Open sets in metric spaces as special unions.

Question

Let $(X,d)$ be a metric space and $O\subseteq X$ be open in the induced topology. Is there always a sequence $(V_n)_{n\in\mathbb{N}}$ of open sets such that $O=\mathop{\bigcup}_{n\in\mathbb{N}} V_n$ and $V_n\subseteq \overline{V_n}\subseteq V_{n+1}$, for all $n\in\mathbb{N}$? If so, please give an argument as to why. If not, please provide a counterexample. Thanks.

Answer 1

Yes, there is. We can suppose $O \neq X$ (or the statement is trivial, take $V_n =X$ for all $n$) and in that case $f(x) = d(x,X\setminus O)$ is well-defined and continuous. This is a continuous function from $X$ to $\mathbb{R}$ and note that $x \in O$ iff $f(x) > 0$, by openness of $O$.

Then define $V_n = \{x: f(x) > \frac{1}{n}\}$ is open by continuity of $f$ (it equals the inverse image under $f$ of the open set $(\frac{1}{n}, +\infty)$ of $\mathbb{R}$) and also $$V_n \subseteq \overline{V_n} \subseteq \{x: f(x) \ge \frac{1}{n} \} \subseteq \{x: f(x) > \frac{1}{n+1} \} = V_{n+1}$$ and $$O = \{x: f(x) > 0 \} = \bigcup_n \{x: f(x) > \frac{1}{n}\} = \bigcup V_n$$