One Centurytag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb0956fc56970d John M posted something http://activitystrea.ms/schema/1.0/post2016-11-26T08:25:09Ztag:api.typepad.com,2009:6a00d83451b33869e201bb0956fc55970dhttp://activitystrea.ms/schema/1.0/comment2016-11-26T08:25:09Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d23cdcb6970c John M posted an entry http://activitystrea.ms/schema/1.0/post2016-11-23T17:04:01Ztag:api.typepad.com,2009:6a00e00995bbf6883301b8d23cdcb4970cMichael Moore Predicted Trump Victory.http://activitystrea.ms/schema/1.0/article2016-11-23T17:04:00Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>Michael Moore predicted this in late October.  I think that he was only partially right, because Hillary Clinton won the popular vote.  There's also the question of how much of our election was hacked by voting computers that don't leave a trail of how a vote actually went.  There are very good computer hackers (<a href="https://stallman.org/articles/on-hacking.html" title="Richard Stallman on Hacking">in the best sense</a>) who are hopefully dedicated to detecting and exposing electronic electoral fraud.</p> <p><a href="https://youtu.be/YKeYbEOSqYc" title="Michael Moore predicts Trump Victory">https://youtu.be/YKeYbEOSqYc</a></p> <p> </p> <p>Michael Moore predicted this in late October.  I think that he was only partially right, because Hillary Clinton won the popular vote.  There's also the question of how much of our election was hacked by voting computers that don't leave a trail of how a vote actually went.  There are very good computer hackers (<a href="https://stallman.org/articles/on-hacking.html" title="Richard Stallman on Hacking">in the best sense</a>) who are hopefully dedicated to detecting and exposing electronic electoral fraud.</p> <p><a href="https://youtu.be/YKeYbEOSqYc" title="Michael Moore predicts Trump Victory">https://youtu.be/YKeYbEOSqYc</a></p> <p> </p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb09556388970d John M posted something http://activitystrea.ms/schema/1.0/post2016-11-21T22:55:41Ztag:api.typepad.com,2009:6a00d83451bd4869e201bb09556387970dhttp://activitystrea.ms/schema/1.0/comment2016-11-21T22:55:41Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307"Presidential Forecast Post-Mortem"tag:api.typepad.com,2009:6a00d83451bd4869e200d8341c5c0553efEnvironmental Economicshttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d23c2bad970c John M posted something http://activitystrea.ms/schema/1.0/post2016-11-21T22:07:01Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d23c2bac970chttp://activitystrea.ms/schema/1.0/comment2016-11-21T22:07:01Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb09555bf2970d John M posted something http://activitystrea.ms/schema/1.0/post2016-11-21T21:44:58Ztag:api.typepad.com,2009:6a00d83451b33869e201bb09555bf1970dhttp://activitystrea.ms/schema/1.0/comment2016-11-21T21:44:57Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307What Size Fiscal Deficits for the United States?tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb0953496b970d John M posted an entry http://activitystrea.ms/schema/1.0/post2016-11-16T06:55:06Ztag:api.typepad.com,2009:6a00e00995bbf6883301bb095348be970dGive Trump a Chance?http://activitystrea.ms/schema/1.0/article2016-11-16T06:55:05Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>Should we give President-Elect Donald Trump a chance, as President Obama urges us?  Or should the Democrats buckle down and obstruct the first bad thing he does?</p> <p>My answer is both.</p> <p>Give him a chance to do the right thing, or at least a good thing.  Fight any bad thing he might do.  Let's not worry about the hypocrisy involved in fighting bad things.  The hypocrisy I'm thinking of is the refusal of mainstream Democrats in Congress to fight President Obama's bad things.  I'm thinking in particular of "Humanitarian Bombings" and other militaristic adventures.  We must fight the evil.</p> <p>Just remember, we have a long history of training and arming "moderate forces" only for them to turn "radical" and against us.  Remember further, that we have a long history of our intelligence agencies lying to us, and having their own agendas contrary to our own policies and fundamental principals.  In fact, our elected officials have had agendas different from those they told us.</p> <p>For example, is a faction of the CIA a major player in the heroin trade in Afghanistan?  The only business the CIA has in any form of the drug production and trade is to have informants.</p> <p>It may turn out that we have to fight ISIS (the "Islamic State") because of how horrible they are.  If so, that should be the one and only battle we take on in the Middle East -- and we should give it our all.  There appears to be a major battle with Iraq trying to take back a major Iraqi city from ISIS.   I wouldn't object to sending a few companies of ground troops in to fight and take over the city -- and hold it, and perhaps go on and further fight ISIS.  Staying in the city should be only done if the residents welcome us.  We are supposed to free the city, not conquer it.  So if they don't want us there, we either withdraw (risking ISIS's return) or go on after ISIS as they withdraw.</p> <p>But remember, our intelligence agencies, our government, and our mainstream media lie about what we are doing in war.  Truth is the first casualty.  Also, as some former military members have argued, war is a scam to enrich the military industry.</p> <p>Should we give President-Elect Donald Trump a chance, as President Obama urges us?  Or should the Democrats buckle down and obstruct the first bad thing he does?</p> <p>My answer is both.</p> <p>Give him a chance to do the right thing, or at least a good thing.  Fight any bad thing he might do.  Let's not worry about the hypocrisy involved in fighting bad things.  The hypocrisy I'm thinking of is the refusal of mainstream Democrats in Congress to fight President Obama's bad things.  I'm thinking in particular of "Humanitarian Bombings" and other militaristic adventures.  We must fight the evil.</p> <p>Just remember, we have a long history of training and arming "moderate forces" only for them to turn "radical" and against us.  Remember further, that we have a long history of our intelligence agencies lying to us, and having their own agendas contrary to our own policies and fundamental principals.  In fact, our elected officials have had agendas different from those they told us.</p> <p>For example, is a faction of the CIA a major player in the heroin trade in Afghanistan?  The only business the CIA has in any form of the drug production and trade is to have informants.</p> <p>It may turn out that we have to fight ISIS (the "Islamic State") because of how horrible they are.  If so, that should be the one and only battle we take on in the Middle East -- and we should give it our all.  There appears to be a major battle with Iraq trying to take back a major Iraqi city from ISIS.   I wouldn't object to sending a few companies of ground troops in to fight and take over the city -- and hold it, and perhaps go on and further fight ISIS.  Staying in the city should be only done if the residents welcome us.  We are supposed to free the city, not conquer it.  So if they don't want us there, we either withdraw (risking ISIS's return) or go on after ISIS as they withdraw.</p> <p>But remember, our intelligence agencies, our government, and our mainstream media lie about what we are doing in war.  Truth is the first casualty.  Also, as some former military members have argued, war is a scam to enrich the military industry.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb0951b6bf970d John M posted something http://activitystrea.ms/schema/1.0/post2016-11-11T18:01:46Ztag:api.typepad.com,2009:6a00d83451b33869e201bb0951b6be970dhttp://activitystrea.ms/schema/1.0/comment2016-11-11T18:01:46Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Paul Krugman: Thoughts for the Horrifiedtag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d238509b970c John M posted an entry http://activitystrea.ms/schema/1.0/post2016-11-11T02:42:07Ztag:api.typepad.com,2009:6a00e00995bbf6883301b8d2385097970cThe "Competitive Enterprise Institute"http://activitystrea.ms/schema/1.0/article2016-11-11T02:42:06Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>Public radio this morning reported a report that Donald Trump has picked a major climate-change denier from the CEI as part of his transition team -- in particular, the part that deals with global warming and climate change.  Send a hired gun to do a scientist's job.</p> <p>Remember, the CEI gave us the ad, "CO<sub>2</sub>: they call it pollution, we call it life."  Remember the parodies: "Water: they call it drowning, we call it life."  "Shit: they call it [random vomit sound], we call it life."