Re: Another stab at Cantor

On 19 Sep 2006 12:05:13 -0700, "georgie" <geo_cant@xxxxxxxxx> wrote:

Virgil wrote:

If L_Omega is not a list and its members cannot be listed, then it is
already uncountable, and Cantor wins.

If black is defined as white, and white is defined as black, then black
is
black and white is whate and black is white and white is black.
Therefore,
Cantor Wins.
QED

But what about the limiting case not being a list and therefore not
subject to diagonalization? Oh yeah. I almost forgot. There's
a bunch of explanations:

1. That's trolling.
2. It doesn't apply because we said it doesn't.
3. It wasn't proven that the final set has all the reals so we don't
want to explain any further.
4. We can come up with personal attacks and diversions that
deflect away from that fact.
5. Cantor Wins.

But if the final result isn't a list, it doesn't contradict the
result of Cantor.

--
Wim Benthem
.

Relevant Pages

  • Re: An uncountable countable set
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  • Re: abundance of irrationals!)
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  • Re: abundance of irrationals!)
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  • Re: Cantor and the binary tree
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