Cardinality of Euclidean SpaceDate: 09/13/2005 at 15:57:18 From: Dejan Subject: Cardinal numbers Cardinal number of a real line is aleph_1 (continuum). Cardinal number of a n-dimensional space is also aleph_1 (Cantor proved it, assuming continuum hypothesis). The Question: What is the cardinal number of a n-dimensional Euclidean space R^n where n tends to infinity? Is it aleph_2? To specify, by "n tends to infinity" I mean "n tends to aleph_0". Also, what do we get if "n tends to aleph_1", or "n tends to aleph_2"...??? Thank you very much, Dejan Date: 09/13/2005 at 16:47:25 From: Doctor Vogler Subject: Re: Cardinal numbers Hi Dejan, Thanks for writing to Dr. Math. The phrase "tends to" especially in the context of "tends to infinity" refers to limits, and a space is not a limit. So that leaves me wondering what you mean by "the cardinal number of a n-dimensional Euclidean space R^n where n tends to infinity". If you mean to get the cardinal number and then take the limit as n goes to infinity, then of course the cardinal number is constant, so the limit is the same as the cardinal number of R. But somehow I don't think that's what you mean. I think you are trying to describe an infinite-dimensional real vector space. There are two infinite-dimensional real vector spaces that are frequently used. One is the countable direct sum of R, and the other is the countable Cartesian product of R. The first space can be thought of as the space of polynomials with real coefficients, or the space of finite sequences of real numbers. The point is that these sequences (or polynomials) can have arbitrarily long length, but always a *finite* length. The second space can be thought of as the space of (formal) power series with real coefficients, or the space of infinite (countable) sequences of real numbers. This is a much bigger space and is generally considered to contain the former by extending a finite sequence by all zeros. As an example of how much bigger it is, the countable direct sum of a countable set (e.g. the space of finite sequences of rational numbers) is a countable set, but the countable Cartesian product of a countable set (e.g. the space of infinite sequences of rational numbers) is uncountable. Given the above, it should not be surprising that the countable direct sum of R (finite sequences of real numbers) has the same cardinality as R. But it leaves you wondering whether the countable Cartesian product of R has the same cardinality or a higher one. In fact, it does have the same cardinality as R. I think the easiest way to see this is to remember that the cardinality of R is the same as the cardinality of the power set (the set of subsets) of the natural numbers, which is the same as maps from the natural numbers into {0,1}. So now consider infinite sequences of subsets of the natural numbers, or infinite sequences of maps from the natural numbers into {0,1}. But an infinite sequence is just a map from the natural numbers too, so that means that we are looking at the set of functions from the natural numbers to the set of functions from the natural numbers into {0,1}, and this is the same as the set of functions from pairs of natural numbers into {0,1}. Finally, since the set of pairs of natural numbers is countable, this is the same as the set of functions from the natural numbers into {0,1}, whose cardinality is the same as the real numbers R. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 09/14/2005 at 14:31:36 From: Dejan Subject: Cardinal numbers Dear Dr. Vogler, Thank you very much for your very comprehensive answer. You were right, I was trying to describe an infinite-dimensional real vector space. In fact, a much better way to formulate my question is: what is the (cardinal) number of all points in a infinite dimensional space where the cardinal number of the basis (the number of vectors in the basis) is (a) aleph_0 (b) aleph_1 (c) aleph_2 etc. ? From your answer I concluded that the countable Cartesian product of R (i.e. the case where the number of basis vectors in a infinite dimensional real vector space is aleph_o) has the same cardinality as R, i.e. aleph_1. But, what is happening if we go one step further, and consider an infinite dimensional real vector space where the cardinality of the basis is aleph_1, or aleph_2, etc? Thank you very much. Dejan Date: 09/14/2005 at 16:18:04 From: Doctor Vogler Subject: Re: Cardinal numbers Hi Dejan, First of all, it is the direct sum (not the Cartesian product) that is the infinite-dimensional vector space with countable basis. The countable Cartesian product is an infinite-dimensional vector space with a basis whose cardinality is the same as R. In any case, if k is a cardinality BIGGER than (or equal to) the cardinality of the real numbers, then either a direct sum or a Cartesian product of k copies of R will have cardinality k. Can you prove that? By the way, the term "aleph one" is usually used to mean "the smallest cardinality bigger than aleph-naught" where aleph-naught means countable (cardinality of the integers). It was conjectured that aleph one is, in fact, the cardinality of the real numbers (this is called the "Continuum Hypothesis"), but it was later shown that this conjecture or hypothesis was neither true nor false but was independent of the usual axioms of mathematical logic. That is, if you assume that it is true, you won't get into any contradictions that weren't there before, or if you assume that it is false, you still won't. So the cardinality of the real numbers is not necessarily aleph one. It is only if you are assuming the Continuum Hypothesis. That is why I was careful to say "the cardinality of R" instead of "aleph one" in each of my responses. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 09/16/2005 at 11:23:40 From: Dejan Subject: Cardinal numbers Dear Dr. Vogler, Thank you very much again for your very comprehensive answer. It helps a lot. So, the answer to my question is that "if k is a cardinality BIGGER than (or equal to) the cardinality of the real numbers, then either a direct sum or a Cartesian product of k copies of R will have cardinality k". I know that a Cartesian product of k copies of R will have cardinality R if k is a finite number (Cantor proved it). I can somehow imagine that this remains true if k is a countable number. However, the non-trivial part for me is to extend it further, i.e. a Cartesian product of k copies of R will have cardinality k (provided that k is a cardinality bigger than or equal to R). Sorry for using the terms "aleph one" and the cardinality of R interchangeably, I was aware of it but was not being very careful. Thank you very much. Dejan Date: 09/17/2005 at 10:43:05 From: Doctor Vogler Subject: Re: Cardinal numbers Hi Dejan, By the way, some people write "c" for the cardinality of the real numbers, as in MathWorld at Continuum http://mathworld.wolfram.com/Continuum.html and Aleph-1 http://mathworld.wolfram.com/Aleph-1.html Others don't like that notation, since c is used for so many other things, but instead write 2^(aleph-0), which is equal to c. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 09/17/2005 at 11:16:57 From: Dejan Subject: Thank you (Cardinal numbers) Dear Dr. Vogler, Thank you very much again for your kind answers. I certainly received satisfactory answers to my questions. I appreciate your time and efforts. Best wishes, Dejan |
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