Why isn't $\{\frac{1}{n}\}_{n \in \mathbb{N}} \cup \{0\}$ a neighborhood retract of any of its neighbourhoods?

Question

I'm trying to understand why there doesn't exist any neighbourhood $U \supset \{\frac{1}{n}\}_{n \in \mathbb{N}} \cup \{0\}$ such that $\{\frac{1}{n}\}_{n \in \mathbb{N}} \cup \{0\}$ is a retract (not necessarily a $\textit{deformation}$ retract, as posted here - although I don't understand the argument given there either: why is "But $U_0$ contains infinitely many points of $X$ onto which it must deformation retract" a contradiction) of $U$. Now, if there were such a retract, then by definition there would exist a neighbourhood $U \supset \{\frac{1}{n}\}_{n \in \mathbb{N}} \cup \{0\}$ and a continuous function $g: U \to \{\frac{1}{n}\}_{n \in \mathbb{N}} \cup \{0\} \doteq V$ such that $g|_{V} = \operatorname{Id}_V$. Intuitively I know such a function cannot be continuous at $0$ and so there is no such neighbourhood, but I haven't been able to prove this. Can anyone help me? I'd appreciate it!

Answer 1

$U$ is open and contains $0$, hence contains some open interval $(-a,a)$ with $a>0$. Pick $n>\frac1a$. Then $I:=[\frac1{n+1},\frac1n]\subset (-a,a)\subseteq U$. From $g(\frac1{n+1})=\frac1{n+1}$ and $g(\frac1n)=\frac1n$, it follows by the Intermediate Value Theorem (viewing $g$ as a function $I\to\Bbb R$) that $g(\xi)=\frac{\frac1{n+1}+\frac1n}{2}$ for some $\xi\in I$, but $\frac{\frac1{n+1}+\frac1n}{2}\notin V$.