</p> <p>The CEI also gave us the phony fit to the scatter plot of corporate income tax rates.  They fit a boundary, not a curve to the data, and called it the Laffer curve.  Those were completely unrelated.</p> <p>Whom do we believe?  Libertarians, Limbaugh, and LaRouche?  Hired guns?  Or working scientists?  In a subject requiring scientific expertise?</p> <p>Public radio this morning reported a report that Donald Trump has picked a major climate-change denier from the CEI as part of his transition team -- in particular, the part that deals with global warming and climate change.  Send a hired gun to do a scientist's job.</p> <p>Remember, the CEI gave us the ad, "CO<sub>2</sub>: they call it pollution, we call it life."  Remember the parodies: "Water: they call it drowning, we call it life."  "Shit: they call it [random vomit sound], we call it life."</p> <p>The CEI also gave us the phony fit to the scatter plot of corporate income tax rates.  They fit a boundary, not a curve to the data, and called it the Laffer curve.  Those were completely unrelated.</p> <p>Whom do we believe?  Libertarians, Limbaugh, and LaRouche?  Hired guns?  Or working scientists?  In a subject requiring scientific expertise?</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb0950f970970d John M posted an entry http://activitystrea.ms/schema/1.0/post2016-11-09T18:44:04Ztag:api.typepad.com,2009:6a00e00995bbf6883301bb0950f96e970dThe Term "Luser"http://activitystrea.ms/schema/1.0/article2016-11-09T18:44:04Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>Trump's election victory and with the use of electronic voting machines, remind me of the story of the creation of the term, "Luser".</p> <p>Many places in the US use electronic voting machines without any kind of provision for auditing: the machines could be either programmed or hacked to wrongly record a person's vote, and we would have no direct evidence.  Normally, I'd be concerned that the machines would be hacked for Republican victories.  This election, I was concerned that they'd be hacked for a Clinton victory.</p> <p>But suppose things were going like this: hackers in the background were working both sides, programming and reprogramming the machines to deliver votes to their side and away from the other side.  At MIT, when they had large multiuser computer systems, everyone logging in was greated with the message of the form, "150 Users Logged In".  Someone had the silly idea of changing "Users" to "Losers", leading to "150 Losers Logged In".  Others didn't like the message, and tried to revert it.  At any given time, it was an open question whether one would see "Users" or "Losers".</p> <p>Someone tried "Lusers", and it stuck.</p> <p>Suppose something similar happened with the voting machines?  Unlike the Luser case, there could be no compromise result.  Also unlike the Luser case, there was a time limit.  So instead of a compromise result, the side that hacked last hacked best -- in a manner of speaking.</p> <p>I seriously don't know whether voting-machine hacking won the final vote.  If it did, the loser was democracy itself.</p> <p>Trump's election victory and with the use of electronic voting machines, remind me of the story of the creation of the term, "Luser".</p> <p>Many places in the US use electronic voting machines without any kind of provision for auditing: the machines could be either programmed or hacked to wrongly record a person's vote, and we would have no direct evidence.  Normally, I'd be concerned that the machines would be hacked for Republican victories.  This election, I was concerned that they'd be hacked for a Clinton victory.</p> <p>But suppose things were going like this: hackers in the background were working both sides, programming and reprogramming the machines to deliver votes to their side and away from the other side.  At MIT, when they had large multiuser computer systems, everyone logging in was greated with the message of the form, "150 Users Logged In".  Someone had the silly idea of changing "Users" to "Losers", leading to "150 Losers Logged In".  Others didn't like the message, and tried to revert it.  At any given time, it was an open question whether one would see "Users" or "Losers".</p> <p>Someone tried "Lusers", and it stuck.</p> <p>Suppose something similar happened with the voting machines?  Unlike the Luser case, there could be no compromise result.  Also unlike the Luser case, there was a time limit.  So instead of a compromise result, the side that hacked last hacked best -- in a manner of speaking.</p> <p>I seriously don't know whether voting-machine hacking won the final vote.  If it did, the loser was democracy itself.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb09506413970d John M posted an entry http://activitystrea.ms/schema/1.0/post2016-11-08T04:53:25Ztag:api.typepad.com,2009:6a00e00995bbf6883301bb09506410970dhttp://activitystrea.ms/schema/1.0/article2016-11-08T04:53:25Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307The following argument in my previous post was meant facetiously, and is invalid.<br/> <br/> "Okay, let's define a validity function v(n), with v(n)=1 meaning valid, and v(n)=0 meaning invalid. We're being told that v(n)=1 for any finite n, but approaches zero as n approaches infinity -- increases without bound. No, "v(n)=1 for any finite n" means that v(n) stays 1 as n approaches infinity.<br/> <br/> This may be a counterexample: any finite set of numbers has a maximum. So let S_n be a finite set for each positive n. The union (for n=1 to N) of S_n is also finite (where N is a positive integer). But the union (for n=1 to infinity) of S_n is not finite, and need not have a maximum. So we might say V(N) is 1 but lim (as N->infinity) of V(N) is 0.<br/> <br/> Now that I think of it, that's not a genuine counterexample, because the union (for n=1 to infinity) of S_n is not the lim (as N->infinity) of the union (for n=1 to N) of S_n -- at least in the sense of any convergence.<br/> The following argument in my previous post was meant facetiously, and is invalid.<br/> <br/> "Okay, let's define a validity function v(n), with v(n)=1 meaning valid, and v(n)=0 meaning invalid. We're being told that v(n)=1 for any finite n, but approaches zero as n approaches infinity -- increases without bound. No, "v(n)=1 for any finite n" means that v(n) stays 1 as n approaches infinity.<br/> <br/> This may be a counterexample: any finite set of numbers has a maximum. So let S_n be a finite set for each positive n. The union (for n=1 to N) of S_n is also finite (where N is a positive integer). But the union (for n=1 to infinity) of S_n is not finite, and need not have a maximum. So we might say V(N) is 1 but lim (as N->infinity) of V(N) is 0.<br/> <br/> Now that I think of it, that's not a genuine counterexample, because the union (for n=1 to infinity) of S_n is not the lim (as N->infinity) of the union (for n=1 to N) of S_n -- at least in the sense of any convergence.<br/> tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d23702fc970c John M posted something http://activitystrea.ms/schema/1.0/post2016-11-07T19:26:32Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d23702fb970chttp://activitystrea.ms/schema/1.0/comment2016-11-07T19:26:32Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Paul Krugman: How to Rig an Electiontag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d2366b86970c John M posted something http://activitystrea.ms/schema/1.0/post2016-11-06T01:17:15Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d2366b85970chttp://activitystrea.ms/schema/1.0/comment2016-11-06T01:17:15Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8acac23970b John M posted something http://activitystrea.ms/schema/1.0/post2016-11-06T00:39:25Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8acac22970bhttp://activitystrea.ms/schema/1.0/comment2016-11-06T00:39:24Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8abf2de970b John M posted something http://activitystrea.ms/schema/1.0/post2016-11-03T20:14:24Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8abf2dd970bhttp://activitystrea.ms/schema/1.0/comment2016-11-03T20:14:24Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8abbc92970b John M posted something http://activitystrea.ms/schema/1.0/post2016-11-03T10:02:48Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8abbc91970bhttp://activitystrea.ms/schema/1.0/comment2016-11-03T10:02:48Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Why We Need Social Insurancetag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb094c53bf970d John M posted an entry http://activitystrea.ms/schema/1.0/post2016-10-29T16:17:32Ztag:api.typepad.com,2009:6a00e00995bbf6883301bb094c53bd970dFurther Reflections on the Proofhttp://activitystrea.ms/schema/1.0/article2016-10-29T16:17:32Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>Since the comments section to <a href="http://johnm2.typepad.com/blog/2016/10/whats-wrong-with-this-proof.html" title="Infinite Set Has Countably Infinite Subset">this post</a> (proof that any infinite set has a countably infinite subset) has been closed, I reiterate my question, what's wrong with that proof?  