# Give a direct simple proof that comb space cannot be strong deformation retracted to $(0,1)$?

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.

• Hint: Consider a small open neighborhood $U$ of $(0,1)$ and note that the compact set $\{(0,1)\}\times I$ is a subset of the open set $H^{-1}(U)$, where $H:X\times I\to X$ is the retracting homotopy. Commented Feb 11, 2015 at 16:26
• How to became a contradiction? Commented Feb 12, 2015 at 12:09
• More hints: By compactness of $\{(0,1)\}\times I$, there is a neighborhood $V$ of $(0,1)$ such that $V\times I\subset H^{-1}(U)$. Choose $U$ small enough and see if such a neighborhood $V$ can exist. Commented Feb 12, 2015 at 12:46
• If $U$ small enough then $V$ is smaller than $U$ that it is not contradiction! Furthermore,what the reason for existence $V$ ? Commented Feb 12, 2015 at 20:19
• The reason why this $V$ exists is the tube lemma. See what happens when $U$ is disjoint from $[0,1]\times\{0\}$. Commented Feb 12, 2015 at 20:22

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 $U$.