Proposition   The sets of real numbers $[0,1]$ , $[0,1)$ , $(0,1]$ , and $(0,1)$ all have the same cardinality.

Proof. Define a map $f:[0,1]\to[0,1]$ by $f(x)=(x+1)/3$ . The map $f$ is strictly increasing, hence injective. Moreover, the image of $f$ is contained in the interval $[\frac{1}{3}, \frac{2}{3}]\subsetneq (0,1)$ , so the maps $f_r:[0,1]\to[0,1)$ and $f_o:[0,1]\to(0,1)$ obtained from $f$ by restricting the codomain are both injective. Since the inclusions into $[0,1]$ are also injective, the Cantor-Schröder-Bernstein theorem can be used to construct bijections $h_r:[0,1]\to[0,1)$ and $h_o:[0,1]\to(0,1)$ . Finally, the map $r:(0,1]\to[0,1)$ defined by $r(x)=1-x$ is a bijection.

Since having the same cardinality is an equivalence relation, all four intervals have the same cardinality.

Proof. Since $[0,1]\cap\mathbb{Q}$ is countable, there is a bijection $a:\mathbb{N}\to[0,1]\cap\mathbb{Q}$ . We may select $a$ so that $a(0)=0$ and $a(1)=1$ . The map $f:[0,1]\cap\mathbb{Q}\to(0,1)\cap\mathbb{Q}$ defined by $f(x)=a(a^{-1}(x)+2)$ is a bijection because it is a composition of bijections. A bijection $h:[0,1]\to(0,1)$ can be constructed by gluing the map $f$ to the identity map on $(0,1)\setminus\mathbb{Q}$ . The formula for $h$ is

The other bijections can be constructed similarly.