The question is:

Let $\displaystyle (X,d)$ be a metric space, and $\displaystyle Y$ be a subset of $\displaystyle X$.

If $\displaystyle G \subset X$ is open, then $\displaystyle G \cap Y$ is open in the metric space $\displaystyle (Y,d)$.

Pf:

$\displaystyle Y \subset X$ implies $\displaystyle (Y,d)$ is a metric space $\displaystyle \rightarrow Y$ is open.

Therefore $\displaystyle Y^{C} $is closed

$\displaystyle G$ is open, thus $\displaystyle G^{C}$ is closed.

So $\displaystyle B = (G^{C} \cup Y^{C})$ is closed, since it is the finite union of closed sets.

Therefore $\displaystyle B^{C} = G \cap Y $ is open.

I seem to follow all the definitions and theorems correct. Is this ok?

There is also a second part to this question.

Conversely, if $\displaystyle E_1 \subset Y$ is open in $\displaystyle (Y,d)$, $\displaystyle \exists $ an open set $\displaystyle E \subset X$ s.t $\displaystyle E_1 = E \cap Y$

I don't get this question. Don't we already have an open set that works in the hypothesis? E = E_1?

Any help is greatly appreciated. Thank you.

-Jame