Letter L and Letter T are not Homeomorphic
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. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
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
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(d)}$