Points outside a closed set are separated from that closed set Let $(X,d)$ be a metric space, $G \subseteq X$ be a closed set, and $x \in X \setminus G$ be a point outside $G$. Then there exists an $\varepsilon \in (0,\infty)$ such that $B_{\varepsilon}(x) \cap G =\emptyset$.
Hint: To prove this aim for a contradiction,

*

*Suppose this wasn't true, and understand what that means.


*Argue that for every $n \in \mathbb{N}$, under this assumption there must be at least one point in $B_{\frac{1}{n+1}}(x) \cap G$.


*By appealing to the Axiom of Choice, define a sequence $a_n$ by requiring that each $a_n \in B_{\frac{1}{n+1}}(x) \cap G$.


*Prove that this sequence converges.
Assume for contradiction that $\forall \varepsilon>0$ we have $B_{\varepsilon}(x) \cap G \ne\emptyset$ whenever $(X,d)$ is a metric space with closed set $G \subseteq X$ and $x \in X \setminus G$.
Suppose $(a_n) \in G \ni (a_n)$ convergent and let $(a_n) \to l$, where $l \in G$. Because $G \subseteq X$, then $(a_n) \in X$ and $l \in X$. Choose $N \in \mathbb{N}$ so that $n>N$ with $n \in \mathbb{N}$, then $d(a_n,l)<\varepsilon \in (0,\infty)$. So for $\varepsilon=\frac{1}{n+1}$ with $n \in \mathbb{N}$, we have $d(a_n,l)<\frac{1}{n+1}$ for every $n \in \mathbb{N}$.
Initially I was thinking about defining a sequence $a: \mathbb{N} \to X$ by $a(n)=x$. But I don't think that'll work since that would mean $x=l$, but $x \in X \setminus G$.
Please let me know what you think.

This follows trivially from the definition of an open set in metric space since the closed sets are defined to be complements of open sets and open sets are defined to be possibly uncountable unions of open balls, each point not in a closed set must be contained in an open ball, then using triangle inequality you can show that there exists a ball centred in x that satisfies the wanted condition exists.
