Closed Unit Interval is Homeomorphic to Letter L

Theorem
Let $\R$ be the real number line under the Euclidean metric.

Let $\Bbb I := \closedint 0 1$ be the closed unit interval.

Let $\mathsf L \subseteq \R^2$ denote the letter $L$:
 * $\mathsf L := \closedint 0 1 \times \set 0 \cup \set 0 \times \closedint 0 1$

Then $\Bbb I$ and $\mathsf L$ are homeomorphic.

Proof

 * Letter-L.png

Consider the mapping $f: \Bbb I \to \mathsf L$ defined as:


 * $\forall x \in \Bbb I: \map f x = \begin {cases} \tuple {0, 1 - 2 x} & : x \in \closedint 0 {\dfrac 1 2} \\ \tuple {2 x - 1, 0} & : x \in \closedint {\dfrac 1 2} 1 \end {cases}$

It is seen that:
 * $f \closedint 0 {\dfrac 1 2} = \set 0 \times \closedint 0 1$

and:
 * $f \closedint {\dfrac 1 2} 1 = \closedint 0 1 \times \set 0$

such that:
 * $\map f {\dfrac 1 2} = \tuple {0, 0}$

By construction, $f$ is a bijection.

Its continuity follows from the Combination Theorem for Continuous Real Functions.