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,2009:6e00e00995bbf6883301b7c8a61bdd970b 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/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb092b6d76970d John M posted something http://activitystrea.ms/schema/1.0/post2016-08-17T14:44:55Ztag:api.typepad.com,2009:6a00d83451b33869e201bb092b6d75970dhttp://activitystrea.ms/schema/1.0/comment2016-08-17T14:44:55Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Paul Krugman: Wisdom, Courage and the Economytag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb092b6c2b970d John M posted something http://activitystrea.ms/schema/1.0/post2016-08-17T14:32:25Ztag:api.typepad.com,2009:6a00d83451b33869e201bb092b6c2a970dhttp://activitystrea.ms/schema/1.0/comment2016-08-17T14:32:25Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Obama Rescued the Economy. Could He Have Done More?tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb092b6966970d John M posted an entry http://activitystrea.ms/schema/1.0/post2016-08-17T14:08:25Ztag:api.typepad.com,2009:6a00e00995bbf6883301bb092b6963970dPunishment for an Unprovoked Nuclear Attackhttp://activitystrea.ms/schema/1.0/article2016-08-17T14:08:24Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>Any first attack with a nuclear weapon is an unprovoked nuclear attack. The best result, as Scott Ritter puts it, is that the nuclear genie will be out of the bottle and some big American city will be taken out. https://www.youtube.com/watch?v=8XQan1qo8T4</p> <p>The worst possible outcome would be all-out nuclear war resulting in the destruction of human life. The movie "Fail Safe" considered the inadvertent launch of a nuke that took out Moscow. The US government nuked Manhattan to avoid an all-out nuclear war. Nuclear retaliation against a nuclear attack is the one and only justified reason for having nuclear weapons.</p> <p>Since nuclear retaliation is a fair response to a nuclear attack, this punishment is very reasonable and should be established for an unprovoked nuclear attack: those who ordered the attack have to sit and watch (and hear) with eyes held open as their loved ones -- spouse, SOs, parents, children, siblings, grandchildren, nephews, nieces, and other close friends and relatives -- are tortured to death.</p> <p>It is hoped that such a punishment would never have to occur. I would make an exception of an attack on a wide-open, sparsely populated area.</p> <p>Any first attack with a nuclear weapon is an unprovoked nuclear attack. The best result, as Scott Ritter puts it, is that the nuclear genie will be out of the bottle and some big American city will be taken out. https://www.youtube.com/watch?v=8XQan1qo8T4</p> <p>The worst possible outcome would be all-out nuclear war resulting in the destruction of human life. The movie "Fail Safe" considered the inadvertent launch of a nuke that took out Moscow. The US government nuked Manhattan to avoid an all-out nuclear war. Nuclear retaliation against a nuclear attack is the one and only justified reason for having nuclear weapons.</p> <p>Since nuclear retaliation is a fair response to a nuclear attack, this punishment is very reasonable and should be established for an unprovoked nuclear attack: those who ordered the attack have to sit and watch (and hear) with eyes held open as their loved ones -- spouse, SOs, parents, children, siblings, grandchildren, nephews, nieces, and other close friends and relatives -- are tortured to death.</p> <p>It is hoped that such a punishment would never have to occur. I would make an exception of an attack on a wide-open, sparsely populated area.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c888132a970b John M posted something http://activitystrea.ms/schema/1.0/post2016-08-17T13:50:05Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c8881329970bhttp://activitystrea.ms/schema/1.0/comment2016-08-17T13:50:05Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Links for 08-17-16tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d211c4e8970c John M posted something http://activitystrea.ms/schema/1.0/post2016-08-17T13:48:36Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d211c4e7970chttp://activitystrea.ms/schema/1.0/comment2016-08-17T13:48:36Ztag: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,2009:6e00e00995bbf6883301bb092108a9970d John M posted something http://activitystrea.ms/schema/1.0/post2016-07-20T13:30:19Ztag:api.typepad.com,2009:6a00d83451b33869e201bb092108a8970dhttp://activitystrea.ms/schema/1.0/comment2016-07-20T13:30:19Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Is Finance a Powerful Driver of Growth?tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d2064825970c John M posted something http://activitystrea.ms/schema/1.0/post2016-07-17T18:57:21Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d2064824970chttp://activitystrea.ms/schema/1.0/comment2016-07-17T18:57:21Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Links for 07-17-16tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c87c6e3a970b John M posted something http://activitystrea.ms/schema/1.0/post2016-07-17T18:19:20Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c87c6e39970bhttp://activitystrea.ms/schema/1.0/comment2016-07-17T18:19:20Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Helicopter Moneytag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c87a127a970b John M posted something http://activitystrea.ms/schema/1.0/post2016-07-11T12:16:54Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c87a1279970bhttp://activitystrea.ms/schema/1.0/comment2016-07-11T12:16:54Ztag: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,2009:6e00e00995bbf6883301b8d200d9b5970c John M posted an entry http://activitystrea.ms/schema/1.0/post2016-07-03T19:00:50Ztag:api.typepad.com,2009:6a00e00995bbf6883301b8d200d9b2970cGoing After ISIShttp://activitystrea.ms/schema/1.0/article2016-07-03T19:00:50Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>As Atrios has pointed out repeatedly, it seems that -- in the Middle East, at least -- the only humanitarian thing we can think of doing is "humanitarian bombing".  