Is it possible for the following to hold in metric spaces? Let (X,d) be a metric space,if A is closed in Y and Y is closed in X then A is closed in X. If possible someone could assist me for a proof. Here I try:Since A is closed in Y, then I want to write A=U∩Y, where U is closed in X. But Y is closed in X,hence A is the intersection of two closed sets in X.Can this be applied here? Thanks
The proof in the OP is correct. Here is a sequence-based proof (which is inferior in that it does not extend to general topological spaces), just for the sake of having a different one.
Suppose $a_n\in A$ for all $n$ and $a_n\to x$. Since $a_n\in Y$ and $Y$ is closed, $x\in Y$. Now the closedness of $A$ in $Y$ implies $x\in A$ as required.