Question

Prove that the boundary of a subset A of a metric space X is always a closed set

Answer 1

Let's fix some notation first.

$\partial A$ is the set denotes the boundary of $A$ .

$M:=X\backslash \partial A= \partial A^{c}$ is the complement of the boundary of $A$ .

We have given that $A\subseteq X$ ,where $X$ is a metric space.

Claim1: No points in $M$ can be the limit points of $\partial A$

Proof: Suppose on the contrary there exist at least one point $q \in M$ is limit point of $\partial A$ .

Thus, for every $\epsilon >0$ there exist an open ball

$$B(q,\epsilon):=\{y \in X |\: d(q,y)<\epsilon\}$$

such that

$$\partial A\cap B(q,\epsilon)\backslash \{q\}\neq \varphi$$

but if we choose $\epsilon$ small enough ,such $\epsilon$ always exist ,we get

$$\partial A\cap B(q,\epsilon)\backslash \{q\}= \varphi$$

Hence contradicting the hypothesis that $q$ is limit point of $A$ . This can be easily seen in the below rough figure

Clearly, the ball $K$ inside $A$ and $F$ outside $A$ does not intersect with boundary $\partial A$ . Hence, proved.

Thus claim1 guaranteed that if $\partial A$ has limit points then it must be on $\partial A$.

Claim2: Every point on the boundary $\partial A$ is limit point of $\partial A$

Proof:let for any arbitrary point $p\in \partial A$ ,thus for every $\epsilon>0$ there exist an open ball $B(p,\epsilon)$ such that

$$\partial A \cap B(p,\epsilon)\backslash \{p\}\neq \varphi$$

which is very clear from the above rough sketch, hence $p$ is the limit point of $\partial A$.

As,$p$ is arbitrary,thus every point of $\partial A$ is limit point of $\partial A \implies$ every limit points of $\partial A$ are contained in $\partial A$ . Hence proved.

Therefore immediately from Claim 2, $\partial A$ is close set.

Hence, we are done.