I will try to imagine what might go wrong if I don't assume any version of the Axiom of Choice.  S is the infinite set (a set with an infinite number of elements)</p> <p>The summary of the proof is that we let <em>x<sub>1</sub></em> be an element in S.  Then assume <em>x<sub>1</sub></em> ... <em>x<sub>n</sub></em> are distinct members of S, and let <em>x<sub>n+1</sub></em> be a member of S with <em>x<sub>1</sub></em> ... <em>x<sub>n</sub></em> removed; <em>x<sub>1</sub></em> ... <em>x<sub>n+1</sub></em> become distinct members of S.  By (strong) induction, for each positive integer <em>n</em>, <em>x<sub>n</sub></em> exists, and they all form a countably infinite set that's a subset of S.</p> <p>So what's wrong here?  The Proof-Wiki <a href="https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset" title="Proof-Wiki Warning">warning</a> says that this works for any finite value of <em>n</em>, but repeating it an infinite number of times requires a version of the Axiom of choice.  Here's a clue: <em>n</em> is finite.  The fact that <em>n</em> can be any positive integer makes the set {<em>x<sub>n</sub></em>} countably infinite.</p> <p>For this to fail, there has to be some (finite) value of <em>n</em> for which the choice of <em>x<sub>n+1</sub></em> fails.  The warning appears to tell us that this method is valid for finite <em>n</em>, but invalid as <em>n</em> approaches infinity.  Okay, let's define a validity function <em>v</em>(<em>n</em>), with <em>v</em>(<em>n</em>)=1 meaning valid, and <em>v</em>(<em>n</em>)=0 meaning invalid.  We're being told that <em>v</em>(<em>n</em>)=1 for any finite <em>n</em>, but approaches zero as <em>n</em> approaches infinity -- increases without bound.  No, "<em>v</em>(<em>n</em>)=1 for any finite <em>n</em>" means that <em>v</em>(<em>n</em>) stays 1 as <em>n</em> approaches infinity.</p> <p>Since the comments section to <a href="http://johnm2.typepad.com/blog/2016/10/whats-wrong-with-this-proof.html" title="Infinite Set Has Countably Infinite Subset">this post</a> (proof that any infinite set has a countably infinite subset) has been closed, I reiterate my question, what's wrong with that proof?  I will try to imagine what might go wrong if I don't assume any version of the Axiom of Choice.  S is the infinite set (a set with an infinite number of elements)</p> <p>The summary of the proof is that we let <em>x<sub>1</sub></em> be an element in S.  Then assume <em>x<sub>1</sub></em> ... <em>x<sub>n</sub></em> are distinct members of S, and let <em>x<sub>n+1</sub></em> be a member of S with <em>x<sub>1</sub></em> ... <em>x<sub>n</sub></em> removed; <em>x<sub>1</sub></em> ... <em>x<sub>n+1</sub></em> become distinct members of S.  By (strong) induction, for each positive integer <em>n</em>, <em>x<sub>n</sub></em> exists, and they all form a countably infinite set that's a subset of S.</p> <p>So what's wrong here?  The Proof-Wiki <a href="https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset" title="Proof-Wiki Warning">warning</a> says that this works for any finite value of <em>n</em>, but repeating it an infinite number of times requires a version of the Axiom of choice.  Here's a clue: <em>n</em> is finite.  The fact that <em>n</em> can be any positive integer makes the set {<em>x<sub>n</sub></em>} countably infinite.</p> <p>For this to fail, there has to be some (finite) value of <em>n</em> for which the choice of <em>x<sub>n+1</sub></em> fails.  The warning appears to tell us that this method is valid for finite <em>n</em>, but invalid as <em>n</em> approaches infinity.  Okay, let's define a validity function <em>v</em>(<em>n</em>), with <em>v</em>(<em>n</em>)=1 meaning valid, and <em>v</em>(<em>n</em>)=0 meaning invalid.  We're being told that <em>v</em>(<em>n</em>)=1 for any finite <em>n</em>, but approaches zero as <em>n</em> approaches infinity -- increases without bound.  No, "<em>v</em>(<em>n</em>)=1 for any finite <em>n</em>" means that <em>v</em>(<em>n</em>) stays 1 as <em>n</em> approaches infinity.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8a61bdd970b John M posted something http://activitystrea.ms/schema/1.0/post2016-10-23T12:54:12Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8a61bdc970bhttp://activitystrea.ms/schema/1.0/comment2016-10-23T12:54:12Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Debt, Diversion, Distractiontag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8a61b68970b John M posted something http://activitystrea.ms/schema/1.0/post2016-10-23T12:41:58Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8a61b67970bhttp://activitystrea.ms/schema/1.0/comment2016-10-23T12:41:58Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Debt, Diversion, Distractiontag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d2284ed4970c John M posted an entry http://activitystrea.ms/schema/1.0/post2016-10-11T06:53:42Ztag:api.typepad.com,2009:6a00e00995bbf6883301b8d2284ed1970cWhat's Wrong With This Proof?http://activitystrea.ms/schema/1.0/article2016-10-11T06:53:42Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>The theorem that I am going to attempt to prove is that every infinite set has a countably infinite subset.  Someone explain to me without a bald assertion what's wrong with this proof.  I use strong induction.  Let S be the infinite set.</p> <p style="padding-left: 30px;">First: let x<sub>1</sub> be a member of S.</p> <p style="padding-left: 30px;">Second: Assume that <em>x<sub>k</sub></em>, for all k from 1 to <em>n</em>, exist, are members of S, and are distinct.  Let <em>x<sub>k+1</sub></em> be a member of S - U<em><sup>n</sup><sub>k=1</sub></em>{<em>x<sub>k</sub></em>} (meaning S without <em>x<sub>1</sub></em>...<em>x<sub>n</sub></em>).  Then <em>x<sub>k</sub></em>, for all <em>k</em> from 1 to <em>n</em>+1, exists, are members of S, and are distinct.</p> <p style="padding-left: 30px;">Conclusion: <em>x<sub>n</sub></em> exists for all positive integers <em>n</em>.  They are distinct members of S.  Their union, U<sub><em>k=1</em></sub><em><sup>infinity</sup></em>{<em>x<sub>k</sub></em>} is a countable subset of S.  QED.</p> <p>Again, someone tell me what's wrong with this proof, without a bald assertion.  The bald assertion I'm thinking of is that some version of the Axiom of Choice is needed to get a full infinite set of elements <em>x<sub>n</sub></em>.  <a href="https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset" title="Proof Wiki on the proof that an infinite set has a countably infinite subset.">Proof Wiki</a> gives just this warning.  However, this method proves that <em>x<sub>n</sub></em> exists for all positive integers <em>n</em>, and only uses only a finite number of <em>x</em> values to get each new <em>x</em> value.</p> <p>At least, they cite the ZF set theory for this assertion.  I should learn and understand the theory.</p> <p>Someone?  I'm listening.</p> <p>The theorem that I am going to attempt to prove is that every infinite set has a countably infinite subset.  Someone explain to me without a bald assertion what's wrong with this proof.  I use strong induction.  Let S be the infinite set.</p> <p style="padding-left: 30px;">First: let x<sub>1</sub> be a member of S.</p> <p style="padding-left: 30px;">Second: Assume that <em>x<sub>k</sub></em>, for all k from 1 to <em>n</em>, exist, are members of S, and are distinct.  Let <em>x<sub>k+1</sub></em> be a member of S - U<em><sup>n</sup><sub>k=1</sub></em>{<em>x<sub>k</sub></em>} (meaning S without <em>x<sub>1</sub></em>...<em>x<sub>n</sub></em>).  Then <em>x<sub>k</sub></em>, for all <em>k</em> from 1 to <em>n</em>+1, exists, are members of S, and are distinct.</p> <p style="padding-left: 30px;">Conclusion: <em>x<sub>n</sub></em> exists for all positive integers <em>n</em>.  They are distinct members of S.  Their union, U<sub><em>k=1</em></sub><em><sup>infinity</sup></em>{<em>x<sub>k</sub></em>} is a countable subset of S.  QED.</p> <p>Again, someone tell me what's wrong with this proof, without a bald assertion.  The bald assertion I'm thinking of is that some version of the Axiom of Choice is needed to get a full infinite set of elements <em>x<sub>n</sub></em>.  <a href="https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset" title="Proof Wiki on the proof that an infinite set has a countably infinite subset.">Proof Wiki</a> gives just this warning.  However, this method proves that <em>x<sub>n</sub></em> exists for all positive integers <em>n</em>, and only uses only a finite number of <em>x</em> values to get each new <em>x</em> value.</p> <p>At least, they cite the ZF set theory for this assertion.  