Letter L and Letter T are not Homeomorphic

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Theorem

Let $\R^2$ denote the real number plane under the Euclidean topology.

Let $\mathsf L \subseteq \R^2$ denote the letter $L$:

$\mathsf L := \closedint 0 1 \times \set 0 \cup \set 0 \times \closedint 0 1$

Let $\mathsf T \subseteq \R^2$ denote the letter $T$:

$\mathsf T := \closedint {-1} 1 \times \set 0 \cup \set 0 \times \closedint 0 1$


Then $\mathsf L$ and $\mathsf T$ are not homeomorphic.


Proof

Aiming for a contradiction, suppose $f: \mathsf T \to \mathsf L$ is a homeomorphism.

Let $g$ be the restriction of $f$ to $\mathsf T \setminus \set \bszero$, where $\bszero := \tuple {0, 0}$ denotes the origin of $\R^2$.

Then from Restriction of Homeomorphism is Homeomorphism, $g$ is also a homeomorphism.

But $\bszero$ is the junction point of $\mathsf T$, which means that $\mathsf T \setminus \set \bszero$ consists of $3$ disjoint half-open intervals.

However, no matter where $\map f x$ is located in $\mathsf L$, the set $\mathsf L \setminus \set {\map f x}$ consists either of $1$ or $2$ half-open intervals.


This needs considerable tedious hard slog to complete it.
In particular: It remains to be shown that spaces consisting of different numbers of half-open intervals are not homeomorphic.
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This contradicts the supposition that such a homeomorphism $f$ exists.

Hence, by Proof by Contradiction, $\mathsf L$ and $\mathsf T$ are not homeomorphic.

$\blacksquare$


Sources

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