Separated set in subspace topology.

Question

If $A,B\subset E\subset X$ where $X$ is a topological space.Then is the following true?

$A,B$ are separated in $X$ $\iff$ $A,B$ are separated in $E$.

i.e. $\overline{A}\cap B=A\cap \overline{B}=\phi$ $\iff $ $\overline{A}^{E}\cap B=A\cap \overline{B}^{E}=\phi$.I have shown that $\implies$ part is true and it is obvious,but is the other one true?

Answer 1

$\overline {A}^{E}$ is nothin but $\overline {A} \ \cap E$. Hence $\overline {A}^{E} \cap B=\emptyset $ implies $\overline {A} \cap E \cap B =\emptyset $ which is same as $\overline {A} \cap B =\emptyset $ since $B \cap E=B$. Similarly for $\overline {B} \cap A$. Hence if the sets are separated in $E$ they are separated in $X$ also.

Answer 2

Note that $$A \cap \bar{B} = \emptyset \iff \text{there is an open set } U \subset X \text{ such that } A\subset U, U \cap B = \emptyset \\ A \cap \bar{B}^E = \emptyset \iff \text{there is an open set } U \subset X \text{ such that } A\subset U, (U\cap E) \cap B = \emptyset$$

Since $B \subset E,$ however, we have $(U \cap E) \cap B = U \cap (E \cap B) = U \cap B$ for any $U \subset X,$ so the right-hand sides are equivalent.