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.
Best Answer
(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.