Cantor Confusion
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From: david petry on 28 Sep 2006 16:17 the_wign(a)yahoo.com wrote: > Cantor's proof is one of the most popular topics on this NG. It > seems that people are confused or uncomfortable with it, so > I've tried to summarize it to the simplest terms: [...] I don't know if anyone in this newsgroup is confused by Cantor's proof, but what you have written misses the whole point of the endless debate that takes place here. The argument of the "anti-Cantorians" is that "real" mathematics is computationally testable. That is, valid, meaningful mathematical statements must make predictions about the outcome of computational experiments. (And note that any mathematics that is potentially applicable must satisfy that requirement) And hence, all objects that exist in the world of "real" mathematics must be computable. And there cannot be more than countably many such objects. If you start with a well defined list of well defined real numbers (so that every digit of every real number can be computed), then the diagonal method gives us a way of constructing a new real number not on that list, but that certainly does not imply that the well defined real numbers are uncountable in Cantor's intended sense of the word.
From: Aatu Koskensilta on 28 Sep 2006 17:19 Virgil wrote: > In article <1159417542.425540.214160(a)i3g2000cwc.googlegroups.com>, > cbrown(a)cbrownsystems.com wrote: >> the_wign(a)yahoo.com wrote: >>> 1. Assume there is a list containing all the reals. >>> 2. Show that a real can be defined/constructed from that list. >>> 3. Show why the real from step 2 is not on the list. >>> 4. Conclude that the premise is wrong because of the contradiction. > > That is a proof by contradiction, which many constructionists object to. No, they don't, presuming you mean constructivists. A direct constructivistic proof of ~A is a proof of contradiction from A. What one can't do constructively is to prove A by proving that ~A leads to a contradiction. -- Aatu Koskensilta (aatu.koskensilta(a)xortec.fi) "Wovon man nicht sprechen kann, daruber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Virgil on 28 Sep 2006 17:58 In article <j%WSg.22037$SY6.17302(a)reader1.news.jippii.net>, Aatu Koskensilta <aatu.koskensilta(a)xortec.fi> wrote: > Virgil wrote: > > In article <1159417542.425540.214160(a)i3g2000cwc.googlegroups.com>, > > cbrown(a)cbrownsystems.com wrote: > >> the_wign(a)yahoo.com wrote: > >>> 1. Assume there is a list containing all the reals. > >>> 2. Show that a real can be defined/constructed from that list. > >>> 3. Show why the real from step 2 is not on the list. > >>> 4. Conclude that the premise is wrong because of the contradiction. > > > > That is a proof by contradiction, which many constructionists object to. > > No, they don't, presuming you mean constructivists. A direct > constructivistic proof of ~A is a proof of contradiction from A. What > one can't do constructively is to prove A by proving that ~A leads to a > contradiction. In my understanding, a "proof by contradiction" is essentially a proof of A by proving ~A false, which requires accepting the law of the excluded middle. I understood that constructivists did not much care for the law of the excluded middle.
From: the_wign on 28 Sep 2006 19:02 Most of the responses point out that a simpler proof or a proof that avoids the problem of the self-reference is to consider ANY list of real numbers. However, that doesn't seem to simplify anything. Now instead of assuming the list contains all reals, we must make no assumptions. But that means we must consider incomplete lists and complete lists, because the construction that occurs in step 2 must be shown to be valid in all cases. In the case of lists that contain all reals, which we must consider unless we are assuming no such list exists (What good is the proof in that case?), then we immediately recognize that the real number described in step #2 as ill-defined and meaningless. So if we consider ANY list we end up with the same problem. In addition we have another case to consider, although I think nobody would object if we neglected to consider that case here.
From: Arturo Magidin on 28 Sep 2006 19:08
In article <virgil-FDA0EC.15584628092006(a)comcast.dca.giganews.com>, Virgil <virgil(a)comcast.net> wrote: >In article <j%WSg.22037$SY6.17302(a)reader1.news.jippii.net>, > Aatu Koskensilta <aatu.koskensilta(a)xortec.fi> wrote: > >> Virgil wrote: >> > In article <1159417542.425540.214160(a)i3g2000cwc.googlegroups.com>, >> > cbrown(a)cbrownsystems.com wrote: >> >> the_wign(a)yahoo.com wrote: >> >>> 1. Assume there is a list containing all the reals. >> >>> 2. Show that a real can be defined/constructed from that list. >> >>> 3. Show why the real from step 2 is not on the list. >> >>> 4. Conclude that the premise is wrong because of the contradiction. >> > >> > That is a proof by contradiction, which many constructionists object to. >> >> No, they don't, presuming you mean constructivists. A direct >> constructivistic proof of ~A is a proof of contradiction from A. What >> one can't do constructively is to prove A by proving that ~A leads to a >> contradiction. > >In my understanding, a "proof by contradiction" is essentially a proof >of A by proving ~A false, which requires accepting the law of the >excluded middle. As I understand it: The "proof by contradiction" blueprint is basically that [A -> (B and ~B)] -> ~A. (or, in some forms, that (P -> ~P) -> ~P ) To be precise, in proof by contradiction, from an assumption A you derive a contradiction, and you deduce ~(A). This is accepted by constructivist. You can assume something exists, for example, deduce a contradiction, and conclude that such a thing does not exist. In the case where A is itself a negation, A = ~B, the constructivist accepts a that from if from A you deduce a contradiction ("deduce" here is restricted to methods allowed by constructivists, of course), then you can conclude ~A. For most mathematicians, this is equivalent to B, since ~A = ~~B, and excluded middle will give you B. For constructivist this is not the case. So: if you assume that there does not exist x such that P(x), and deduce a contradiction, the constructivist will accept that "it is absurd that there does not exist x such that P(x)". He will not, however, accept that there ->exists<- an x such that P(x). i.e., he will accept ~(~(Ex P(x))), but he will not accept Ex P(x). >I understood that constructivists did not much care for the law of the >excluded middle. -- ====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org |