I should learn and understand the theory.</p> <p>Someone?  I'm listening.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d22419c0970c John M posted something http://activitystrea.ms/schema/1.0/post2016-10-02T02:44:03Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d22419bf970chttp://activitystrea.ms/schema/1.0/comment2016-10-02T02:44:03Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Paul Krugman: How the Clinton-Trump Race Got Closetag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c898df6c970b John M posted something http://activitystrea.ms/schema/1.0/post2016-09-29T05:33:33Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c898df6b970bhttp://activitystrea.ms/schema/1.0/comment2016-09-29T05:33:33Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d222bf5a970c John M posted something http://activitystrea.ms/schema/1.0/post2016-09-29T05:32:56Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d222bf59970chttp://activitystrea.ms/schema/1.0/comment2016-09-29T05:32:56Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d2220a97970c John M posted something http://activitystrea.ms/schema/1.0/post2016-09-28T03:21:02Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d2220a96970chttp://activitystrea.ms/schema/1.0/comment2016-09-28T03:21:02Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Why Study Economics?tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d2213e22970c John M posted something http://activitystrea.ms/schema/1.0/post2016-09-26T19:54:54Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d2213e21970chttp://activitystrea.ms/schema/1.0/comment2016-09-26T19:54:54Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb093ad98a970d John M posted something http://activitystrea.ms/schema/1.0/post2016-09-26T19:29:26Ztag:api.typepad.com,2009:6a00d83451b33869e201bb093ad989970dhttp://activitystrea.ms/schema/1.0/comment2016-09-26T19:29:26Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d221306d970c John M posted something http://activitystrea.ms/schema/1.0/post2016-09-26T17:40:23Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d221306c970chttp://activitystrea.ms/schema/1.0/comment2016-09-26T17:40:23Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Trump Picks Top Climate Skeptic to Lead EPA Transition tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d21afdc5970c John M posted something http://activitystrea.ms/schema/1.0/post2016-09-09T17:12:50Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d21afdc4970chttp://activitystrea.ms/schema/1.0/comment2016-09-09T17:12:50Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c890d47a970b John M posted something http://activitystrea.ms/schema/1.0/post2016-09-08T15:36:51Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c890d479970bhttp://activitystrea.ms/schema/1.0/comment2016-09-08T15:36:51Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb093447fe970d John M posted something http://activitystrea.ms/schema/1.0/post2016-09-08T15:17:52Ztag:api.typepad.com,2009:6a00d83451b33869e201bb093447fd970dhttp://activitystrea.ms/schema/1.0/comment2016-09-08T15:17:52Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c890d0de970b John M posted something http://activitystrea.ms/schema/1.0/post2016-09-08T15:01:13Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c890d0dd970bhttp://activitystrea.ms/schema/1.0/comment2016-09-08T15:01:13Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb09344602970d John M posted something http://activitystrea.ms/schema/1.0/post2016-09-08T14:59:13Ztag:api.typepad.com,2009:6a00d83451b33869e201bb09344601970dhttp://activitystrea.ms/schema/1.0/comment2016-09-08T14:59:13Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d219c1fd970c John M posted something http://activitystrea.ms/schema/1.0/post2016-09-06T15:18:06Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d219c1fc970chttp://activitystrea.ms/schema/1.0/comment2016-09-06T15:18:06Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Paul Krugman: Hillary Clinton Gets Goredtag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb09337734970d John M posted something http://activitystrea.ms/schema/1.0/post2016-09-06T14:52:11Ztag:api.typepad.com,2009:6a00d83451b33869e201bb09337733970dhttp://activitystrea.ms/schema/1.0/comment2016-09-06T14:52:11Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb09337580970d John M posted something http://activitystrea.ms/schema/1.0/post2016-09-06T14:40:59Ztag:api.typepad.com,2009:6a00d83451b33869e201bb0933757f970dhttp://activitystrea.ms/schema/1.0/comment2016-09-06T14:40:59Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Paul Krugman: Hillary Clinton Gets Goredtag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c88eb083970b John M posted something http://activitystrea.ms/schema/1.0/post2016-09-02T14:25:09Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c88eb082970bhttp://activitystrea.ms/schema/1.0/comment2016-09-02T14:25:09Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Links for 09-02-16tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d21719c2970c John M posted an entry http://activitystrea.ms/schema/1.0/post2016-08-30T02:12:52Ztag:api.typepad.com,2009:6a00e00995bbf6883301b8d21719c0970cThe Cartesian Producthttp://activitystrea.ms/schema/1.0/article2016-08-30T02:12:52Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>The Cartesian product of an infinite number of non-empty sets can't be non-empty. It may sometimes be ill-defined (or even undefined) for an uncountable number of sets, but it can't be empty. The proof is simple: if every set has one element, then the Cartesian product has one element.<br /> <br /> Increase the number of elements in any set, the number in the Cartesian product increases as well. If the number is already infinity, the cardinality may remain unchanged, but it can't decrease.</p> <p>The Cartesian product of an infinite number of non-empty sets can't be non-empty. It may sometimes be ill-defined (or even undefined) for an uncountable number of sets, but it can't be empty. The proof is simple: if every set has one element, then the Cartesian product has one element.<br /> <br /> Increase the number of elements in any set, the number in the Cartesian product increases as well. If the number is already infinity, the cardinality may remain unchanged, but it can't decrease.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d216c6f4970c John M posted an entry http://activitystrea.ms/schema/1.0/post2016-08-29T11:36:05Ztag:api.typepad.com,2009:6a00e00995bbf6883301b8d216c6f2970cOn Gödel's Theoremhttp://activitystrea.ms/schema/1.0/article2016-08-29T11:36:05Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>I came across this <a href="http://mathlair.allfunandgames.ca/godelstheorem.php" title="Gödel's Theorem">elementary introduction to Gödel's theorem</a> on incompleteness in math.  It contained this statement:</p> <blockquote> <p>[C]onsider a system that consists of the <a href="http://mathlair.allfunandgames.ca/axiom.php">axioms</a> of arithmetic excluding the axiom of <a href="http://mathlair.allfunandgames.ca/induction.php">mathematical induction</a>. In this system, it would be impossible to prove the following statement (among others), even though it is true:</p> <p style="text-align: center;">1 + 2 + 3 + 4 + 5 + ... + <em>n </em>=<em> n</em>(<em>n+1</em>)/2</p> </blockquote> <p>Oh really?  I put it to you, the left side of the equality can't even be defined without mathematical induction.</p> <p>Let's be clearer.  I'll give my axioms defining the natural numbers (positive integers) <strong>N</strong>:</p> <ul> <li><strong>N</strong> contains <strong>one</strong>.</li> <li>There exists a injective successor function <em>s</em>: <strong>N</strong> --> <strong>N</strong>-{<strong>one</strong>}.  (An injective function is defined: if <em>s</em>(<em>m</em>) = <em>s</em>(<em>n</em>) then <em>m </em>= <em>n</em>.  An injective function is also known as "one-to-one".)</li> <li>Suppose we have a theorem (or claim) for the natural numbers (say T<sub><em>n</em></sub>).  We prove T<sub><strong>one</strong></sub>, and then prove if T<sub><em>n</em></sub> then T<sub><em>s</em>(<em>n</em>)</sub> (equivalently, T<sub><em>n</em></sub> implies T<sub><em>s</em>(<em>n</em>)</sub>). Then T<sub><em>n</em></sub> is guaranteed true for all natural numbers <em>n</em>.</li> </ul> <p><strong>N</strong>-{<strong>one</strong>} means <strong>N</strong> without <strong>one</strong>.  I use that (and the injective requirement) to ensure that every number is followed by a new and different number.  The final axiom is the axiom of mathematical induction -- or the special case starting with <strong>one</strong>.  Remove it, and what can happen?  Here are some sets that will satisfy the beginning and middle axioms:</p> <ul> <li>The positive integers: 1, 2, 3, ...</li> <li>The positive integers and half-integers: 0.5, 1, 1.5, 2, 2.5, 3, 3.5 ...