A theorem on restriction of a metric Consider $(X,d)$ be a metric space. Let $Y\subseteq X$ be a metric on itself and $E\subseteq Y$. Then there is a theorem which says that $E$ is open in $Y$ iff $E=Y\cap G$ for some open subset $G$ of the original space $G$. The intuition behind behind this condition is not clear. Why is that intersection is so special? I understand that $E$ being a subset of some open set of $X$ is not enough because a subset of a open set need not be an open set ?  
 A: This is not something peculiar to metric spaces. If $\langle X,\tau\rangle$ is any topological space, and $Y\subseteq X$, we define the subspace (or relative) topology on $Y$ to be $\tau_Y=\{U\cap Y:U\in\tau\}$. This $\tau_Y$ is easily shown to be a topology on $Y$, and it is related to that of $X$ in the simplest way: in effect it is exactly what we can see of $\tau$ if our vision is limited to the set $Y$.
Now suppose that the topology $\tau$ is generated by a metric $d$; $d$ is a function from $X\times X$ to $\Bbb R$, so we can look at the restriction of $d$ to $Y\times Y$. Let $d_Y=d\upharpoonright Y\times Y$ be this restriction; then it’s easily verified that $d_Y$ is a metric on $Y$. This is the simplest way to get a metric on $Y$ that is naturally related to $d$; in fact, it is exactly what we can see of $d$ if our vision is limited to $Y$.
Now to tie the two notions together: it turns out (and is not hard to show) that the topology generated on $Y$ by $d_Y$ is exactly $\tau_Y$, the subspace topology on $Y$ defined from $\tau$.
