# metric spaces proving the boundary of A is closed

here is my question:

Let $A\subset X$, with $(X,d)$ being a metric space.

Prove $\partial A$ is closed

and prove that A is open iff $A\cap \partial A = \emptyset$

I've found lots about showing A is closed but not much on the boundary of A or A being open. Appreciate any help.

• What is your definition of a closed set? There are plenty of equivalent conditions and different books use different definitions. – Sahiba Arora Jun 29 '17 at 23:07
• $\partial A=\overline{A}\cap\overline{A^{c}},$ so it is an intersection of two closed sets. If $A\cap\partial A=\varnothing,$ then $A=\overline{A^{c}}^{c},$ which is the complement of a closed set. If $A$ is open, then $A^{c}$ is closed, so $\varnothing=A\cap\overline{A^{c}}=A\cap\partial A.$ You'll need to work out the details, but this gives you an outline. – RideTheWavelet Jun 29 '17 at 23:13
• "A subset of a metric space is called closed if it contains all of its cluster points" – grapevine Jun 29 '17 at 23:16

(I). Let $(x_n)_n$ be a sequence, of members of $\partial A,$ converging to $x.$ For each $n$ there is (i) a sequence of members of $A$ that converges to $x_n$ and (ii) a sequence of members of $A^c$ that converges to $x_n.$

So for each $m\in \mathbb N$ take some (any) $n_m$ such that $d(x,x_{n_m})<1/2m.$ By (i) we may take some $y_m\in A$ with $d(x_{n_m},y_m)<1/2m$ and by (ii) we may take some $z_m\in A^c$ with $d(x_{n_m},z_m)<1/2m.$

Now for each $m$ we have $d(x,y_m)\leq d(x,x_{n_m})+d(x_{n_m},y_m)<1/m.$ And similarly $d(x,z_m)<1/m.$ So $x$ is the limit of a sequence $(y_m)_m$ of members of $A$ and $x$ is the limit of a sequence $(z_m)_m$ of members of $A^c.$

That is, $x\in \partial A.$

(II). Note that we can replace $A$ and $A^c$ with any two closed sets in the above argument, and deduce that the intersection of any two closed sets is closed. And by induction that the intersection of finitely many closed sets is closed: For $n\geq 2$ and closed sets $A_1,...,A_{n+1},$ if $B=\cap_{i=1}^nA_i$ is closed, then $B\cap A_{n+1}$ is closed; that is, $\cap_{i=1}^{n+1}A_i$ is closed.

There are many ways; staying close to your definition in the comments:

Suppose that $x$ is a (cluster/limit) point of $\partial A$. Suppose $x \notin \partial A$. This means that for some $r>0$, $B(x,r) \cap A = \emptyset$ or $(B(x,r) \cap (X \setminus A) = \emptyset$. But by the definition of cluster point, there is some $y \in B(x,r) \cap \partial A$, with $y \neq x$. As open balls are open, there is $s>0$ such that $B(y,s) \subseteq B(x,r)$. But then $B(y,s)$ also shows that $y \notin \partial A$ (as it misses $A$ or its complement). Contradiction. So $X \in \partial A$ and $\partial A$ contains all of its cluster points, hence is closed.

Suppose $\partial A \cap A = \emptyset$. Let $x \in A$. Then $x \notin \partial A$, but every balls $B(x,r)$ intersects $A$ (in $x$) so the only way it can fail to be in $\partial A$ if there is some $r>0$ with $B(x.r) \cap (X\setminus A) = \emptyset$ which is equivalent to $B(x,r) \subseteq A$. So this witnessing ball for $x\notin \partial A$ also shows $x \in \operatorname{int}(A)$. So $A$ is open, as all its points are interior points.

The reverse implication is the same idea: an interior point of $A$ can never be a boundary point..

Use the theorem or definition that bd A = cl A - int A.
Thus if A is open, A cap cl A - int A = A - A.
Conversly A - int A = A cap cl A - int A is empty.
Whence A subset int A subset A showing A = int A is open.
(bd, cl, int for boundary, closure, interior, resp.)
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