Does any uncountable subset of $\mathbb{R^n}$ have the same cardinality as $\mathbb{R}^m$?

Question

We can map $[0,1]$ to $\mathbb{R}$ (in fact any closed interval of $\mathbb{R}$, there's nothing special about $[0, 1]$) bijectively, but is the same true for $[0,1]\times[0,1] \times[0,1]$ and $\mathbb{R}^3$? Also, do $\mathbb{R}^\mathrm{n}$ and $\mathbb{R}^\mathrm{m}$ always have the same cardinality? If so, how would one go about proving it and what's some helpful intuition there?

If anyone's wondering, I thought of this question just now when I saw that beautiful visual proof that any segment of a line can be mapped to a whole line by stereographic projection. It would be interesting to know how far that can be generalized.

Answer 1

We can map $[0,1]$ to $\mathbb{R}$ (in fact any closed interval of $\mathbb{R}$, there's nothing special about $[0, 1]$) bijectively, but is the same true for $[0,1]\times[0,1] \times[0,1]$ and $\mathbb{R}^3$? Also, do $\mathbb{R}^\mathrm{n}$ and $\mathbb{R}^\mathrm{m}$ always have the same cardinality? If so, how would one go about proving it and what's some helpful intuition there?

Saucy O'Path's answer does a pretty good job of answering for these particular examples: all of the sets that you mentioned have the same cardinality.

The relevant theorem here is:

• If $\kappa$ is an infinite cardinal number, then $\kappa \cdot \kappa = \kappa$.
• Therefore, if $\kappa$ is an infinite cardinal number and $n$ is a positive integer, then $\kappa^n = \kappa$.

But your title question is: "Does any uncountable subset of $\mathbb{R^n}$ have the same cardinality as $\mathbb{R}^m$?" By the above theorem, this question is equivalent to "Does every uncountable subset of $\mathbb{R}$ have the same cardinality of $\mathbb{R}$?" This question is the continuum hypothesis, and both a "yes" answer and a "no" answer are consistent with ZFC.

Answer 2

• Once you have a bijection $f:A\to B$, then you automatically have a bijection $f_n:A^n\to B^n$ in $f_n(a_1,\cdots, a_n)=(f(a_1),\cdots, f(a_n))$.

• $\Bbb R^n$ indeed has the same cardinality as $\Bbb R^m$ for al $n,m\ne 0$, and even as $\Bbb R^{\Bbb N}$, the set of functions which have domain $\Bbb N$ and codomain $\Bbb R$. The quickest way to see it is by observing that $\Bbb R\cong \{0,1\}^{\Bbb N}$ and so $\Bbb R^{A}\cong \{0,1\}^{\Bbb N\times A}$; since $\Bbb N\times A\cong \Bbb N$ for all countable non-empty $A$, this results in a bijection $\Bbb R^A\cong\{0,1\}^{\Bbb N}\cong \Bbb R$.

• Of course, $\Bbb R$ and $\Bbb R^n$ contain subsets which are not in bijection with $\Bbb R$. Namely, $\Bbb N$. On the other hand, the statement $$CH\stackrel{\text{def}}\equiv \forall S\subseteq \Bbb R,\ \lvert S\rvert\le\lvert \Bbb N\rvert\lor \lvert S\rvert=\lvert\Bbb R\rvert$$ is a famous instance of a statement such that $ZFC+CH$ proves $[1=0]$ if and only if $ZFC+\neg CH$ proves $[1=0]$, if and only if $ZFC$ proves $[1=0]$.

  In more English terms, $CH$ means "every uncountable subset of $\Bbb R$ is in bijection with $\Bbb R$", and $\neg CH$ is of course its negation. It has been shown that the following are equivalent:

  1. $ZFC$ proves a contradiction (I picked $[1=0]$, but perhaps I should have picked $[\emptyset\ne\emptyset]$)
  2. $ZFC+CH$ proves a contradiction
  3. $ZFC+\neg CH$ proves a contradiction

Answer 3

One way to see that $\mathbb{R}^2 \cong \mathbb{R}$ is to take the decimal expansions of the pair $(x,y)$ and simply alternate the digits of both beginning at the decimal point and working outward. With a little care you can make this a well-defined and invertible map so it's a bijection and we have that the cardinality is the same. This argument can be easily generalized to show $\mathbb{R}^n \cong \mathbb{R}$ as well and then you have that all of them have the same cardinality.