Closed Set in a Metric Space is G-Delta

Theorem

Let $\left({X, d}\right)$ be a metric space, and let $F \subset X$ be a closed set.

Then $F$ is a $G_\delta$ set of X.

Proof

Let $n \in \N$.

Let $\displaystyle F_{\frac 1 n} = \bigcup \limits _{x \in F} B \left({x, \dfrac 1 n}\right)$, where $B \left({x, \dfrac 1 n}\right)$ is the open ball around $x$ with radius $\dfrac 1 n$.

$F_{\frac 1 n}$ is an open set by definition of open ball.

Also, $F_{\frac 1 n}$ contains $F$ as it is the union of open balls around every element of $F$.

$\displaystyle F \subseteq \bigcap \limits _{n=1}^{\infty} F_{\frac 1 n}$ because $F \subseteq F_{\frac 1 n}$ for each $n \in \N$.

Note that:

$\displaystyle \lim_{n \to \infty} B \left({x, \dfrac 1 n}\right) = \left\{{x}\right\}$

so the above statement holds even in the limit.

Let $\displaystyle y \in \bigcap \limits _{n=1}^{\infty} F_{\frac 1 n}$.

Then $y$ is a Limit Point of $F$.

Since $F$ is closed, then $y \in F$.

Therefore:

$\displaystyle F \supseteq \bigcap \limits_{n=1}^{\infty} F_{\frac 1 n}$

We conclude that $\displaystyle F = \bigcap \limits _{n=1}^{\infty} F_{\frac 1 n}$, a countable intersection of open sets.

Therefore $F$ is a $G_\delta$ set by definition.

$\blacksquare$