Re: Logarithm of transfinite numbers

Virgil said:
In article <MPG.1e80bc50bd90201a98aad1@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

But, if you have aleph_0 bits, then you have an uncountable set of strings,
right? How many bits, exactly, gives you a countably infinite set of strings?
You can't answer that in your theory.

Nor can TO in his, until he can cobble up some cardinal X such that
2^X = aleph_0.

First of all, countable is not infinite in my theory. Second, I've already done
this, when I defined log2(N) as a possible limit point in the T-riffics. Just
as one can define infinite descending chains by decrementation or by division,
one can do it with logs as well.


But if a set S of cardinality X does not allow injections into proper
subsets, then 2^X < aleph_0, and if it does then 2^X > aleph_0.


Only in the system of cardinalities, which have nothing to do with my theory.
You complaint hat I impose my theory on yours, but that's what you're doing
here. Cardinality and aleph_0 are both schlock, and have no place in a proper
theory regarding infinite sets and quantities.

--
Smiles,

Tony
.

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