# Prove that $int(A)=A\setminus bd(A)$

$A$ is a subset of a metric space $M$.

I know I will need to prove $A$ is a subset of $M$. As well as $M$ is a subset of $A$. So for, $A$ is a subset of $A$: $int(A)$ implies that it is a subset of $A$ itself. Thus if $x$ is in $int(A)$ it must also be in $A$. Not sure if I'm off to right track any help is great

Suppose $x\in int(A)$. Then there exists an open neighbourhood $U_{x}$ of $x$ such that $x\in U_{x}\subseteq A$.

So we have $U_{x}\cap A^{c}=\varnothing$.

Now assume that $x\in bd(A)$.

Then for all neighbourhood $U$ of $x$, $U\cap A\neq \varnothing$ and $U\cap A^{c}\neq \varnothing$.

But this is a contradiction, since $U_{x}\cap A^{c}=\varnothing$.

So $x\notin bd(A)$ and hence $x\in A\setminus bd(A)$.

That is $int(A)\subseteq A\setminus bd(A)$.

Now conversely suppose $y\in A\setminus bd(A)$.

Then $y\in A$ and $y\notin bd(A)$.

Then there exist a neighbourhood $U_{y}$ of $y$ such that $U_{y}\cap A^{c}= \varnothing$.

So $y\in U_{y}\subseteq A$ and hence $y\in int(A)$.

Therefore $A\setminus bd(A)\subseteq int(A)$.

Thus $int(A)= A\setminus bd(A)$. $\square$

Hint: $x\in bd (A)$ if for every open neighborhood $V$ of $x$, we have that

$$A\cap V\neq \emptyset \text{ and } (M-A)\cap V\neq\emptyset.$$