Let X be topological space and let $Y\subset X$ have subspace topology if A is closed in Y and Y is Closed in X then A is closed in X

Question

I'm trying to prove this problem

First

Suppose A is closed in Y the A Contain all limit point of Y. But I'm stuck since I don't know what condition that gaurantee A will still contain all limit point when in X

Then I use different method Suppose A is closed I Y then complement of A is open in Y

Now consider X since X is topological space,then it's open and consider X-A since X is open and A is closed X-A is open therefore the complement of A in X is open then A is closed in X

Is the second method is valid proof ?

Answer 1

In order to prove this limit points can be left out.

Just realize that $A\subseteq Y$ is closed in $Y$ iff there is a set $F$ closed in original $X$ such that $A=Y\cap F$.

(If this is new for you then realize that $Y-A$ is open in $Y$ which means exactly that $Y-A=Y\cap U$ for a set $U$ that is open in $X$. Now note that $A=Y\cap F$ for the closed set $F=U^{\complement}$).

Now if $Y$ is also closed in the original $X$ then $Y\cap F$ can be recognized as a binary intersection of closed sets. The collection of closed sets is by definition closed under the formation of finite intersections, so we conclude immediately that $A=Y\cap F$ is a closed set in $X$.

Answer 2

What you did is not correct, because you assume the very thing that you want to prove: that $A$ is closed in $X$.

Consider the set $X\setminus A$. You want to prove that it is an open subset of $X$. But $$X\setminus A=(X\setminus Y)\cup(Y\setminus A).\tag1$$Since $Y$ is closed in $X$, $X\setminus Y$ is open in $X$. And, since $A$ is closed in $Y$, $Y\setminus A$ is open in $A$. And since $Y\setminus A$ is open in $Y$ and $Y$ is open in $X$, $Y\setminus A$ is open in $X$. So, it follows from $(1)$ that $X\setminus A$ is open in $X$.