# Interior of an non-open set

Hi I was working on an exercise from topology without tears from the section neighborhoods in chapter 3. I was trying to prove the following statement ($$\overline{S}$$ meaning the closure of S (a closure meaning the union of a set and all of its limit points) ) :

Let (X,T) be any topological space and A any subset of X. The largest open set contained in A is called the interior of A and is denoted by Int(A). [It is the union of all open sets in X which lie wholly in A.]

Show that if A is any subset of a topological space (X,T), then $$Int(A) = X\setminus \overline{(X \setminus A)}$$.

I wanted to prove this by making a breakdown by distinguishing between open sets and non-open sets. My "proof" for open sets (please correct me) essentially states that if A is an open set $$Int(A) = A$$. Thus : $$A = X\setminus (\overline{X\setminus A})$$ this makes sense as $$\overline{X\setminus A} = X\setminus A$$ because all points A occur in A without any element of $$X\setminus A$$.

Does anyone know how to prove this for non-open sets? Is my proof correct so far?

• Note. Use $X \setminus A$ to get $X \setminus A$. Jul 29 at 19:18
• thx I tought it looked off Jul 29 at 19:20
• What's your definition for $\overline S$? Jul 29 at 19:44
• it is the closure Jul 29 at 19:47
• Please don't include images of math on this site—it messes up search engines and makes questions harder to edit and streamline. Jul 29 at 19:48

$$\overline{X\setminus A}=\bigcap_{X\setminus A\subset C\\ C: closed}C$$

By negation and letting $$U=X\setminus C$$,

$$X\setminus \overline{X\setminus A}=\bigcup_{U\subset A\\ U: open} U=Int(A).$$

By your given definition of the interior of $$A$$, we have to prove three things. We have to prove that $$X\setminus\overline{X\setminus A}$$ is open, contained in $$A$$, and every open set contained in $$A$$ is also contained in $$X\setminus\overline{X\setminus A}$$.

Let $$A$$ be a subset of $$X$$. By the definition of closure, $$\overline{X\setminus A}$$ is a closed set in $$X$$. Then, $$X\setminus \overline{X\setminus A}$$ is open in $$X$$.

Now, if $$x \in X\setminus\overline{X\setminus A}$$, then $$x \not\in \overline{X\setminus A}$$. Since $$X\setminus A \subseteq \overline{X\setminus A}$$, we have that $$x\not\in X\setminus A$$. Hence, $$x \in A$$. It follows that $$X\setminus \overline{X\setminus A}$$ is an open set contained in $$A$$.

Finally, suppose that $$U$$ is an open set in $$X$$ such that $$U \subseteq A$$. We have that $$U \subseteq A \implies X\setminus U \supseteq X\setminus A \implies \overline{X\setminus U} \supseteq \overline{X\setminus A} \implies X\setminus\overline{X\setminus U} \subseteq X\setminus\overline{X\setminus A}$$ We can conclude that $$U \subseteq X\setminus\overline{X\setminus U} \subseteq X\setminus\overline{X\setminus A}$$.

It follows that $$\text{Int}(A) = X\setminus\overline{X\setminus A}$$