TITLE: Is every compact subset of $\mathbb{R}^n$ a deformation retract of some open neighborhood?
QUESTION [5 upvotes]: Suppose $A \subset X=\mathbb{R}^n $ is compact. Is it necessary that $ \exists$ an open set $U \supset A$ such that $A$ is a deformation retract of $U$? If yes, is there a concrete construction of the retraction homotopy? I am unable to come up with a proof or a counter example. The statement above holds in all examples that I can think of.

REPLY [6 votes]: No. Consider the set $X = \{0\}\cup\{ \frac{1}{n} \mid n \in \mathbb{N}\} \subset \mathbb{R}$. As it is closed and bounded, $X$ is compact. If $U$ is an open set containing $X$, then $U_0$, the connected component of $U$ containing $0$, also contains $\frac{1}{n}$ for all $n \geq N$ for some $N \in \mathbb{N}$. But the connected open sets in $\mathbb{R}$ are open intervals which are homotopy equivalent to a point, but $U_0$ contains infinitely many points of $X$ onto which it must deformation retract. This is a contradiction.