Let $Y$ be a subspace of a topological space $X$, and let $A$ and $B$ be two disjoint closed subsets of $Y$ in the subspace topology. Show that $\bar A \cap B=\varnothing$ and $A \cap \bar B=\varnothing$.
Attempt:Let $A$ and $B$ be two disjoint closed sets in $Y$. Then $A=\bar A \cap Y$,where $\bar A$ is the closure of $A$ in $X$. $B=\bar B \cap Y$, where $\bar B$ is the closure of $B$ in $X$.Then if $y \in \bar A,y \notin B$ since $X-\bar A$ is an open set containing $B$ disjoint from $A$. Similarly if $y \in \bar B$, $y \notin A$. Thus $\bar A \cap B=\varnothing$ and $A \cap \bar B=\varnothing$.
Comment: Unfortunately I think it is ok to assume if $A$ is closed in $Y$ $A=\bar A \cap Y$ for $\bar A$ the closure of $A$ in X, similarly for $B$, but I cannot understand why. Is this justified, if so what is the reason this is justified? Also, I am unsure whether it is correct to say that $X-\bar A$ contains $B$ and similarly for $X-\bar B$. Is this ok? If so why is this justified?
Best Answer
Yes, it’s correct.
Let $A$ be a relatively closed subset of $Y$. Then $A=Y\cap F$ for some $F$ that is closed in $X$, and $\operatorname{cl}_XA$ is the intersection of all closed subsets of $X$ that contain $A$, so $\operatorname{cl}_XA\subseteq F$, and therefore
$$A\subseteq Y\cap\operatorname{cl}_XA\subseteq Y\cap F=A\,,$$
which immediately implies that $A=Y\cap\operatorname{cl}_XA$.