Set in a Metrizable Space is Closed

Question

Let $X$ be some metrizable space, $U$ some open set in $X$, and $n \in \Bbb{N}$. Define $S_n(U) := \{x \in X \mid B(x,\frac{1}{n}) \subseteq U \}$. I am trying to show that this set is closed. Here is one of many abysmal attempts:

Suppose that $z \in \overline{S_n(U)}$. I'm not sure what to choose $\epsilon > 0$ to be, but it seems that the only thing that would work is to take $\epsilon < \frac{1}{n}$. Then there exists an $x \in B(z,\frac{1}{n}) \cap S_n(U)$, which means that $d(x,z) < \epsilon < \frac{1}{n}$ and $B(x,\frac{1}{n}) \subseteq U$. If we try to show that $B(z,\frac{1}{n})$, this won't work: suppose $y \in B(z,\frac{1}{n})$; then $d(x,y) \le d(x,z) + d(z,y) < \epsilon + \frac{1}{n}$, where $\epsilon + \frac{1}{n} > \frac{1}{n}$ for every $\epsilon > 0$, which indicates that it isn't necessarily the case that $B(z,\frac{1}{n}) \subseteq B(x,\frac{1}{n})$, which was my only hope of showing $B(z,\frac{1}{n}) \subseteq U$.

As I said, this was a pretty abysmal attempt. I tried out several other ideas (e.g., proof by contradiction, trying to write $S_n(U)$ as the preimage of a closed set under a continuous map, and others), but all of them were unsuccessful. I could use a hint.

Answer 1

Hint: Suppose that $U$ is not empty, and not equal to $X$. Let $V=X-U$ the complement of $U$ in $X$. Put $f(x)=d(x,V)={\rm inf}\{d(x,v), v\in V\}$. Then your set is $f^{-1}([\frac{1}{n},+\infty[)$.

Answer 2

Have you tried this?

$$S_n(U) = \left\{x \in X \mid d(x,U^c) \ge \frac{1}{n}\right\} = d(\cdot,U^c)^{-1}\left(\left[\frac{1}{n},\infty\right)\right)$$

Answer 3

Let $x$ be a limit point of $S_n(U)$ and let $(x_k)_k$ be a sequence with $x_k\in S_n(U)$ for every $k$ and $\lim_{k\to\infty}x_k=x$.

Assume that $x\notin S_n(U)$.

Then some $y$ exists with $y\notin U$ and $d(x,y)<\frac1n$.

Then, since $d(-,y):X\to\mathbb R$ is a continuous map, for $k$ large enough we also get $d(x_k,y)<\frac1n$ contradicting that $x_k\in S_n(U)$.

So the assumption $x\notin S_n(U)$ must be wrong, and we conclude that $x\in S_n(U)$.

Proved is now that $S_n(U)$ contains its limit points hence is closed.

Answer 4

Prove that the following set is closed

I’ve answered your question with a different proof in the above thread.