In Munkres's Topology (section 17 Closed Sets and Limit Points, 2nd edition), Theorem 17.4 shows that (below, $\bar{A}$ is the closure of $A$):
Theorem 17.4: Let $Y$ be a subspace of $X$; let $A$ be a subset of $Y$; let $\bar{A}$ denote the closure of $A$ in $X$. Then the closure of $A$ in $Y$ equals $\bar{A} \cap Y$.
Let $B$ denote the closure of $A$ in $Y$. The proof is to show that both $B \subset (\bar{A} \cap Y)$ and $(\bar{A} \cap Y) \subset B$ are satisfied. I can follow the first part; however, I am confused about the proof of $(\bar{A} \cap Y) \subset B$:
$B$ is the closure of $A$ in $Y$, so $B$ is closed in $Y$. By Theorem 17.2, $B = C \cap Y$ for some set $C$ closed in $X$. Then $C$ is a closed set of $X$ containing $A$; because $\bar{A}$ is the intersection of all such closed sets, we conclude that $\bar{A} \subset C$. Then $(\bar{A} \cap Y) \subset (C \cap Y) = B$.
My problem is that (as indicated in bold) why $C$ here is a closed set of $X$ containing $A$, especially why is $A \subset C$?
Best Answer
Since $B$ is the closure of $A$ (in $Y$), $A\subset B=C\cap Y\subset C$, so $A\subset C$.