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Cardinality of Euclidean Space

Date: 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
Associated Topics:
College Logic
College Modern Algebra

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