I had to show that a closed subspace of a normal space is normal.
To show that, I let $X$ be a normal space, and $C$ be a closed subset of $X$.
And take disjoint $A$ and $B$ such that closed subsets of $C$.
Since $A$, $B$ are also closed in $X$, by normality of $X$, there exist disjoint open $U$, $V$ s.t. contain $A$, $B$ respectively.
Then $U\cap C$, $V\cap C$ are open in $C$, disjoint and contain $A$, $B$ respectively.
Hence $C$ is also normal.
In this proccess, I don't know why $C$ has to be closed. I can't find error in this proof either.
What am I missing here?
Best Answer
You used it when you stated that $A,B$ are also closed in $X$. By assumption they are closed in $C$, so there are closed $F,G$ closed in $X$, such that $A = F \cap C$ and $B = G \cap C$. But then $A$ and $B$ are indeed closed in $X$ as the intersection of two closed sets of $X$.
In short, closed in closed is closed. (And the same holds for open in open, same proof.)