</li> <li>The following numbers: 0.2, 1, 1.2, 2, 2.2, 3, 3.2 ...</li> <li>The rational numbers excluding zero and the negative integers</li> <li>The real numbers excluding zero and the negative integers</li> </ul> <p>Now consider any of the sets which include 13.5.  How do we even define 1 + 2 + 3 + 4 + 5 + ... + 13.5?  Clearly the theorem, valid for positive integers (okay, possibly zero as well) is invalid in general on sets satisfying the other axioms.  The induction axiom excludes all but the first example.</p> <p>In my definition of the natural numbers, the induction axiom is part of the definition.  Remove it, and we have so many other possibilities satisfying the other axioms that statements proved using induction in general aren't true.</p> <p>I came across this <a href="http://mathlair.allfunandgames.ca/godelstheorem.php" title="Gödel's Theorem">elementary introduction to Gödel's theorem</a> on incompleteness in math.  It contained this statement:</p> <blockquote> <p>[C]onsider a system that consists of the <a href="http://mathlair.allfunandgames.ca/axiom.php">axioms</a> of arithmetic excluding the axiom of <a href="http://mathlair.allfunandgames.ca/induction.php">mathematical induction</a>. In this system, it would be impossible to prove the following statement (among others), even though it is true:</p> <p style="text-align: center;">1 + 2 + 3 + 4 + 5 + ... + <em>n </em>=<em> n</em>(<em>n+1</em>)/2</p> </blockquote> <p>Oh really?  I put it to you, the left side of the equality can't even be defined without mathematical induction.</p> <p>Let's be clearer.  I'll give my axioms defining the natural numbers (positive integers) <strong>N</strong>:</p> <ul> <li><strong>N</strong> contains <strong>one</strong>.</li> <li>There exists a injective successor function <em>s</em>: <strong>N</strong> --> <strong>N</strong>-{<strong>one</strong>}.  (An injective function is defined: if <em>s</em>(<em>m</em>) = <em>s</em>(<em>n</em>) then <em>m </em>= <em>n</em>.  An injective function is also known as "one-to-one".)</li> <li>Suppose we have a theorem (or claim) for the natural numbers (say T<sub><em>n</em></sub>).  We prove T<sub><strong>one</strong></sub>, and then prove if T<sub><em>n</em></sub> then T<sub><em>s</em>(<em>n</em>)</sub> (equivalently, T<sub><em>n</em></sub> implies T<sub><em>s</em>(<em>n</em>)</sub>). Then T<sub><em>n</em></sub> is guaranteed true for all natural numbers <em>n</em>.</li> </ul> <p><strong>N</strong>-{<strong>one</strong>} means <strong>N</strong> without <strong>one</strong>.  I use that (and the injective requirement) to ensure that every number is followed by a new and different number.  The final axiom is the axiom of mathematical induction -- or the special case starting with <strong>one</strong>.  Remove it, and what can happen?  Here are some sets that will satisfy the beginning and middle axioms:</p> <ul> <li>The positive integers: 1, 2, 3, ...</li> <li>The positive integers and half-integers: 0.5, 1, 1.5, 2, 2.5, 3, 3.5 ...</li> <li>The following numbers: 0.2, 1, 1.2, 2, 2.2, 3, 3.2 ...</li> <li>The rational numbers excluding zero and the negative integers</li> <li>The real numbers excluding zero and the negative integers</li> </ul> <p>Now consider any of the sets which include 13.5.  How do we even define 1 + 2 + 3 + 4 + 5 + ... + 13.5?  Clearly the theorem, valid for positive integers (okay, possibly zero as well) is invalid in general on sets satisfying the other axioms.  The induction axiom excludes all but the first example.</p> <p>In my definition of the natural numbers, the induction axiom is part of the definition.  Remove it, and we have so many other possibilities satisfying the other axioms that statements proved using induction in general aren't true.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c8896bed970b John M posted an entry http://activitystrea.ms/schema/1.0/post2016-08-20T19:24:39Ztag:api.typepad.com,2009:6a00e00995bbf6883301b7c8896beb970bMore on the Axiom of Choicehttp://activitystrea.ms/schema/1.0/article2016-08-20T19:24:39Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>I found something at <a href="http://www.math.vanderbilt.edu/~schectex/ccc/choice.html" title="Vanderbilt's Axiom of Choice">Vanderbilt College</a> about the "Axiom of Choice".  The author poses this problem: he wants to define "small" sets of positive integers.  He wants the definition to including the following properties:</p> <ol style="list-style-type: lower-alpha;"> <li>Any set with zero or one members is "small". </li> <li>Any union of two "small" sets is "small". </li> <li>A set is "small" if and only if its complement isn't "small."</li> </ol> <p>The "complement" of a set S means the set of positive integers not in S.  I might mention that the sets of positive integers is uncountably infinite.  If a proof is needed, here's a simple one:</p> <p style="padding-left: 30px;">Suppose we have sets of positive integers A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>,...  Let <em>n</em> be a positive integer.  Either <em>n</em> is in A<sub><em>n</em></sub> or it's not.  Define N = {<em>n</em>:<em>n</em> is not in A<sub><em>n</em></sub>}.  N is a defined set given A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>,...  Then <em>n</em> is in N if and only if <em>n</em> is not in A<sub><em>n</em></sub>.  That means that N can't be A<sub><em>n</em></sub> for any <em>n</em>.</p> <p>Any alleged one-to-one mapping of the positive integers onto the sets of positive integers is defeated by the existence of another set outside of the list of sets.  Getting a one-to-one mapping for all sets of integers is literally impossible.  The sets of positive integers must be uncountable.  As I commented in my previous post, things get squirrelly with the uncountable.</p> <p>Back to the author's problem, it's important to disconnect our own notion of "small" with "small" as he would like to define it.  If we keep to our intuitive notion of small, property a satisfies it.  Property a and b together mean that any finite set must be "small".  That includes a set with a googleplex of elements.  But a googleplex is tiny compared with infinity.  A finite set may or may not be intuitively small.  But necessarily, it must be "small" if properties a and b are satisfied.</p> <p>The author gives examples where any two of the three properties are satisfied:</p> <ol> <li>A "small" set is defined as a finite set.  That fails to satisfy property c, because neither the evens nor the odds are "small".</li> <li>A "small" set is a set not containing the number 1.  This satisfies properties b and c, but not a: {1} isn't small by this definition.</li> <li>A "small" set contains at most one of the first three numbers.  This satisfies a and c, but fails b: {1} and {2} are small, but {1,2} isn't small.</li> </ol> <p>The author's claim is that a definition exists, but it is impossible to come up with it as an example.  Proofs of the existence are non-constructive, and cannot be made constructive.  He does give an apparent summary of the proof, which was incomprehensible to me.  I don't have the background for even the terminology, let alone the substance.  But from what he says, the Axiom of Choice is required to prove the existence of a definition of "small" satisfying all three properties.<br /><br />Right now, I'm still very uncertain about the Axiom of Choice.  For this particular problem, I'd rather conclude that no such solution exists, rather than that such a solution exists but we can't find it even in approximation.</p> <p>I found something at <a href="http://www.math.vanderbilt.edu/~schectex/ccc/choice.html" title="Vanderbilt's Axiom of Choice">Vanderbilt College</a> about the "Axiom of Choice".  The author poses this problem: he wants to define "small" sets of positive integers.  He wants the definition to including the following properties:</p> <ol style="list-style-type: lower-alpha;"> <li>Any set with zero or one members is "small". </li> <li>Any union of two "small" sets is "small". </li> <li>A set is "small" if and only if its complement isn't "small."</li> </ol> <p>The "complement" of a set S means the set of positive integers not in S.  I might mention that the sets of positive integers is uncountably infinite.  If a proof is needed, here's a simple one:</p> <p style="padding-left: 30px;">Suppose we have sets of positive integers A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>,...  Let <em>n</em> be a positive integer.  Either <em>n</em> is in A<sub><em>n</em></sub> or it's not.  Define N = {<em>n</em>:<em>n</em> is not in A<sub><em>n</em></sub>}.  N is a defined set given A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>,...  Then <em>n</em> is in N if and only if <em>n</em> is not in A<sub><em>n</em></sub>.  That means that N can't be A<sub><em>n</em></sub> for any <em>n</em>.