Re: induction vs Cantor

From: Poker Joker (Poker_at_wi.rr.com)
Date: 11/28/04 

Date: Sun, 28 Nov 2004 18:52:02 GMT

"J.E." <troubled6man@yahoo.com> wrote in message
news:39d6e584.0411271230.3f762cf3@posting.google.com...

<snipped>

> If you take the union of all the L_n, then you get a countable set, so
> you can construct a NEW list L_{omega} that lists every element in L_1
> and all the D_n. But if you apply the diagonal arguement to L_{omega}
> then you get D_{omega} which isn't in the union, so it isn't in L_1 or
> any of the D_n. If you made a new list L_{omega+1} which had
> D_{omega} in it, then I could use the diagonal arguement again to get
> D_{omega+1}. The whole point is that for any list L_X, there is a
> real D_X that isn't on L_X. So you can make as many L's as you want,
> but there is always another D. It's like the proof of an infinite
> number of primes. One way to do it is to assume that there are a
> finite number p1 to pn then multiple them together and add one. This
> new number is not divisible by any of the others, and so it's a new
> prime. The diagonal arguement takes a countable number of things and
> produces "another one" not matter how many "countable" things there
> were.

Yes, its like the primes. There are only countably many primes.

>> > Have you followed through with your process a little
>> > bit with a particular L_0? All of the D_n's have the
>> > property that, if k > n, then the first (n/2)-1 digits
>> > of D_n match the first (n/2)-1 digits of D_k.
>>
>> No I haven't but I would have to add that there are
>> countably many ways of creating "anti-diagonal" numbers
>> so I'm not certain how you are doing that analysis.
>
> The thing is that there are as many ways to make anti-diagonal numbers
> as there are lists, because for every list there is another "new"
> antidiagonal number. So as in my example above if you union L_n
> together to get a "new" list (which we'll call L_{omega}, then there
> is a "new" anti-diagonal.

All the descriptions of the methods for creating a number not in the
list forms a countable set. You can call one of the lists "new" but
it's just one list. We always have room for one more in our
countable set.

Relevant Pages

  • Re: induction vs Cantor
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  • Re: Proof by contradiction
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