I'm reading the Appendix of Hatcher's Algebraic Topology about CW complexes. He defines a subcomplex $A\subset X$ of a CW complex $X$ as a union of open cells such that the closure of each cell is contained in $A$. The subcomplex $A$ will itself be a CW complex. Then he says that the the subspace topology on $A$ inherited from $X$ is the same as the topology on $A$ as a CW complex. I understand why the two topologies agree for each finite dimensional case $A^n=A\cap X^n$ for each $n$. But why is the weak topology on the union $$\bigcup A^n=A\cap X$$ the same as the subspace topology?
I can see why open sets in the subspace topology would also be open in the weak topology. To do the other direction, let $U\subseteq \bigcup A^n$ be open. Then each $U\cap A^n\subseteq A^n$ is open. Since $A^n=A\cap X^n$ has the subspace topology inherited from $X^n$, there exist open sets $V^n\subseteq X^n$ such that $V^n\cap A=U\cap A^n$. We might like to construct an open set $V\subseteq X$ such that $V\cap A=U$, to show that $U$ is open in the subspace topology of $A\subseteq X.$ It is potentially relevant that $X^n\subseteq X^{n+1}$ is always a closed subspace. Where can I go from here?
Best Answer
Almost always when dealing with questions like these, it is easier to work with closed sets instead of open sets. If $C\subseteq A$ is closed in the weak topology on $A$, then $C\cap A^n$ is closed in $A^n$ for each $n$. Since $A^n$ is closed in $X^n$, this means $C\cap A^n=C\cap X^n$ is also closed in $X^n$. So, since $X$ has the weak topology with respect to its skeleta $X^n$, this means $C$ is closed in $X$, and thus also in the subspace topology on $A$.