$\text{int}(\text{cl}\,A) \subset A$

Question

Let $(X,\mathcal{T})$ be a topological space, and $A\in\mathcal{T}$.

I'm trying to prove that the interior of the boundary of an open set $A$ is empty.

Note: We define the boundary of $A$ as $\partial A = (\text{cl}\,A) \setminus \text{int}A$

What I've tried is:

$$\text{int}(\partial A) = \text{int}(\text{cl}\,A \setminus\text{int}\,A) = \text{int}(\text{cl}\, A) \setminus A \quad(A \text{ is an open set} \Rightarrow \text{int}\,A = A)$$

Now my question is: $\text{int}(\text{cl}\,A) \subset A$ ? If that's true then $\text{int}(\partial A) = \text{int}(\text{cl}\, A) \setminus A \subset A\setminus A = \varnothing$ and that's it.

If not, how can I approach this problem?

Answer 1

It is not true that $int(cl(A))\subset A$ (even if $A$ is open) as the example $A=\mathbb R-\{0\}\subset \mathbb R$ shows. The interior of the closure is everything.

As for the overarching question: Let $B:=int(\partial A)$. Put $C:=B\cap A$. Note that $C$ is open (as intersection of two opens) and $C\subset A$ (by construction), hence $C\subset int(A)$. This means $C= C\cap int(A) = B\cap A\cap int(A)=B\cap int(A)=\emptyset$ (where the last equality holds because $B\subset \partial A$). Hence $B\cap A=\emptyset$. But this means that $cl(A)-B$ is a closed set (since $B$ is open) which contains $A$; this can only happen if $B$ is empty.

Answer 2

As mentioned, your approach does not work. Try arguing by contradiction. Assume $x\in int(\partial A)$. Then $x$ has some neighborhood $U$ contained in $\partial A$. Since $x$ is a boundary point of $A$, there is some other point $y\in U$ such that $y\in A$. Now derive a contradiction using $y$.