That and training and arming "moderates" who wind up fighting against us as one of the "extremists".  In fact ISIS may have come from the "moderate" rebels against Syria's government that we supported.<br /><br />It may turn out that we have to go after ISIS, given their propensity for brutality and overt bragging about it.  However, if we do, we must satisfy three requirements.  I add that those requirements must be satisfied for any military engagement in the Middle East.  ISIS is just the most plausible example.<br /><br />1.  Pick one military objective, and abandon all others.  In the case of ISIS, abandon things such as preventing Iran from getting nuclear weapons.  (BTW, these days, the US with nukes is scarier than Iran with nukes.  It is all too plausible to see the US nuking Iran, motivated to prevent the possibility of Iran nuking someone.)  Stop all military action in Pakistan, Yemen, and elsewhere not a part of going after ISIS.<br /><br />2.  Audit the CIA, the NSA, the DoD, and other intelligence and national security organizations, to make sure that we aren't being fed propaganda, lied to, etc.  We must also check for involvement with organized crime, and alliances contrary to our own foreign policies.<br /><br />Example of this: the CIA rendering people to prisons in Syria and Libya, despite their being opponents of ours.<br /><br />3.  Take the kingpins who decided to invade Iraq, who issued the orders, and who pushed the invasion in the media, pick out (say) five of those at random, and chop them to pieces or otherwise torture them to death on national television.<br /><br />About pushing the Iraq invasion in the media: I'm not talking about those who advocated the invasion in their articles.  As the word "kingpins" implies, I am aiming at the top people who gave orders to slant coverage in favor of the invasion, or who required reporters or editors to support the invasion.<br /><br />If we had a free press, I wouldn't even be concerned about the media issue.  If one media faction had orders to promote the invasion, we would have seen numerous others opposing it.  Instead of a free press, we have one under <strong>potentially</strong> complete control.<br /><br />News blackouts can't occur under a free press.</p> <p>As Atrios has pointed out repeatedly, it seems that -- in the Middle East, at least -- the only humanitarian thing we can think of doing is "humanitarian bombing".  That and training and arming "moderates" who wind up fighting against us as one of the "extremists".  In fact ISIS may have come from the "moderate" rebels against Syria's government that we supported.<br /><br />It may turn out that we have to go after ISIS, given their propensity for brutality and overt bragging about it.  However, if we do, we must satisfy three requirements.  I add that those requirements must be satisfied for any military engagement in the Middle East.  ISIS is just the most plausible example.<br /><br />1.  Pick one military objective, and abandon all others.  In the case of ISIS, abandon things such as preventing Iran from getting nuclear weapons.  (BTW, these days, the US with nukes is scarier than Iran with nukes.  It is all too plausible to see the US nuking Iran, motivated to prevent the possibility of Iran nuking someone.)  Stop all military action in Pakistan, Yemen, and elsewhere not a part of going after ISIS.<br /><br />2.  Audit the CIA, the NSA, the DoD, and other intelligence and national security organizations, to make sure that we aren't being fed propaganda, lied to, etc.  We must also check for involvement with organized crime, and alliances contrary to our own foreign policies.<br /><br />Example of this: the CIA rendering people to prisons in Syria and Libya, despite their being opponents of ours.<br /><br />3.  Take the kingpins who decided to invade Iraq, who issued the orders, and who pushed the invasion in the media, pick out (say) five of those at random, and chop them to pieces or otherwise torture them to death on national television.<br /><br />About pushing the Iraq invasion in the media: I'm not talking about those who advocated the invasion in their articles.  As the word "kingpins" implies, I am aiming at the top people who gave orders to slant coverage in favor of the invasion, or who required reporters or editors to support the invasion.<br /><br />If we had a free press, I wouldn't even be concerned about the media issue.  If one media faction had orders to promote the invasion, we would have seen numerous others opposing it.  Instead of a free press, we have one under <strong>potentially</strong> complete control.<br /><br />News blackouts can't occur under a free press.