open and closed intervals have the same cardinality

Proposition.

The sets of real numbers $[0,1]$ , $[0,1)$ , $(0,1]$ , and $(0,1)$ 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 $[\frac{1}{3},\frac{2}{3}]\subsetneq(0,1)$ , 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 (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

h(x)=\begin{cases}f(x),&x\in\mathbb{Q}\\ x,&x\notin\mathbb{Q}.\end{cases}

The other bijections can be constructed similarly. ∎

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

Title	open and closed intervals have the same cardinality
Canonical name	OpenAndClosedIntervalsHaveTheSameCardinality
Date of creation	2013-03-22 15:43:32
Last modified on	2013-03-22 15:43:32
Owner	mps (409)
Last modified by	mps (409)
Numerical id	8
Author	mps (409)
Entry type	Result
Classification	msc 26A03
Classification	msc 03E10