More thouhts on infinity

Aug 29, 2009 11:31

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electrichobbit August 29 2009, 18:44:22 UTC
Your mom's uncomfortably infinite.

So that would be Aleph_{your mom}.

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alteredhistory August 30 2009, 20:47:24 UTC
Tell me... do you have the same unsettled feeling that I do of having one of those start with the letter "C" instead of the letter "U"? I've been trying to come up with a U word that means "Countable." It just seems wrong, hanging out there like that.

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electrichobbit August 29 2009, 18:43:01 UTC
Now, let me preface this by saying I'm not a set theoretician. But I don't think this works. For one thing, saying that the cardinality of R is aleph-1 is the continuum hypothesis, which is independent of the traditional notions of set theory (Zermelo-Frankel). But that's somewhat ok...you can add it as a hypothesis. Lord knows we use the axiom of choice enough.

But I don't think the rest follows, even assuming some generalization of the continuum hypothesis. I can't really give you any concrete reason on the train working through my iPhone. But basically it's this: we can think of R as the set of all possibly infinite subsets of Z. Why is your next step the set of all possibly infinite subsets of R? Or, I guess, why is what you have also a valid generalization of the continuum hypothesis?

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alteredhistory August 30 2009, 00:42:32 UTC
Keep in mind I'm just toying... I don't have enough foundations to prove out anything. But the set of all functions is different than the possible subsets of R. It's the possible ordered subsets of R, which themselves are not orderable. In other words, in the same way that R is not countable using integers, the set of functions is not countable using R. For every individual element of set R, there are an infinite number of functions passing through that point, in much the same way that for every given integer there are an infinite number of reals between it and the next.

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electrichobbit August 30 2009, 13:36:40 UTC
I think I see what you're saying now. You're thinking of R as the set of all functions from Z to itself, and then your next level is the set of all functions from R to itself - call it F(R). I think that is Aleph_2, assuming the continuum hypothesis.

But then your next level should be the set of all maps from F(R) to F(R) (we'll call this F^2(R)). I don't see how that jives with the language you're using.

And you could repeat this process ad infinitum - F^3(R) would be maps from F^2(R) to F^2(R), etc. So if anything, you could call your final continuum Aleph_{infinity}...but that would be some kind of direct limit, and I still don't think it would really be "the set of all sets".

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alteredhistory August 30 2009, 20:30:54 UTC
> But then your next level should be the set of all maps from
> F(R) to F(R) (we'll call this F^2(R)). I don't see how that
> jives with the language you're using.

Except that, I think, the map of F(R) to F(R) is already an element of F(R), or, at the very least, doesn't extend cardinality any more than adding dimensions extends the cardinality of the real numbers. I say this because any given F represents some traversal through the real numbers, and if we map that to another traversal through the real numbers, we end up with just a more complex traversal through the real numbers, which should already be somewhere in the set. Function composition falls to the same argument. To truly expand cardinality, you need a new operator, one that maps from F to something *not* F, where each F maps to more than 1 item in the new set, preferably each one maps to infinitely more items in the new set. Thus, concepts.

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