</p> <p>Any alleged one-to-one mapping of the positive integers onto the sets of positive integers is defeated by the existence of another set outside of the list of sets.  Getting a one-to-one mapping for all sets of integers is literally impossible.  The sets of positive integers must be uncountable.  As I commented in my previous post, things get squirrelly with the uncountable.</p> <p>Back to the author's problem, it's important to disconnect our own notion of "small" with "small" as he would like to define it.  If we keep to our intuitive notion of small, property a satisfies it.  Property a and b together mean that any finite set must be "small".  That includes a set with a googleplex of elements.  But a googleplex is tiny compared with infinity.  A finite set may or may not be intuitively small.  But necessarily, it must be "small" if properties a and b are satisfied.</p> <p>The author gives examples where any two of the three properties are satisfied:</p> <ol> <li>A "small" set is defined as a finite set.  That fails to satisfy property c, because neither the evens nor the odds are "small".</li> <li>A "small" set is a set not containing the number 1.  This satisfies properties b and c, but not a: {1} isn't small by this definition.</li> <li>A "small" set contains at most one of the first three numbers.  This satisfies a and c, but fails b: {1} and {2} are small, but {1,2} isn't small.</li> </ol> <p>The author's claim is that a definition exists, but it is impossible to come up with it as an example.  Proofs of the existence are non-constructive, and cannot be made constructive.  He does give an apparent summary of the proof, which was incomprehensible to me.  I don't have the background for even the terminology, let alone the substance.  But from what he says, the Axiom of Choice is required to prove the existence of a definition of "small" satisfying all three properties.<br /><br />Right now, I'm still very uncertain about the Axiom of Choice.  For this particular problem, I'd rather conclude that no such solution exists, rather than that such a solution exists but we can't find it even in approximation.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb092c1c28970d John M posted an entry http://activitystrea.ms/schema/1.0/post2016-08-19T07:24:41Ztag:api.typepad.com,2009:6a00e00995bbf6883301bb092c1c26970dThe Axiom of Choicehttp://activitystrea.ms/schema/1.0/article2016-08-19T07:24:41Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>The "Axiom of Choice" is a feature of theoretical math, and appears to be accepted by the mathematical community.  It is one of several claims that I will have to be dragged, kicking and screaming, into accepting as necessary.  I've seen several statements of the Axiom that I think I understand, as well as some incomprehensible versions.</p> <p>Let's be clear: for now, I believe the Axiom to be valid.  My problem is in calling it an axiom.  As an Axiom, one can choose not to include it.  This gives us a larger span of possibilities, including situations where the Axiom is false.  Including the Axiom narrows our possibilities.  In general, adding an axiom to a definition requires more, and therefore narrows the range of possibilities.  As an illustration how adding an axiom narrows our possibilities, consider the definition of a group:</p> <p style="padding-left: 30px;">A group is a set (say, G) with a binary operation (generically called multiplication, although any operation serves) satisfying the following requirements (axioms) (<em>x,</em> <em>y, z</em> are members of G):</p> <ol> <li>Closure: <em>xy</em> (<em>x</em> multiplied by <em>y</em>) is a member of G.</li> <li>Associativity: (<em>xy</em>)<em>z</em> = <em>x</em>(<em>yz</em>), and can be written <em>xyz</em>.</li> <li>Right Identity: G has at least one member (<em>e</em>) satisfying <em>xe = x</em>.</li> <li>Right Inverses: for (at least) one such right identity, there exists <em>x</em><sup>-1</sup> satisfying <em>x</em><em>x</em><sup>-1</sup> = e.</li> </ol> <p>An Abelian (or commutative) group has one extra defining property (or axiom): xy = yx.  Group multiplication commutes.  There are fewer (non-isomorphic) Abelian groups of a given order (number of elements) than groups in general (again, non-isomorphic and of the same order).  On the other hand, what would happen if we didn't require associativity?  I don't know, but many more set-operations would qualify.</p> <p>The point is, if we needed the Axiom of Choice as an axiom, it would be possible to define everything in set theory without the Axiom, and have types of set that don't satisfy the axiom.  One version goes like this:</p> <p style="padding-left: 30px;">Given non-empty sets S<em><sub>n</sub></em>, it is possible to let x<em><sub>n</sub></em> be a member of S<em><sub>n</sub></em>, for each <em>n</em>.</p> <p>Why do we need an axiom to make such a statement?  S<em><sub>n</sub></em> is non-empty, and therefore has a member; let x<em><sub>n</sub></em> be a member.  I'm going to consider only a countable number of sets -- finite or countably infinite.  We can always take our index <em>n</em> as a positive integer.  Things get squirrelly once we go beyond the countable.  At least countably infinite is semi-sane.  (We still have to get used to things like infinity+1 = infinity, infinity+1000 = infinity, 2*infinity=infinity, 100*infinity=infinity, infinity<sup>2</sup>=infinity, infinity<sup>20</sup>=infinity, and the like.  Note, however, that 2<sup>infinity</sup> is uncountable.)</p> <p>Perhaps it's some solace that the Axiom is only allegedly needed if the number of sets is infinite.  Another statement of the Axiom is that the Cartesian product (or direct product) of an infinite number of non-empty sets is non-empty.  Here are some examples of Cartesian products:</p> <p>The Cartesian product of two sets is a set of ordered pairs:</p> <p style="padding-left: 30px;">S<sub>1</sub> = {1,2} and S<sub>2</sub> = {7,8,9}: S<sub>1</sub> x S<sub>2</sub> = {(1,7),(1,8),(1,9),(2,7),(2,8),(2,9)}</p> <p style="padding-left: 30px;">S<sub>1</sub> = {1,2} and S<sub>2</sub> = {1,2,3}: S<sub>1</sub> x S<sub>2</sub> = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)}</p> <p style="padding-left: 30px;">The standard 52-card deck is the Cartesian product of 13 ranks and four suits.  (Example from Wikipedia.)</p> <p>The numbers in an ordered pair may be duplicated, although not in a set.  The Cartesian product of three sets is an ordered triple.  In general, the Cartesian product of <em>n</em> sets is an ordered <em>n</em>-tuple, which may be defined inductively.  (Using the inductive definition, one might consider an element of the Cartesian product of 3 sets to be ((a,b),c) rather than (a,b,c).  But we can declare them to mean the same thing.)</p> <p style="padding-left: 30px;">S<sub>1</sub> = {1,2},  S<sub>2</sub> = {1,2}, and S<sub>3</sub> = {7,8,9}: S<sub>1</sub> x S<sub>2</sub> x S<sub>3</sub> = {(1,1,7),(1,1,8),(1,1,9),(1,2,7),(1,2,8),(1,2,9),(2,1,7),(2,1,8),(2,1,9),(2,2,7),(2,2,8),(2,2,9)}</p> <p>Since S<sub>1</sub> = S<sub>2</sub>, we could have called the example above, S<sub>1</sub> x S<sub>1</sub> x S<sub>3</sub>.  One can take the Cartesian product of a set with itself.</p> <p>The number of elements in the Cartesian product of <em>N</em> sets is the product of the numbers of elements from each set.  The number of elements in each example Cartesian product above is 6, 6, 52 and 12, respectively.  As long as sets with more than one element continue to exist as we increase <em>N</em>, the number of elements in their Cartesian product increases rapidly.  In fact, it approaches at least the cardinality of the power set of integers -- that limit is uncountable, the same infinity as the real numbers, the same infinity as the power set of the integers, or 2<sup>infinity,</sup> where "infinity" refers to countable infinity.</p> <p>The Cardinal product of a countably infinite number of sets consists of all elements of the form (x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, x<sub>5</sub>, x<sub>6</sub>, x<sub>7</sub>, ...).</p> <p>I may concede troubles in defining the Cartesian product of an infinite number of sets, but if we surmount that issue, the Cartesian product of an infinite number of sets (of more than one element each) can't be empty, since it's order (number, cardinality) is infinite.  Perhaps I should be less arrogant, and instead ask, if we don't assume the Axiom of Choice, how we can get an infinite number of non-empty sets, but their Cartesian product be the empty set?  That is, how does the Axiom of Choice restrict the sets that exist?</p> <p>A shorter statement: if all sets are non-empty, then the Cartesian product of them has at least as many elements as the largest set.