# Is $[0;1]$ equinumerous to $\mathbb{R}^{\omega}$?

Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space.

I was wondering why is it the case for finite-dimensional spaces, but not for $\mathbb{R}^{\omega}$. Can't a PMI proof of a bijection between $[0;1]$ and $\mathbb{R}^{\omega}$ be established, or would that proof be showing that one can establish a bijection between $[0;1]$ and $\mathbb{R}^n$ for any given $n\in \mathbb{R}/\mathbb{N}$ but not $\mathbb{R}^{\omega}$? Is there something special about infinitely-dimensional $R$ that makes it have more points than $\mathbb{R}$ or any segment on $\mathbb{R}$? Thanks a lot.

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You should be careful with writing "$R^\infty$". Some authors - e.g. Munkres in his Topology - take $\mathbb{R}^\infty$ to mean the set of sequences in $\mathbb{R}$ that are eventually 0 and $\mathbb{R}^\omega$ to be the set of sequences in $\mathbb{R}$ –  kahen Dec 27 '10 at 0:38
Should I keep using $\lim_{x \to \infty}$ $R^x$ in that case, if I want to be talking about infinitely-dimensional R-space? –  user5135 Dec 27 '10 at 0:39
"$\displaystyle \lim_{x \to \infty} \mathbb{R}^x$" leaves a lot of room for interpretation which is a bad thing. For example it isn't clear whether you're talking about $\mathbb{R}^\mathbb{N}$ or $\mathbb{R}^{[0,\infty)}$ or some other set –  kahen Dec 27 '10 at 0:43
@user5135 : For any $A$, ${\mathbb R}^A$ is the set of functions from $A$ to ${\mathbb R}$. Since cardinals in set theory are sets, we can define ${\mathbb R}^\kappa$ for any cardinal $\kappa$, be it finite, countably infinite, or uncountable. For example, in set theory, $n=\{0,1,\dots,n-1\}$, and there there is an obvious bijection between ${\mathbb R}^n$ (understood as a set of $n$-tuples, as usual) and ${\mathbb R}^n$ in this new sense. The advantage of the functional notation is that we can talk about any dimension $\kappa$ we want, and ${\mathbb R}^\kappa$ is well-defined & unambiguous. –  Andres Caicedo Dec 27 '10 at 1:46
@user5135 : The question of whether ${\mathbb R}^\kappa$ has the same size as ${\mathbb R}$ for uncountable $\kappa$ is a tricky one. Cantor's theorem shows that if ${\mathfrak c}$ is the size of the reals, then ${\mathbb R}^{\mathfrak c}$ is strictly larger than ${\mathbb R}$. On the other hand, $\omega_1$ is the least uncountable cardinal, and it is independent of the usual axioms of set theory whether ${\mathbb R}^{\omega_1}$ has the same size as ${\mathbb R}$ or is strictly larger. –  Andres Caicedo Dec 27 '10 at 1:49

Whoever told you that $[0, 1]$ is smaller than $\mathbb{R}^{\omega}$ is wrong; they both have the same cardinality as $\{ 0, 1 \}^{\omega} = \{ 0, 1 \}^{\omega \cdot \omega}$. By the way, something about your post suggests to me that you are not convinced that this set has a rigorous definition, and it does: $\mathbb{R}^{\omega}$ is precisely the set of functions from $\omega$ to $\mathbb{R}$.

There are several other ways one might imagine putting infinitely many copies of $\mathbb{R}$ together. The above is the infinite direct product; there is also the infinite direct sum $\bigoplus_{n=1}^{\infty} \mathbb{R}$, which consists of all sequences of real numbers which are eventually zero; unlike the infinite direct product, the infinite direct sum is countable-dimensional. And also one can consider the sequences $c_0$ of real numbers which converge to zero; this is not countable-dimensional but has nice topological properties.

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Could you elaborate a little? So it's possible to construct a bijection between [0;1] and infinitely dimensional R-space (which from what I understand is different from $R^n$ for any $n \in N$)? –  user5135 Dec 27 '10 at 0:41
Yes it is. First one uses the fact that R has the same cardinality as {0, 1}^{\omega} (you can think of this as because of binary expansion), and then it follows by formal properties of cardinal arithmetic and the fact that omega*omega has the same cardinality as omega. (Here omega denotes the smallest infinite cardinal.) –  Qiaochu Yuan Dec 27 '10 at 0:44
I see, thanks a lot! So I guess the wikipedia article didn't tell all the truth. –  user5135 Dec 27 '10 at 0:45
@user5135: if by that you mean that the wikipedia article didn't include every true statement about cardinalities of sets...well, in order to do that it would have to be infinitely long, which would be hard to achieve (not to mention very stressful on your browser). But I take your point that it stops short of stating the strongest true fact in this particular case. It might be worthwhile for someone to edit the article to include the observation that also a countable direct product of copies of $\mathbb{R}$ has continuum cardinality. –  Pete L. Clark Dec 27 '10 at 1:04
"But I take your point that it stops short of stating the strongest true fact in this particular case." Yes, that's what I meant -- it seemed strange to stop so short of it. –  user5135 Dec 27 '10 at 1:07

The cleanest way of seeing the bijection from $\mathbb{R}^{\omega}$ to $\mathbb{R}$ is probably by taking advantage of your favorite bijection from $\mathbb{R}$ to $\left[0,1\right]$, then treating the numbers in that range as a sequence of decimals, 'stacking' them to get a $\omega\times\omega$ grid of digits, and using a diagonal sweep to compose them into a single decimal; so far instance, if the first three elements of your sequence of reals were .314159, .271828, and .161803, then take the first digit of the first, then the second digit of the first and the first digit of the second, then the third digit of the first, the second digit of the second, and the first digit of the first, continuing to interleave this way: .312471... It's easy to convince yourself that this is a one-to-one bijection; you can go the other way by writing the digits of your single number one by one diagonally into a grid, and then taking the result as a series of reals.

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Thanks a lot, this makes sense. –  user5135 Dec 27 '10 at 1:08
No, I disagree: the argument using cardinal arithmetic is much cleaner than this. Your way is probably the most immediately perspicuous to a general audience, though. (+1) –  Pete L. Clark Dec 27 '10 at 2:33
...except that it isn't a one-to-one bijection: you run into problems having to do with uniqueness of decimal expansions. If you had said "your favorite bijection from R to {0, 1}^w," of course, everything would work out fine (and this would just be a more concrete way of stating the argument using cardinal arithmetic). –  Qiaochu Yuan Dec 27 '10 at 4:32
Good point, Qiaochu - when you're dealing with only one real it's easy to massage out the uniqueness issue, but there's no getting around it for this particular mapping. I hadn't even considered that aspect of it; mea culpa... –  Steven Stadnicki Dec 27 '10 at 6:04