TITLE: A is open iff A is the set of interior points
QUESTION [1 upvotes]: Let $X$ be an nonempty set. Consider a topological space $(X,T)$, let $A$ be a set. Prove that $A \in T $ iff $A = Int(A)$ where $Int (A)$ means the set containing all interior points.
My attempt: $(\Leftarrow)$ $\forall x \in A, \exists V_x \in T$ containing $x$ s.t. $V_x \subset A$. Then $A=\bigcup_{x\in A}V_x$ and by definition $A \in T$
But how can I prove the reverse direction?
Thank you!

REPLY [0 votes]: You are done $\Leftarrow:$ 
For $\Rightarrow:$ Assume that $A\in T$. By definition, it is clear $int(A)\subset A$. It remains to show that $A\subset int(A)$. Let $x\in A$. Because $A$ is open and $x\in A\subset A$, it follows that $x\in int(A)$. This proves that $A\subset int(A)$ and equality follows.
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