# Closure subspace proof easy, but I am overlooking something

Theorem: 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: 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 $Y$. Since $\bar{A} \cap Y$ contains $A$, and since by definition $B$ equals the intersection of all closed sets in $Y$ containing $A$, we must have $B \subset \bar{A} \cap Y$. On the other hand, we know that $B$ is closed in $Y$. Then $B = C \cap Y$ for some closed set $C$ in $X$. Then $C$ is a closed set of $X$ containing $A$; because $\bar{A}$ is the intersection of all such closed sets. Then we have $\bar{A} \cap Y \subset C \cap Y = B.$

Would someone explain the bolded part to me? I don't understand why $B$ being the smallest closed set containing $A$ in $Y$, implies the inclusion, and just how do we know $C$ contains $A$? Is it because by definition $B$ is in both $C$ and $Y$ and by the definition of $B$ we must contain $A$?

Why is it not a good idea to prove $B = B \cup B' = \bar{A} \cap Y$?

Note that $A\subseteq B$ and $A\subseteq\overline A$ and $A\subseteq Y$. Therefore $A\subseteq\overline A\cap Y$. But since $B$ is the smallest set which is closed in $Y$ and contains $A$, it follows that $B\subseteq\overline A\cap Y$.
On the other hand, by definition of relatively closed sets, there is a set $C\subseteq X$ which closed and $C\cap Y=B$. But since $A\subseteq B$ we have that $A\subseteq C$. And again since the closure of $A$ is the smallest closed set containing it we have that $\overline A\subseteq C$; and therefore $\overline A\cap Y\subseteq C\cap Y=B$.
• Wait how does $B$ being the smallest closed set in $Y$ and containing $A$ imply $B \subset \bar{A}$ also? – Hawk Jun 17 '14 at 1:21
• Recall that $\overline A\cap Y$ is a closed set in $Y$ which contains $A$. – Asaf Karagila Jun 17 '14 at 1:25
• Yes, $A \subset \bar{A} \cap Y$ as you have wrote, but the line that follow doesn't make sense to me still... – Hawk Jun 17 '14 at 1:26
• The definition I am getting right now is that $A \subset B \subset Y$, I am not seeing why $B \subset \bar{A}$ also. – Hawk Jun 17 '14 at 1:31