</p> <p>Here's Wikipedia quoting Bertrand Russell on the subject: "The Axiom of Choice is necessary to select a set from an infinite number of pairs of socks, but not an infinite number of pairs of shoes."  Wikipedia's comment on Russell's quote: "The observation here is that one can define a function to select from an infinite number of pairs of shoes by stating for example, to choose a left shoe. Without the axiom of choice, one cannot assert that such a function exists for pairs of socks, because left and right socks are (presumably) indistinguishable."</p> <p>We are supposed to abstract the "pairs of socks" example to represent pairs of indistinguishable objects, because of course, physical socks are distinguishable -- by position if nothing else.  In fact, it may be impossible to abstract away the distinguishability of the socks.  If the socks are truly indistinguishable, perhaps one literally can't choose between them -- not even at random.  In this case, we can't pick out one from a single pair, let alone one from each of an infinite number of pairs.</p> <p>It's possible that if the socks are truly indistinguishable, then the "pair of socks" are in fact the same sock.</p> <p>How would this apply to sets of numbers?  First of all, we can't have a set of two fives: {5,5}.  Either 5 is a member of the set or it isn't.  (This should be distinguished from the ordered pair of two fives: (5,5).  A single 5 can be plotted on a number line.  The ordered pair would be plotted on a plane.)  All numbers in a set of numbers are different.  The closest thing to a pair of identical numbers is {i,-i} (with i being the square root of -1).  But this is the same as the left-shoe, right-shoe example.  (Left and right have the same conceptual issue as i and -i.)</p> <p>If we have finite sets of real numbers, where for each positive integer <em>n</em>, S<sub><em>n</em></sub> contains a finite number of real numbers, then even by the standards of the Axiom of Choice, we need not invoke the Axiom.  Each set has a minimum number, so we can define <em>x<sub>n</sub></em> = min(S<sub>n</sub>).  If we allow for a finite number of infinite sets, then one can pick a number from each of those.  Then the remaining finite sets all have <em>x<sub>n</sub></em> = min(S<sub>n</sub>).  So now, we need to deal with an infinite number of infinite sets.  In fact, if we have finite sets among the infinite sets, we can always choose for those <em>x<sub>n</sub></em> = min(S<sub>n</sub>).  So we can limit ourselves to all sets being infinite.</p> <p>Let's once and for all dispose of the notion that the Cartesian product of the sets is the empty set, simply because the number of elements in the Cartesian product explodes as the number of sets approaches infinity.  At worst, the Cartesian product is ill-defined.</p> <p style="padding-left: 30px;">So what's to prevent us from saying, "Suppose <em>x<sub>n</sub></em> is a member of S<sub><em>n</em></sub>, for all positive integers <em>n</em>?"</p> <p>I've heard it claimed that one needs the Axiom of Choice (or a weaker version, the Axiom of Countable Choice) to prove that any infinite set has a countably infinite subset.  See, for example, <a href="https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset" title="Any infinite set has a countably infinite subset.">https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset</a>.  They first give an intuitive proof, then warn against it.  I'm going to formalize their intuitive proof, and challenge anyone to tell me the mistake I've made.  The proof uses strong induction.  (Prove something for<em> k=1</em>, assume it's true for <em>k=1...n</em>, then prove it for <em>n+1</em>.  This proves it for all positive integers.)</p> <p style="padding-left: 30px;">S is the infinite set.  Let <em>x<sub>1</sub></em> be a member of S.  Now assume that <em>x<sub>1</sub> ... x<sub>n</sub></em> are distinct members of S.  Let <em>x<sub>n</sub></em><sub>+1</sub> be a member of S - U<sub>k=1</sub><sup>n</sup> {<em>x<sub>k</sub></em>} -- where U refers to the union, and the "subtraction" refers to S without any element from the union.  Then <em>x<sub>1</sub> ... x<sub>n</sub></em><sub>+1</sub> are distinct members of S.  Therefore, by strong induction, <em>x<sub>n</sub></em> are all distinct members of S, for positive integers <em>n</em>.  U<sub>k=1</sub><sup>infinity</sup> {<em>x<sub>k</sub></em>} is a countable subset of S.  QED.</p> <p>If something is actually wrong with this, would they kindly explain with more than something like it takes forever to make such a choice, or that we need to invoke the Axiom of (Countable) Choice to do this.  It's clear that the subtracted set is nonempty, because we are removing a finite number of elements from an infinite set.  The subtracted set is well-defined by the induction assumption.  Therefore, "Let <em>x<sub>n</sub></em><sub>+1</sub> be a member..." has to be valid.  If we're talking about infinite time, let's avoid this by letting the <em>n</em><sup>th</sup> step take 1/2<em><sup>n</sup></em> seconds<em>.  </em>The total time is one second.</p> <p>In the spirit of the infinite pairs of socks, I've been considering the two endpoints of a line segment, and three vertices of an equilateral triangle.  In both cases, the points are indistinguishable.  I might discuss those situations later.</p> <p>In theoretical math, axioms are used to define objects.  See the axioms defining a group above.  On the other hand, "axioms" have been thought of as things that are so obvious that they are assumed true, and used as starting points.  It's possible that the Axiom of Choice (and the Axiom of Countable Choice) is used in that latter sense of an assumption rather than as part of a definition.</p> <p>The "Axiom of Choice" is a feature of theoretical math, and appears to be accepted by the mathematical community.  It is one of several claims that I will have to be dragged, kicking and screaming, into accepting as necessary.  I've seen several statements of the Axiom that I think I understand, as well as some incomprehensible versions.</p> <p>Let's be clear: for now, I believe the Axiom to be valid.  My problem is in calling it an axiom.  As an Axiom, one can choose not to include it.  This gives us a larger span of possibilities, including situations where the Axiom is false.  Including the Axiom narrows our possibilities.  In general, adding an axiom to a definition requires more, and therefore narrows the range of possibilities.  As an illustration how adding an axiom narrows our possibilities, consider the definition of a group:</p> <p style="padding-left: 30px;">A group is a set (say, G) with a binary operation (generically called multiplication, although any operation serves) satisfying the following requirements (axioms) (<em>x,</em> <em>y, z</em> are members of G):</p> <ol> <li>Closure: <em>xy</em> (<em>x</em> multiplied by <em>y</em>) is a member of G.</li> <li>Associativity: (<em>xy</em>)<em>z</em> = <em>x</em>(<em>yz</em>), and can be written <em>xyz</em>.</li> <li>Right Identity: G has at least one member (<em>e</em>) satisfying <em>xe = x</em>.</li> <li>Right Inverses: for (at least) one such right identity, there exists <em>x</em><sup>-1</sup> satisfying <em>x</em><em>x</em><sup>-1</sup> = e.</li> </ol> <p>An Abelian (or commutative) group has one extra defining property (or axiom): xy = yx.  Group multiplication commutes.  There are fewer (non-isomorphic) Abelian groups of a given order (number of elements) than groups in general (again, non-isomorphic and of the same order).  On the other hand, what would happen if we didn't require associativity?  I don't know, but many more set-operations would qualify.</p> <p>The point is, if we needed the Axiom of Choice as an axiom, it would be possible to define everything in set theory without the Axiom, and have types of set that don't satisfy the axiom.  One version goes like this:</p> <p style="padding-left: 30px;">Given non-empty sets S<em><sub>n</sub></em>, it is possible to let x<em><sub>n</sub></em> be a member of S<em><sub>n</sub></em>, for each <em>n</em>.</p> <p>Why do we need an axiom to make such a statement?  S<em><sub>n</sub></em> is non-empty, and therefore has a member; let x<em><sub>n</sub></em> be a member.  I'm going to consider only a countable number of sets -- finite or countably infinite.  We can always take our index <em>n</em> as a positive integer.  Things get squirrelly once we go beyond the countable.  At least countably infinite is semi-sane.  (We still have to get used to things like infinity+1 = infinity, infinity+1000 = infinity, 2*infinity=infinity, 100*infinity=infinity, infinity<sup>2</sup>=infinity, infinity<sup>20</sup>=infinity, and the like.  Note, however, that 2<sup>infinity</sup> is uncountable.)</p> <p>Perhaps it's some solace that the Axiom is only allegedly needed if the number of sets is infinite.  Another statement of the Axiom is that the Cartesian product (or direct product) of an infinite number of non-empty sets is non-empty.  