Topology question interior of a set

Question

Let $(X, \tau)$ be a topological space and suppose $A\subset X$ Show that $A$ is open in $X\iff\operatorname{int}A=A$

The definition of open is that it is an element of the topology. I was able to prove that if $A$ is open in $X$, then $\operatorname{int}A=A$, that follows from the fact that $\operatorname{int}A \subset A $ and that setting $A$ to be the open set containing $X$, we have $A\subset \operatorname{int}A$ and so $A=\operatorname{int}A$ Now for the reverse direction, I understand that $\operatorname{int}A=A$ means that for every element in $A$ there exists an open set containing $x$ that is in $A$. But shouldn’t noting that $A \subset X$ suffices to conclude that $A$ is open?

Answer 1

It seems that your definition of the interior is the following:

$$\operatorname{int}(A)=\{x\in A\ \mid \text{ there is an open neighbourhood }U\text{ of }x\text{ such that }U\subseteq A\}$$

By the definition $\operatorname{int}(A)\subseteq A$ regardless of what $A$ is. Now the proper proof goes like this:

"$\Rightarrow$" as you've noted yourself, if $A$ is open then for any element $x\in A$ we can choose $A$ to be such neighbourhood and so $\operatorname{int}(A)=A$ immediately.

"$\Leftarrow$" if $\operatorname{int}(A)=A$ then it means that for any $x\in A$ there is an open neighbourhood $U_x$ of $x$ such that $U_x\subseteq A$. Therefore $\bigcup_{x\in A}U_x\subseteq A$. But the "$\supseteq$" inclusion also follows due to the fact that any $x\in A$ belongs to $U_x$. And so $\bigcup_{x\in A}U_x= A$ and since we have a union of open subsets on the left side then $A$ has to be open.

But shouldn’t noting that $A \subset X$ suffice to conclude that $A$ is open?

No, such implication is just wrong. Ultimately it is true but for other reasons.