I am trying to prove that for any set A $\subset X$ where $(X,d)$ is a metric space, $\partial A$ is closed without using definitions of closure or interior.
I tried to show the equivalent problem of showing that X\ $\partial A$ is open.
$ x\in \partial A \Rightarrow x $ is an interior point of A or an exterior point of A
Assume for a contradiction that $\exists \epsilon > 0 $ s.t. $B_{\epsilon}(x) \cap \partial A \neq \phi$.
However I am unsure what to do from here to achieve the contradiction.
Best Answer
$x$ is in $\partial A$ iff
$$\forall \varepsilon>0: (B_\varepsilon(x) \cap A \neq \emptyset) \land (B_\varepsilon(x) \cap A^\complement \neq \emptyset)$$
so logically, $x \notin \partial A$ iff
$$\exists \varepsilon>0: (B_\varepsilon(x) \subseteq A^\complement) \lor (B_\varepsilon(x) \subseteq A)$$
In both cases, any point of $B_\varepsilon(x)$ will also not be in $\partial A$ (as balls are open), so in fact
$$\exists \varepsilon>0: (B_\varepsilon(x) \subseteq (\partial A)^\complement$$
showing that the complement of the boundary is open, so the boundary is closed.