- 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 $\bar{A} \cap Y .$
Proof. Proof. Let $B$ denote the closure of $A$ in $Y$. The set $\bar{A}$ is closed in $X,$ so $\bar{A} \cap Y$ is
closed in $Y$ by Theorem 17.2 . since $\bar{A} \cap Y$ contains $A,$ and since by definition $B$ equals
the intersection of all closed subsets of $Y$ containing $A,$ we must have $B \subset(\bar{A} \cap Y) .$
On the other hand, we know that $B$ is closed in $Y$. Hence 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 question is:''…because $\bar{A}$ is the intersection of all such closed sets, we conclude that $\bar{A} \subset C$. Then
$(\bar{A} \cap Y) \subset(C \cap Y)=B$.'' I couldn't understand that how did the writer get $\bar{A} \subset C$? Can you explain? Thanks…
Best Answer
The closure of $A$ in $X$ is precisely the intersection of all closed subsets $S$ of $X$ such that $A \subseteq S$. Then since the $C$ in the proof above is both closed and contains $A$, $\overline{A}$, the closure of $A$, is the intersection of $C$ with all other closed sets containing $A$. In particular then, $\overline{A} \subseteq C$.