Definition of relatively open sets in Rudin

Question

Let E be a subset of Y. E is open relative to Y if to each p$\in$E there is associated an r greater than 0 such that q$\in$E whenever d(p,q) is less than r and q$\in$Y.

For this to be valid, mustn’t Y be open?

Answer 1

No, not necessarily. Suppose $Y$ is the closed interval $[0,2]$. Then the set $E = (1/2,3/2)$ is relatively open in Y. To see this, let $p \in E$ and let $r = \min(p-1/2,3/2-p)$. Then for any point $q$ satisfying $d(p,q) < r$, we see that $q \in E$ (and as $E$ is a subset of $Y$, we also have $q \in Y$). Thus $E$ is relatively open in $Y$ even though $Y$ itself is not open.

Answer 2

You should also note that the metric under consideration is just the metric on the mother space restricted on to the subspace. That's why the ball $B_Y(p, r) $ is just $B_X(p, r) \cap Y$ hence the set Y needn't be open it's just a subset of X, the mother space. This also Satisfies the open set conditions for subspace topology.

The condition is precisely saying that in order E to be open for each $p \in E$ there is a r>0 such that $B_Y(p, r) \subset E$ . Note that [0, 1) is open in [0, 1] under Euclidean topology even when both the sets are not open in $\Bbb R$.