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?
Best Answer
$$\overline{X\setminus A}=\bigcap_{X\setminus A\subset C\\ C: closed}C$$
By negation and letting $U=X\setminus C$,
$$X\setminus \overline{X\setminus A}=\bigcup_{U\subset A\\ U: open} U=Int(A).$$