$| \mathbb{R} |^2 = \mathfrak{c} $ - is this true?

Question

In my set theory lecture notes this is one of the corollaries given along with the brief proof: $$ \mathbb{R} ∼ \left[ 0,1 \right] ∼ \left[ 0,1 \right]^2 ∼ \mathbb{R} ^2 $$ Where the notation is given by $A ∼ B$ if there is a bijection $f : A \to B$, and the only of these of these equivalences directly shown in the notes is $ \left[ 0,1 \right] ∼ \left[ 0,1 \right]^2 $ which I understand - however the other 2 equivalences don't immediately seem to be true to me. I am unable to see how a bijection can be formed when for example $ \left[ 0,1 \right]$ has a maximal element but clearly $\mathbb{R}$ does not, maybe there are some steps missing that I am unaware of.

If anybody would be able to enlighten me it would be greatly appreciated, thanks!

Answer 1

You can't find a continuous bijection between $\mathbb{R}$ and $[0,1]$. Its proof requires some topology. I guess it may the reason why you have a trouble to find a bijection between $\mathbb{R}$ and $[0,1]$.

However, we can find a non-continuous bijeection between them; for example, $$f(x) = \begin{cases} \pm 1 & \text{if }x=\pm1 \\ \arctan(\tfrac{1}{n-\operatorname{sgn}(n)\cdot1})& \text{if $x=1/n$ for some integer $n$ such that $|n|>2$,}\\ \arctan(x)&\text{otherwise}\end{cases}$$ is a bijection between $\mathbb{R}$ and $[-1,1]$. Moreover, the function $(x,y)\mapsto (f(x),f(y))$ is a bijection between $\mathbb{R}^2$ and $[-1,1]^2$.

Answer 2

It is perhaps more instructive to build up the correspondence in stages:

• Any two open intervals are equinumerous.

• Any two half-open intervals are equinumerous.

• $(0,1]\sim (0,0)$ by the bijection $$ f(x) = \begin{cases} 1/(n+1) & \text{if $x=1/n$ for some $n\in\mathbb N_+$} \\ x & \text{otherwise}\end{cases}$$

  So every half-open interval is equinumerous with every open interval.

• $(0,1)\sim (0,\infty)$ by the bijection $g(x) = 1/x-1$.

So to map $[0,1]$ to $\mathbb R$, first partition $[0,1]$ into two half-open intervals and a single point. Map the point to $0$, and the half-open intervals to $(-\infty,0)$ and $(0,\infty)$.