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Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 8:49 pm    Post subject: 41 Of course you can do something that's impossible if you don't actually have to do it to consider it to be done.
MTGAP
Daedalian Member

 Posted: Sun Aug 17, 2008 4:14 am    Post subject: 42 Is it true that if you have 2^infinity objects, you cannot line them up, no matter how long the line is? And that you need an infinite number of dimensions to place all the objects? I read somewhere that 2^infinity is equal to aleph one, and all alephs above one. Is that true? It doesn't seem right to me._________________This statement is false.
Bicho the Inhaler
Daedalian Member

 Posted: Sun Aug 17, 2008 4:33 am    Post subject: 43 I think at this point you should just get a good discrete mathematics textbook. Your questions seem a bit ill-posed, but reading about cardinal numbers would go a long way toward addressing whatever you have in mind. (I'd recommend the book I had in high school, but I don't remember the exact title or the author. )
ralphmerridew
Daedalian Member

 Posted: Sun Aug 17, 2008 10:35 am    Post subject: 44 MTGAP: The proposition that aleph-one is the power set of aleph-null (the "Continuum Hypothesis") is undecidable within standard set theory.
MTGAP
Daedalian Member

Posted: Sun Aug 17, 2008 8:41 pm    Post subject: 45

 ralphmerridew wrote: MTGAP: The proposition that aleph-one is the power set of aleph-null (the "Continuum Hypothesis") is undecidable within standard set theory.

Ah, so I'm answering questions that there's no answer to.

Wonderful.
_________________
This statement is false.
Chuck
Daedalian Member

 Posted: Thu Aug 21, 2008 3:44 pm    Post subject: 46 It seems like you should be able to represent all of the subsets of the integers as all of the binary real numbers from zero to one. Each binary number would represent one subset. If an integer is in the subset then the binary digit in that position in the corresponding real number would be one, otherwise it would be zero. That way each real number from zero to one corresponds to a different subset of the integers since each such point has a different binary expansion. None would be left out. No subsets of the integers would be left out since each would convert to a different binary number. They'd be the same order of infinity.
alphatango
Daedalian Member

 Posted: Thu Aug 21, 2008 5:31 pm    Post subject: 47 You can indeed do such a thing -- see this everything2 article, for example. In particular, it deals with the problem under your scheme where two different subsets map to the same thing; 0.0111... and 0.1000..., for example, would represent the sets {2, 3, 4, ...} and {1}, respectively, but the two binary numbers are equal._________________"Hanging is too good for a man who makes puns; he should be drawn and quoted." -- Fred Allen Keeper of the Eternal Flame of the Inner Geek.
ralphmerridew
Daedalian Member

 Posted: Thu Aug 21, 2008 6:19 pm    Post subject: 48 Chuck: That is possible. Forming a bijection beween the power set of the integers and the integers is not possible.
MTGAP
Daedalian Member

 Posted: Sat Sep 27, 2008 3:54 pm    Post subject: 49 Some more questions: 1. If there are infinite infinities, how many is that? Are there aleph null infinities? Are there aleph (aleph null) infinities? 2. If there are infinite infinities, that means there are numbers between irrationals, right? Or does it mean that all the infinities above aleph one are not representable with numbers? If so, are they representable at all?_________________This statement is false.
Nsof
Daedalian Member

 Posted: Sat Sep 27, 2008 10:53 pm    Post subject: 50 I think your first question depends on the Continuum hypothesis_________________Will sell this place for beer
extro...*
Guest

Posted: Sun Sep 28, 2008 3:28 am    Post subject: 51

 Nsof wrote: I think your first question depends on the Continuum hypothesis

A bit rusty, but I think:
1) CH (continuum hypothesis) implies aleph null infinities
2) "CH is false" does not imply aleph null infinities, nor does it imply more than aleph null.

With CH, infinities are ordered like the natural numbers, each having one that is greater, with none in between the two. Without CH, infinities may be ordered like the non-negative rational numbers - aleph null, but for any two that are not equal, there are others between them.
MTGAP
Daedalian Member

 Posted: Mon Oct 13, 2008 5:29 pm    Post subject: 52 Apparently, it's been proved that there are more curves than there are real numbers. But I can't find this proof anywhere; no one seems to care about it, and instead they bother with things like proofs that there are more irrationals than rationals. So what's the proof?_________________This statement is false.
extro...*
Guest

Posted: Mon Oct 13, 2008 7:52 pm    Post subject: 53

 MTGAP wrote: Apparently, it's been proved that there are more curves than there are real numbers. ... So what's the proof?

Define "curve".

If you mean functions from reals to reals, it's pretty straightforward.
extro...*
Guest

 Posted: Tue Oct 14, 2008 10:55 pm    Post subject: 54 If by curve you mean a continuous function, it sounds like an interesting problem - to prove that the size of the set of all such functions is the same as the size of the set of reals ... or greater? For non-continuous functions, one need only consider that for every set R of reals, there is some "characteristic" function f such that f(x) = 1 if x is in R, and f(x) = 0 if x is not in R. It's trivial to prove the powerset of an infinite set S is of greater cardinality than S itself, so similarly the set of functions from R to {0,1} is of greater cardinality than the set R. But many of these characteristic functions are non-continuous (with some exceptions). Any ideas about the continuous case? How to show the set of continuous functions from R to R is larger than R, or the same as R? (it's at least as big as R) ... The problem sounds simple enough.
extro...*
Guest

 Posted: Tue Oct 14, 2008 11:58 pm    Post subject: 55 OK, so according to a wikipedia page, http://en.wikipedia.org/wiki/Cardinality_of_the_continuum#Sets_with_cardinality_c , Sets with cardinality c A great many sets studied in mathematics have cardinality equal to c. Some common examples are the following: * the real numbers R * any (nondegenerate) closed or open interval in R (such as the unit interval [0,1]) * the irrational numbers * the transcendental numbers * Euclidean space Rn * the complex numbers C * the power set of the natural numbers (the set of all subsets of the natural numbers) * the set of sequences of integers (i.e. all functions N → Z, often denoted ZN) * the set of sequences of real numbers, RN * the set of all continuous functions from R to R * the Cantor set * the Euclidean topology on Rn (i.e. the set of all open sets in Rn) * the Borel σ-algebra on R (i.e. the set of all Borel sets in R)
ralphmerridew
Daedalian Member

 Posted: Wed Oct 15, 2008 1:06 am    Post subject: 56 I think I have a proof of that statement: Let g:Q->Z+ be a bijection (possible since rationals are known to be countable). Let h:R->[0,1] be a bijection Consider the set of functions f:Q->[0,1] i: The set can be injected into the reals. Associate a function f with the real number sum(i=1 to inf: d_i * 10^-i) where d_i = the m-th digit of the decimal expansion of f(g(k)) if i = ((2m+1)*2^(k+1)) for positive m and k d_i = 0 otherwise ii: The continuous functions k:R->R can be injected into those functions f. (Calculate h(k(x)) at each rational x. Continuity allows at most one k to map to any f.) iii: Putting those two injections together gives an injection from functions k to R. iv: And the reals can be injected into the continuous functions trivially.
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