Why does $A_j \subseteq cl ( \bigcup_{m=1}^n A_m )$ imply $cl(A_j) \subseteq cl ( \bigcup_{m=1}^n A_m )$?

Question

Why does $A_j \subseteq cl \bigg( \cup_{m=1}^n A_m \bigg)$ imply $cl(A_j) \subseteq cl \bigg( \cup_{m=1}^n A_m \bigg)$?

This is intuitive, but I was thinking of whether one can be sure that there are no counterexamples.

Answer 1

It's clear that $\operatorname{cl} (\bigcup_{n=1}^m A_m)$ is a closed set that contains $A_j$. So the closure of $A_j$, which is the smallest closed set containing $A_j$ is a subset of that set. Hence the second inclusion.