Give a direct simple proof that comb space cannot be strong deformation retracted to $(0,1)$ where comb space is $\bigl(\bigcup_{n\in \mathbb{N}} \{\frac{1}{n}\}\times[0,1]\bigr)\cup \bigl([0,1]\times\{0\}\bigr)\cup\bigl(\{0\}\times [0,1]\bigr)$?
Thanks.
Let $x_0=(0,1)$. Consider a neighborhood $U$ of $x_0$ which is disjoint from $I\times\{0\}$. Suppose $H:X\times I\to X$ is the homotopy starting with the identity on $X$ and ending with the constant map $X\to\{x_0\}$ such that $H(x_0,t)=x_0$ for all times $t$. That means $\{x_0\}\times I$ has a neighborhood $H^{-1}(U)$. By the tube lemma, there is an open set $V$ such that $\{x_0\}\times I\subset V\times I\subset H^{-1}(U)$. That means every point in $V$ stays in $U$ during the entire deformation. However a point $y=(a,1)$ must traverse a path to $x_0$ and no such path exists within $U$.