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July 19th, 2016, 06:44 PM   #21
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Quote:
Originally Posted by JeffM1 View Post
Nothing in Cantor's proof depends on a "list" however defined.
Well, I would say that both an enumeration and a sequence are types of list and that Cantor uses both, but this is now becoming a semantic argument of little importance to people who understand the proof (among whom I include you).

To me, the most elegant formulation of the proof uses explicit indices in both the enumeration and the sequences and does away entirely with the proof by contradiction which is the root of many misunderstandings in favour of a subsidiary theorem that no countable set of sequences/real numbers contains every such sequence/real number. The contrapositive of this is the statement that the reals are not countable.

Quote:
Originally Posted by JeffM1 View Post
You do not believe that Zylo ever has in mind an infinite set, let alone a set of all the real numbers in (0, 1). If you are correct, his argument is infantile because a finite set cannot by definition be put into one-to-one correspondence with an infinite one. On the other hand, I believe that Zylo has in mind (at least some of the time) the set of all real numbers in the interval [0, 1) or (0, 1). He does not, however, even attempt (and would necessarily fail if he did attempt) to show that the set is in one-to-one correspondence with the natural numbers.
There is a difference of opinion here, yes. But you characterise my belief incorrectly. He has in mind the set of reals (sometimes over an interval as in this case). His "proof" then consists of constructing that set in the form of a list. Were he to be successful, this would constitute the required 1-1 mapping with the natural numbers. Of course, he always fails and the reason is that he always works with an countably infinite number of finite subsets of the reals: those consisting of $n \in \mathbb N$ digits after the decimal point. As his most recent post in this thread shows, he is unable to understand that the "limit" as $n$ grows without bound does not exist in his formulation and that his formulation doesn't work for non-terminating decimals and that as a result every one of his real numbers is of the terminating variety.

So his "proof" fails not because there is no mapping with the naturals: because it is a list the mapping exists and the set is countably infinite; but because the set he creates is not the set he believes it to be: the real numbers. $\frac13$, $(\pi-3)$ and $\sqrt2 -1$ are not in it. He is unable or unwilling to see this fact.
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Last edited by v8archie; July 19th, 2016 at 06:51 PM.
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