Vtyjkyuk

Let $(X,mathcal T)$ a topological space. Let $Bsubset X$. Then the induced topology of $B$ is $$mathcal T_B=Ucap Bmid Uin mathcal T.$$
In other word, a set $A$ is open in $B$ if there is $Uinmathcal T$ such that $$A=Ucap B.$$

But what is a closed set in $B$ ? I natural definition would be $A$ is closed in $B$ if $A^c$ is open in $B$. But the problem is that $A^c$ is not in $B$. Maybe $A$ is closed in $B$ if $A^ccap B$ is closed in $B$ ? But I'm not really sure that it makes sense.

In every topology a set is closed iff it is the complement of an open set.

So in this case we find that sets of the form $B-(Ucap B)$ are closed.

Observe that $$B-(Ucap B)=B-U=U^complementcap Btag1$$ where $U^complement$ denotes the complement of $U$ as a subset of $X$.

So actually $(1)$ shows us that a set that is closed in $B$ can be written as $Fcap B$ where $F$ is a closed set in the original topological space $X$.

Also the converse of this is true: whenever $F$ is a closed set in $X$ then $Fcap B$ is a closed set in $B$.

This because $B-Fcap B=F^complementcap B$ where $F^complement$ denotes the complement of $F$ in $X$, hence is open in $X$.