If $(X,E)$ is a CW complex, then a subset $A$ of $X$ is closed iff $A \cap X ' $ is closed in $X'$ for every finite CW subcomplex $X'$ in $X$.

Question

Prove: If $(X,E)$ is a CW complex, then a subset $A$ of $X$ is closed iff $A \cap X ' $ is closed in $X'$ for every finite CW subcomplex $X'$ in $X$.

The $\Rightarrow$ proof in Rotman's Algebraic Topology says: if $A$ is closed in $X$, then $A \cap X'$ is certainly closed in $X ' $.

Whys is this true? I don't see how this implies that $A \cap X' $ is closed in $X'$.

Answer 1

If $A \subset X$ is closed, $X \setminus A$ is open in $X$. So by the definition of the subspace topology on $X'$, $(X\setminus A) \cap X'$ is open in $X'$. But elementary set theory tells us that:

$$X' \cap A = X' \setminus ((X\setminus A) \cap X')$$

and so $X' \cap A$ is closed in $X'$ as the relative complement of a relatively open set.

This is often one of the first lemma's to show after introducing the subspace topology.

Another way: suppose $A$ is closed in $X$ and let $x \in X'$ be a point of the closure of $X' \cap A$ (in the subspace topology).
Let $O$ be any open set in $X$ that contains $x$. Then $O \cap X'$ is open in $X'$ and still contains $x$, so this open set must intersect $X' \cap A$, so in particular, $O \cap A \neq \emptyset$. As $O$ was arbitrary, $x \in \overline{A} = A$, so that $\overline{A \cap X'}^{(X')} = A \cap X'$.