</p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c871cfbf970b John M posted something http://activitystrea.ms/schema/1.0/post2016-06-22T04:38:49Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c871cfbe970bhttp://activitystrea.ms/schema/1.0/comment2016-06-22T04:38:49Ztag: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,2009:6e00e00995bbf6883301bb091173d2970d John M posted something http://activitystrea.ms/schema/1.0/post2016-06-15T07:50:07Ztag:api.typepad.com,2009:6a00d83451b33869e201bb091173d1970dhttp://activitystrea.ms/schema/1.0/comment2016-06-15T07:50:07Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Links for 06-15-16tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301bb09116c30970d John M posted an entry http://activitystrea.ms/schema/1.0/post2016-06-15T05:26:58Ztag:api.typepad.com,2009:6a00e00995bbf6883301bb09116c2c970dThe Rules of Logic and Real Lifehttp://activitystrea.ms/schema/1.0/article2016-06-15T05:26:58Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307<p>I believe that the formal rules of logic have to be adjusted to account for the real world.  There is a reason that we have "the Mathematician's Answer" to questions such as, "Would you like sugar or cream in your coffee?"</p> <p>For example "A or B" is true if one or both are true, right?  So you go to Physics Department and ask the office administrator if Professor Jones is available.  The administrator answers, "Either he's in his office, or he's on his way to class."  Her answer is true if one is true, right?  Shouldn't we expect a third requirement?  For example, that she not know which it is?  If she knows perfectly well that Professor Jones is in his office, then her statement was deceptive.</p> <p>Also, ask a question in court.  If the witness answers (something) "or 2 + 2 = 4", he's not going to immunize himself from perjury if that (something) is false.  In fact, the witness might be penalized for the statement.</p> <p>Consider this situation in physics, such as a string tied between two walls.  One derives the equation, A sin(kL) cos(wt) = 0.  That means at least one of A, sin(kL), and cos(wt) is zero.  The variable t is time, which progresses forward, so we rule out cos(wt) = 0.  This leaves us A=0 or sin(kL)=0.  Either could be true, and are perfectly fine solutions.  We consider the interesting situation where A doesn't equal zero, so sin(kL) = 0 leading to k=n*pi/L.</p> <hr /> <p> </p> <p>Now consider "A and B".  True if both statements are true, false otherwise?  I put it to you that in the real world, (A and B) could be true, while A without B mentioned somewhere is patently false.  That's called, "Omitting critical information".  Stating that someone broke into a house at night might get him convicted of burglary.  Include the facts that the house was on fire, and a child was trapped inside and the person went in to rescue him, and a conviction of burglary becomes a miscarriage of justice.  The accuracy of an account is more than just the accuracy of the individual statements.</p> <p>We now can see good reason for psychological results showing that people often say that "Sue works as a bank teller and is involved in the feminist movement" is more likely than "Sue works as a bank teller."  The issue is an accurate description rather than the literal truth of the statements.</p> <p>There are situations where specifically avoiding an actual false statement is proof that one is trying to deceive, while making the false statement might be consistent with ignorance but good faith.</p> <hr /> <p>"A false statement implies anything."  In other words, when A is false, "If A then B" is automatically true.  This notion should be discarded in favor of this alternative: "The truth or falsity of A is irrelevant to the truth of `if A then B'."  I'm not concerned about statements like, "If 2+2=5 then I am the Pope."  I'm thinking of statements like, "If Superman flies at 500 mph, he travels 1000 miles in five minutes."  In a science class (or any class, for that matter) we wish to say that that statement is false, regardless that Superman doesn't exist and therefore doesn't fly at 500 mph.</p> <p>Likewise, a trick question goes, "If a peacock lays 5 eggs on Tuesday and 3 eggs on Wednesday (and no other eggs), how many eggs does he lay?"  The correct answer is 8 eggs, even though peacocks don't lay eggs.  The reason is the same as above: we disregard the actual falsity of the statement, and instead see what follows from the truth of the statement.</p> <hr /> <p>"If A then B" is equivalent to "If B is false, then A is false."  This should stand pretty much as it is.  It's the heart of being able to figure things out, and the heart of comprehending what is said.  The rules of logic are based on the meaning of words, and should be used to understand things, rather than play games and play dumb about what is said.</p> <p>Communication should be for understanding, not obfuscation.</p> <p> </p> <p> </p> <p>I believe that the formal rules of logic have to be adjusted to account for the real world.  There is a reason that we have "the Mathematician's Answer" to questions such as, "Would you like sugar or cream in your coffee?"</p> <p>For example "A or B" is true if one or both are true, right?  So you go to Physics Department and ask the office administrator if Professor Jones is available.  The administrator answers, "Either he's in his office, or he's on his way to class."  Her answer is true if one is true, right?  Shouldn't we expect a third requirement?  For example, that she not know which it is?  