Re: Logarithm of transfinite numbers

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

The 2^x=aleph_0 and aleph_0 is finite. Is it aleph_0?

Yes.

Not if that makes 2^aleph_0 strings.

No. You don't need ALL of the length aleph_0 strings to
make the natural numbers. Just some of them.

Ahem! How many is "some"? (sigh) That's not very mathematically precise.

Actually, to make the naturals, you don't need any infinite strings.

That's because it's not an infinite set. N=S^L is only infinite if either S is
infinite (it's 2) or L is infinite (it's finite for every n in N).


Is zero precise enough for you, TO?

Zero bits is sifficient to denote every element in a singleton set, period.
That's because 2^0=1.


You just need that there is no finite bound on lengths.


Well, so maybe the set doesn't need to be infinite, but just to have no finite
bound on its size. how is that different?
--
Smiles,

Tony
.

Relevant Pages

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