Cardinality of $[0,1]$ and $\mathbb{R}$

Question

As far as I remember the closed set $[0,1] \subset \mathbb{R}$ contains every real number being that the cardinality of $[0,1]$ is the same of $\mathbb{R}$. If so can someone post an explicit biijection and in such case what is the antiimage of $\pi$ and $e$?

Answer 1

A bijection from $[0,1]$ to $(0,1)$, is the following $f(0)=1/2$, $f(1)=1/4$, for $n\geq 1$, $f(1/2^n)=1/2^{n+2}$ and $f(x)=x$ otherwise. A bijection from $(0,1)$ to $\mathbb{R}$ is $g(x)=\tan(\pi(x-1/2))$. Hence a bijection from $[0,1]$ to $\mathbb{R}$, if $h:=g\circ f$. Therefore $h^{-1}(e)=g^{-1}(e)=\frac{1}{2}+\frac{\arctan(e)}{\pi}$.

Answer 2

The Cantor-Bernstein Theorem shows - explicitly! - that if I have injections from $A$ to $B$ and from $B$ to $A$, then there is a bijection from $A$ to $B$. By "explicitly," I mean that the proof actually constructs a bijection from the two injections given.

To apply this here, note that $[0, 1]$ clearly injects into $\mathbb{R}$, being a subset of it; in the other direction, consider the map $x\mapsto {2(\arctan(x)+{\pi\over 2})\over\pi}.$ This map injects $\mathbb{R}$ into $[0, 1]$ (with image $(0, 1)$).

Note, however, that there is no continuous bijection between $[0,1]$ and $\mathbb{R}$; so any bijection is going to be a bit messy.

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Here's an outline of a specific example.

• First, let $f$ be the map given above which injects $\mathbb{R}$ into $[0, 1]$. Now, this fails to be a bijection because it missed two points . . .

• So let's fix it! Pick a countable subset $A$ of $(0, 1)$ - e.g. $\mathbb{Q}\cap (0, 1)$. Write this subset as $A=\{a_1, a_2, a_3, . . .\}$.

• Now, consider the map $g$ sending $a_1$ to $0$, $a_2$ to $1$, and $a_{i+2}$ to $a_i$. (Think Hilbert's hotel.)

• The map $g\circ f$ is a bijection from $\mathbb{R}$ to $[0, 1]$.

You asked about what happens to specific real numbers under such a bijection. Well, obviously that depends on the bijection used - there are lots of bijections from $\mathbb{R}$ to $[0, 1]$, and we can arrange for $e$ to go to whatever real number we want it to!

In the example I've given here, the choice you have is basically what countable set you pick, and how you enumerate it. For example, since $f(e)$ is irrational, in the specific case above $g$ wouldn't move $f(e)$, so $g\circ f(e)=f(e)={2(\arctan(e)+{\pi\over 2})\over\pi}$. Similarly with $f(\pi)$.