open and closed intervals have the same cardinality

Proposition.

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

We give two proofs of this proposition.

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 $[\frac{1}{3},\frac{2}{3}]⊊(0,1)$, 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 (http://planetmath.org/SchroederBernsteinTheorem) 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.∎

The reasoning above can be extended to show that any two arbitrary intervals in $ℝ$ have the same cardinality.