This is Theorem 17.3 on page 95 in Chapter 16 of James R. Munkers’ Topology 2nd Edition textbook.

Theorem

Let \(Y\) be a subspace of \(X\). If \(A\) is closed in \(Y\) and \(Y\) is closed in \(X\), then \(A\) is closed in \(X\).

Proof

Let \(A\) be closed in \(Y\). Let \(B\) be closed in \(X\). Because \(Y\) is closed in \(X\), \(X - Y\) is open in \(X\). Furthermore, from Theorem 17.2, \(A = B \cap Y\).

Taking the compliment of both sides, the equation \(X - A = X - (B \cap Y)\) is formed, simplified into \(X - A = (X - B) \cup (X - Y)\). Since \((X - B)\) and \((X - Y)\) are both open in \(X\), by definition of a topology, \(X - A\) must be an open set in \(X\). Thus \(A\) is a closed set in \(X\).

\[\tag*{$\blacksquare$}\]

Extras

link to the mathjax LaTeX specification: https://treeofmath.github.io/tex-commands-in-mathjax/TeXSyntax.htm