In a metric space $(X,d)$, if $G \subseteq X$ is closed, then its complement $G^c$ is open.
Lemma: 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 $B_{\varepsilon}(x) \cap G=\emptyset$.
Proof. As $G$ is a closed set, then $X \setminus G$ is open. This mean that $\forall x \in X \setminus G, \exists \varepsilon>0$ such that $B_{\varepsilon}(x) \subseteq X \setminus G$. Because $B_{\varepsilon}(x) \subseteq X \setminus G$, we have $\forall x (x\in B_{\varepsilon}(x) \Rightarrow x \in X \setminus G)$. Thus $B_{\varepsilon}(x) \cap G=\emptyset$.
Proof. Suppose $G$ is closed, and let $x \in G^c$. To show that $G^c$ is open, we need to find an $\varepsilon \in (0,\infty)$ for which $B_{\varepsilon}(x) \subseteq G^c$.
Because $x \in G^c, x \in X \setminus G$ is a point outside $G$. By the lemma, this means that $\exists \varepsilon \in (0,\infty)$ such that $B_{\varepsilon}(x) \cap G=\emptyset$. Now, $B_{\varepsilon}(x) \cap G =\emptyset \iff B_{\varepsilon}(x) \subseteq G^c$. Thus $\exists \varepsilon >0$ such that $B_{\varepsilon}(x) \subseteq G^c$. As $x$ arbitrarily chosen, $G^c$ is an open set.
How would you show that $B_{\varepsilon}(x) \cap G =\emptyset \iff B_{\varepsilon}(x) \subseteq G^c$?
I'm not sure this would work here since I tried to prove the lemma using the proposition and the proposition using the lemma.
Also, please note that closed sets are defined based on the limits of convergent sequences, and not in terms of limit points.
Any insight would be appreciated!
Edit
Suppose $G$ is closed, so let $(a_n) \in G$ such that $(a_n) \to x$, where $x \in G$. To prove that $G^c$ is open, it suffices to show that for every $\tilde x \in G^c, \exists \tilde \varepsilon >0$ such that $B_{\tilde \varepsilon}(\tilde x) \subseteq G^c$. Because $(a_n) \to x$, then by definition of every open neighborhood $B_{\varepsilon}(x)$, $\exists$ at least one element $a_n \in B_{\varepsilon}(x)$.
I'm stuck here. Right now my approach is to negate the conclusion and aim for a contradiction.
Best Answer
Your lemma already assumes the theorem on line 1 is true: but you say that closedness is defined by convergent sequences (probably like this: $A$ is closed iff whenever we have a sequence $a_n \to a$ in $(X,d)$ and all $a_n \in A$ then $a \in A$ as well) but I see nothing of that in the proof.
If $G$ is open then $G':=X\setminus G$ is closed: let $x_n \to x$ and all $x_n \in G'$. Then suppose (for a contradiction) that $x \in G$. As $G$ is open, there is some $\varepsilon>0$ so that $B_\varepsilon(x) \subseteq G$, and as $x_n \to x$ there is some $N$ so that for all $n \ge N$ we have $x_n \in B_\varepsilon(x)$. But note that $x_N \in G'$ so $x_n \notin G$ but by the convergence $x_n \in B_\varepsilon(x)$ so $x_n \in G$. This contradiction proves that $x \notin G$ so $x \in G'$, as required. By the sequence definition of closedness, $X\setminus G$ is indeed closed. QED.
Now for the complementary fact: if $G$ is closed, $X\setminus G$ is open.
So let $G$ be closed (in the sequence sense), and let $x \in X\setminus G$. Suppose that there were no $\varepsilon >0$ so that $B_\varepsilon(x) \subseteq X\setminus G$. Then for each $n \in \Bbb N$ we have that in particular $B_{\frac1n}(x) \nsubseteq X\setminus G$ so we can pick $x_n \in B_{\frac1n}(x)$ such that $x_n \notin X\setminus G$; so we know two things about $x_n$:
$$x_n \in G \text{ and } d(x_n, x) < \frac1n$$
Now it's easy to see by the definition of convergence that $x_n \to x$ in $(X,d)$: having $\varepsilon >0$ we find $N \in \Bbb N$ so that $\frac{1}{N} < \varepsilon$ and then for all $n \ge N$ we have $d(x_n, x) < \frac1n \le \frac1N < \varepsilon$ and we're done. So $x_n \to x$ and all $x_n \in G$ so as $G$ is closed we conclude $x \in G$ and we have a contradiction as $x \in X\setminus G$ from the start.. So the assumption we made was wrong and we do have $\varepsilon>0$ so that $B_\varepsilon(x) \subseteq X\setminus G$. QED.