# Closed sets in induced metric space

Let $$(X,d)$$ be a metric space, let $$Y$$ be a subset of $$X$$ and let $$E$$ be a subset of $$Y$$, then

$$(i)$$ $$E$$ is relatively open with respect to $$Y$$ if and only if $$E=V \cap Y$$ for some set $$V \subseteq X$$ which is open in $$X$$.

$$(ii)$$ $$E$$ is relatively closed with respect to $$Y$$ if and only if $$E=K \cap Y$$ for some set $$k \subseteq X$$ which is closed in $$X$$.

I need to prove $$(ii)$$. I looked at this solution for a similar problem in a topological space, but it uses $$(i)$$. I tried to write a proof that doesn't depend on (i), but I don't know if it works. I am particularly suspicious about the bold text:

$$Proof.$$ Let $$E$$ be closed in $$(Y,d)$$ (with the metric $$d$$ restricted to $$Y\times Y$$) and let $$K = \bar E$$, where $$\bar E$$ is the closure of $$E$$ with respect to $$(X,d)$$. Then $$K$$ is closed in $$(X,d)$$ and since $$K$$ is the union of adherent points to $$E$$ in $$(Y^c,d)$$ and adherent points to $$E$$ in $$(Y,d)$$, then $$K \cap Y =$$ adherent points to $$E$$ in $$(Y,d)$$. Since $$E$$ is closed, it contains all its adherent points, so $$K \cap Y = E$$. Similarly, let $$K$$ be a subset of $$X$$ closed in $$(X,d)$$ such that $$K \cap Y = E$$ , then $$E$$ = all adherent points of $$K$$ in $$Y$$. Thus $$E$$ is closed. $$q.e.d.$$

Here is another way to write the same argument; perhaps looking at the statements from a different angle will be more convincing.

First, observe that if $$x$$ is a limit point of a set $$A$$ in $$Y \subset X$$, then $$x$$ is also a limit point of $$A$$ in $$X$$.

Now, let $$K \subset X$$ be the closure of $$E$$ in $$X$$, as you have defined it. We want to show that $$K \cap Y = E$$; that is, if $$x$$ is a limit point of $$E$$ in $$X$$ that does not lie in $$E$$, then in fact $$x$$ does not lie in $$Y$$. In other words, we want to prove the following:

Let $$x\in X$$ be a limit point of $$E \subset X$$. Then, $$x \not\in E \implies x \not\in Y$$.

But now, the contrapositive of this statement makes things clear:

Let $$x\in X$$ be a limit point of $$E \subset X$$. Then, $$x \in Y \implies x \in E$$.

But, this is true simply because, as we observed, if $$x \in X$$ is a limit point of $$E$$ and $$x$$ also belongs to $$Y$$, then $$x$$ is a limit point of $$E$$ in $$Y$$. Since $$E$$ is closed in $$Y$$, this implies that $$x \in E$$. $$\tag{\blacksquare}$$