Let \(A\subset X\) with \(A\) closed and \(X\) compact, and show that \(A\) is compact.
Solution
Let \(A\) be a compact subset of \(X\) a Hausdorff space, we will show \(X\setminus A\) is open
Fix \(x\in X\setminus A\).
Since \(X\) is Hausdorff, for every \(y\in A\) we can find \(U_y \ni y\) and \(V_x(y) \ni x\) depending on \(y\) such that \(U_x(y) \intersect U_y = \emptyset\).
Then \(\theset{U_y \suchthat y\in A} \covers A\), and by compactness of \(A\) there is a finite subcover corresponding to a finite collection \(\theset{y_1, \cdots, y_n}\).
Set \(U = \union U_{y_i}\) and \(V = \intersect V_x(y_i)\);
- We have \(A\subset U\): each \(y \in A\) lies in some \(U_{y_i}\) (the \(U_{y_i}\) cover \(A\)), so \(y \in \bigcup_i U_{y_i} = U\).
- We have \(x\in V\): \(x \in V_x(y_i)\) for every \(i\) by construction, so \(x \in \bigcap_i V_x(y_i) = V\).
- We have \(U\intersect V = \emptyset\): if \(z \in U \cap V\), then \(z \in U_{y_i}\) for some \(i\) and \(z \in V_x(y_i)\) for that same \(i\), contradicting \(U_{y_i} \cap V_x(y_i) = \emptyset\).
Done: for every \(x\in X\setminus A\), we have found an open set \(V\ni x\) such that \(V\intersect A = \emptyset\), so \(x\) is an interior point and a set is open iff every point is an interior point.
