Closure of $B$ in $A$ is closure of $B$ in $X \cap A$ First course in topology, and up until now I have been able to hang using my real analysis knowledge, but open and closed sets have always been hard for me.
Want to show:
Let $X$ be a topological space, $A\subseteq X$ and $B\subseteq A$. Show that $\mathrm{cl}_A (B)=\mathrm{cl}_X (B) \cap A$.
Defintion:
$\mathrm{cl}_X (A)=\bigcap \{C|A\subseteq C$ and $C $ is closed in $X$$\}$
Proof:
By definition we have $cl_{A}(B)\subseteq A$ and $cl_{A}(B)={\displaystyle \bigcap_{\overset{B\subseteq C}{C\mbox{ Closed in }A}}C}\subseteq{\displaystyle \bigcap_{\overset{B\subseteq C}{C\mbox{ Closed in }X}}C}=cl_{X}(B)$. Thus, $cl_{A}\subseteq cl_{X}(B)\cap A$.
Conversely, let $C$ now be closed in $X$ such that $cl_{A}(B)=C\cap A$. Then $B\subseteq C$ gives $cl_{X}(B)\subseteq C$ since $C$ is closed in $X$, and this yields $cl_{X}(B)\cap A\subseteq C\cap A=cl_{A}(B)$. Thus, $cl_{A}\supseteq cl_{X}(B)\cap A$. Therefore, $cl_{A}(B)=cl_{X}(B)\cap A$.
This is what I have finished with, any corrections would be great.
Thanks.
 A: Recall that we define, given $A\subset X$; the subspace topology of $A$ as $$\mathscr T_A=\mathscr T\cap A=\{A\cap O:O\in \mathscr T\}$$
With this topology, $(A,\mathscr T_A)$ is a topological space in its own right. Morever, given $B\subset X$, $B$ is closed (open) in $A$ if and only if $B=A\cap O$ ($A\cap F$) with $O$ open ($F$ closed) in $X$. Thus, we have  $${\operatorname{cl}}_AB=\bigcap \{A\cap C:C\text{ is closed in }X \text{ and } B\subset A\cap C\}$$
while 
$${\operatorname{cl}}_X B=\bigcap \{ C:C\text{ is closed in }X \text{ and } B\subset  C\}$$
Note that assuming $B\subseteq A$ $$B\subseteq C\iff B\subseteq A\cap C$$
A: One way to approach this problem is to use the definition of the closure of $B$ as the smallest closed set containing it.  In other words, the closure of $B$ in $A$ is the unique closed subset $cl_A(B)$ in $A$ with $B\subset cl_A(B)$ and $cl_A(B)\subset C$ for all closed $C$ in $A$ containing $B$.  Similarly for $cl_X(B)$.
Now $cl_X(B)\cap A$ is closed in $A$ and contains $B$, so $cl_A(B)\subset cl_X(B)\cap A$ by definition.  By definition of the subspace topology, we also know that there is some closed set $C\subset X$, such that $cl_A(B)=C\cap A$.  It follows that $B\subset C$, so by definition of closure in $X$, $cl_X(B)\subset C$.  This gives us $cl_X(B)\cap A\subset C\cap A=cl_A(B)$, which is the other inclusion.