If she knows perfectly well that Professor Jones is in his office, then her statement was deceptive.</p> <p>Also, ask a question in court.  If the witness answers (something) "or 2 + 2 = 4", he's not going to immunize himself from perjury if that (something) is false.  In fact, the witness might be penalized for the statement.</p> <p>Consider this situation in physics, such as a string tied between two walls.  One derives the equation, A sin(kL) cos(wt) = 0.  That means at least one of A, sin(kL), and cos(wt) is zero.  The variable t is time, which progresses forward, so we rule out cos(wt) = 0.  This leaves us A=0 or sin(kL)=0.  Either could be true, and are perfectly fine solutions.  We consider the interesting situation where A doesn't equal zero, so sin(kL) = 0 leading to k=n*pi/L.</p> <hr /> <p> </p> <p>Now consider "A and B".  True if both statements are true, false otherwise?  I put it to you that in the real world, (A and B) could be true, while A without B mentioned somewhere is patently false.  That's called, "Omitting critical information".  Stating that someone broke into a house at night might get him convicted of burglary.  Include the facts that the house was on fire, and a child was trapped inside and the person went in to rescue him, and a conviction of burglary becomes a miscarriage of justice.  The accuracy of an account is more than just the accuracy of the individual statements.</p> <p>We now can see good reason for psychological results showing that people often say that "Sue works as a bank teller and is involved in the feminist movement" is more likely than "Sue works as a bank teller."  The issue is an accurate description rather than the literal truth of the statements.</p> <p>There are situations where specifically avoiding an actual false statement is proof that one is trying to deceive, while making the false statement might be consistent with ignorance but good faith.</p> <hr /> <p>"A false statement implies anything."  In other words, when A is false, "If A then B" is automatically true.  This notion should be discarded in favor of this alternative: "The truth or falsity of A is irrelevant to the truth of `if A then B'."  I'm not concerned about statements like, "If 2+2=5 then I am the Pope."  I'm thinking of statements like, "If Superman flies at 500 mph, he travels 1000 miles in five minutes."  In a science class (or any class, for that matter) we wish to say that that statement is false, regardless that Superman doesn't exist and therefore doesn't fly at 500 mph.</p> <p>Likewise, a trick question goes, "If a peacock lays 5 eggs on Tuesday and 3 eggs on Wednesday (and no other eggs), how many eggs does he lay?"  The correct answer is 8 eggs, even though peacocks don't lay eggs.  The reason is the same as above: we disregard the actual falsity of the statement, and instead see what follows from the truth of the statement.</p> <hr /> <p>"If A then B" is equivalent to "If B is false, then A is false."  This should stand pretty much as it is.  It's the heart of being able to figure things out, and the heart of comprehending what is said.  The rules of logic are based on the meaning of words, and should be used to understand things, rather than play games and play dumb about what is said.</p> <p>Communication should be for understanding, not obfuscation.</p> <p> </p> <p> </p>tag:api.typepad.com,2009:6a00e00995bbf688330148c8449d70970cSome Random Nobodyhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b7c86d6dd4970b John M posted something http://activitystrea.ms/schema/1.0/post2016-06-14T05:13:21Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c86d6dd3970bhttp://activitystrea.ms/schema/1.0/comment2016-06-14T05:13:21Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307Tim Duy’s Five Questions for Janet Yellentag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d1f743b6970c John M posted something 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http://activitystrea.ms/schema/1.0/post2016-05-10T14:51:24Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d1e1b2f7970chttp://activitystrea.ms/schema/1.0/comment2016-05-10T14:51:24Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307As Jobs Vanish, Forgetting What Government Is Fortag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d1e1ae36970c John M posted something http://activitystrea.ms/schema/1.0/post2016-05-10T14:27:50Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d1e1ae35970chttp://activitystrea.ms/schema/1.0/comment2016-05-10T14:27:50Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307The Economic "Disease" Eating away at the U.S.tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's Viewhttp://activitystrea.ms/schema/1.0/collectiontag:api.typepad.com,2009:6e00e00995bbf6883301b8d1cd11ea970c John M posted something http://activitystrea.ms/schema/1.0/post2016-04-22T04:34:32Ztag:api.typepad.com,2009:6a00d83451b33869e201b8d1cd11e9970chttp://activitystrea.ms/schema/1.0/comment2016-04-22T04:34:32Ztag: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,2009:6e00e00995bbf6883301b7c842c8ed970b John M posted something http://activitystrea.ms/schema/1.0/post2016-04-22T04:05:24Ztag:api.typepad.com,2009:6a00d83451b33869e201b7c842c8ec970bhttp://activitystrea.ms/schema/1.0/comment2016-04-22T04:05:23Ztag:api.typepad.com,2009:6p00e00995bbf68833John Mhttp://profile.typepad.com/johnm307tag:api.typepad.com,2009:6a00d83451b33869e200d83451b33b69e2Economist's 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