Show that if \(A\) is closed in \(Y\) and \(Y\) is closed in \(X\), then \(A\) is closed in \(X\).
Solution
Since \(A\) is closed in the subspace \(Y\), there is a closed set \(C\) of \(X\) with \[ A=Y\cap C. \] Because \(Y\) is closed in \(X\), both \(Y\) and \(C\) are closed in \(X\), hence so is their intersection \(A\).