Here are some examples of Cartesian products:</p> <p>The Cartesian product of two sets is a set of ordered pairs:</p> <p style="padding-left: 30px;">S<sub>1</sub> = {1,2} and S<sub>2</sub> = {7,8,9}: S<sub>1</sub> x S<sub>2</sub> = {(1,7),(1,8),(1,9),(2,7),(2,8),(2,9)}</p> <p style="padding-left: 30px;">S<sub>1</sub> = {1,2} and S<sub>2</sub> = {1,2,3}: S<sub>1</sub> x S<sub>2</sub> = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)}</p> <p style="padding-left: 30px;">The standard 52-card deck is the Cartesian product of 13 ranks and four suits.  (Example from Wikipedia.)</p> <p>The numbers in an ordered pair may be duplicated, although not in a set.  The Cartesian product of three sets is an ordered triple.  In general, the Cartesian product of <em>n</em> sets is an ordered <em>n</em>-tuple, which may be defined inductively.  (Using the inductive definition, one might consider an element of the Cartesian product of 3 sets to be ((a,b),c) rather than (a,b,c).  But we can declare them to mean the same thing.)</p> <p style="padding-left: 30px;">S<sub>1</sub> = {1,2},  S<sub>2</sub> = {1,2}, and S<sub>3</sub> = {7,8,9}: S<sub>1</sub> x S<sub>2</sub> x S<sub>3</sub> = {(1,1,7),(1,1,8),(1,1,9),(1,2,7),(1,2,8),(1,2,9),(2,1,7),(2,1,8),(2,1,9),(2,2,7),(2,2,8),(2,2,9)}</p> <p>Since S<sub>1</sub> = S<sub>2</sub>, we could have called the example above, S<sub>1</sub> x S<sub>1</sub> x S<sub>3</sub>.  One can take the Cartesian product of a set with itself.</p> <p>The number of elements in the Cartesian product of <em>N</em> sets is the product of the numbers of elements from each set.  The number of elements in each example Cartesian product above is 6, 6, 52 and 12, respectively.  As long as sets with more than one element continue to exist as we increase <em>N</em>, the number of elements in their Cartesian product increases rapidly.  In fact, it approaches at least the cardinality of the power set of integers -- that limit is uncountable, the same infinity as the real numbers, the same infinity as the power set of the integers, or 2<sup>infinity,</sup> where "infinity" refers to countable infinity.</p> <p>The Cardinal product of a countably infinite number of sets consists of all elements of the form (x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, x<sub>5</sub>, x<sub>6</sub>, x<sub>7</sub>, ...).</p> <p>I may concede troubles in defining the Cartesian product of an infinite number of sets, but if we surmount that issue, the Cartesian product of an infinite number of sets (of more than one element each) can't be empty, since it's order (number, cardinality) is infinite.  Perhaps I should be less arrogant, and instead ask, if we don't assume the Axiom of Choice, how we can get an infinite number of non-empty sets, but their Cartesian product be the empty set?  That is, how does the Axiom of Choice restrict the sets that exist?</p> <p>A shorter statement: if all sets are non-empty, then the Cartesian product of them has at least as many elements as the largest set.</p> <p>Here's Wikipedia quoting Bertrand Russell on the subject: "The Axiom of Choice is necessary to select a set from an infinite number of pairs of socks, but not an infinite number of pairs of shoes."  Wikipedia's comment on Russell's quote: "The observation here is that one can define a function to select from an infinite number of pairs of shoes by stating for example, to choose a left shoe. Without the axiom of choice, one cannot assert that such a function exists for pairs of socks, because left and right socks are (presumably) indistinguishable."</p> <p>We are supposed to abstract the "pairs of socks" example to represent pairs of indistinguishable objects, because of course, physical socks are distinguishable -- by position if nothing else.  In fact, it may be impossible to abstract away the distinguishability of the socks.  If the socks are truly indistinguishable, perhaps one literally can't choose between them -- not even at random.  In this case, we can't pick out one from a single pair, let alone one from each of an infinite number of pairs.</p> <p>It's possible that if the socks are truly indistinguishable, then the "pair of socks" are in fact the same sock.</p> <p>How would this apply to sets of numbers?  First of all, we can't have a set of two fives: {5,5}.  Either 5 is a member of the set or it isn't.  (This should be distinguished from the ordered pair of two fives: (5,5).  A single 5 can be plotted on a number line.  The ordered pair would be plotted on a plane.)  All numbers in a set of numbers are different.  The closest thing to a pair of identical numbers is {i,-i} (with i being the square root of -1).  But this is the same as the left-shoe, right-shoe example.  (Left and right have the same conceptual issue as i and -i.)</p> <p>If we have finite sets of real numbers, where for each positive integer <em>n</em>, S<sub><em>n</em></sub> contains a finite number of real numbers, then even by the standards of the Axiom of Choice, we need not invoke the Axiom.  Each set has a minimum number, so we can define <em>x<sub>n</sub></em> = min(S<sub>n</sub>).  If we allow for a finite number of infinite sets, then one can pick a number from each of those.  Then the remaining finite sets all have <em>x<sub>n</sub></em> = min(S<sub>n</sub>).  So now, we need to deal with an infinite number of infinite sets.  In fact, if we have finite sets among the infinite sets, we can always choose for those <em>x<sub>n</sub></em> = min(S<sub>n</sub>).  So we can limit ourselves to all sets being infinite.</p> <p>Let's once and for all dispose of the notion that the Cartesian product of the sets is the empty set, simply because the number of elements in the Cartesian product explodes as the number of sets approaches infinity.  At worst, the Cartesian product is ill-defined.</p> <p style="padding-left: 30px;">So what's to prevent us from saying, "Suppose <em>x<sub>n</sub></em> is a member of S<sub><em>n</em></sub>, for all positive integers <em>n</em>?"</p> <p>I've heard it claimed that one needs the Axiom of Choice (or a weaker version, the Axiom of Countable Choice) to prove that any infinite set has a countably infinite subset.  See, for example, <a href="https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset" title="Any infinite set has a countably infinite subset.">https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset</a>.  They first give an intuitive proof, then warn against it.  I'm going to formalize their intuitive proof, and challenge anyone to tell me the mistake I've made.  The proof uses strong induction.  (Prove something for<em> k=1</em>, assume it's true for <em>k=1...n</em>, then prove it for <em>n+1</em>.  This proves it for all positive integers.)</p> <p style="padding-left: 30px;">S is the infinite set.  Let <em>x<sub>1</sub></em> be a member of S.  Now assume that <em>x<sub>1</sub> ... x<sub>n</sub></em> are distinct members of S.  Let <em>x<sub>n</sub></em><sub>+1</sub> be a member of S - U<sub>k=1</sub><sup>n</sup> {<em>x<sub>k</sub></em>} -- where U refers to the union, and the "subtraction" refers to S without any element from the union.  Then <em>x<sub>1</sub> ... x<sub>n</sub></em><sub>+1</sub> are distinct members of S.  Therefore, by strong induction, <em>x<sub>n</sub></em> are all distinct members of S, for positive integers <em>n</em>.  U<sub>k=1</sub><sup>infinity</sup> {<em>x<sub>k</sub></em>} is a countable subset of S.  QED.</p> <p>If something is actually wrong with this, would they kindly explain with more than something like it takes forever to make such a choice, or that we need to invoke the Axiom of (Countable) Choice to do this.  It's clear that the subtracted set is nonempty, because we are removing a finite number of elements from an infinite set.  The subtracted set is well-defined by the induction assumption.  Therefore, "Let <em>x<sub>n</sub></em><sub>+1</sub> be a member..." has to be valid.  If we're talking about infinite time, let's avoid this by letting the <em>n</em><sup>th</sup> step take 1/2<em><sup>n</sup></em> seconds<em>.  </em>The total time is one second.</p> <p>In the spirit of the infinite pairs of socks, I've been considering the two endpoints of a line segment, and three vertices of an equilateral triangle.  In both cases, the points are indistinguishable.  I might discuss those situations later.</p> <p>In theoretical math, axioms are used to define objects.  See the axioms defining a group above.  On the other hand, "axioms" have been thought of as things that are so obvious that they are assumed true, and used as starting points.  It's possible that the Axiom of Choice (and the Axiom of Countable Choice) is used in that latter sense of an assumption rather than as part of a definition.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collection