Properties of topological spaces with respect to subspace topology

Question

Let $B \subset A \subset X$ where $X$ is a topological space.

prove that if $B$ is closed in $X$, then $B$ is closed with respect to the subspace topology on $A$.

I'm guessing, since $B$ is closed in $(X,\mathcal T)$ then $B$ is closed in $\mathcal T _A$ since $\mathcal T_A \subset (X,\mathcal T)$

Am I on the right track?

Answer 1

No. It is just this: since $B$ is closed in $X$, $B^\complement$ is open in $X$. But the complement of $B$ in $A$ is $B^\complement\cap A$. Since $B^\complement$ is open in $X$, $B^\complement\cap A$ is open in $A$. In other words, $B$ is closed in $A$.

Answer 2

For $B$ to be closed in $A$, then there must be a closed set $V$ in the topology of $X$ such that $B=A\cap V$. If $B$ is closed in $X$, you can let $V=B$.