Closed Unit Interval is Homeomorphic to Letter L

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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.

$\blacksquare$


Sources