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?

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

2 Answers 2


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

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


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