$A$ 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

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


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