Proposition

Let $M$ be a metric space, and let $S \subset M$ be any subset. Show that the subspace topology on $S$ is the same as the metric topology obtained by restricting the metric of $M$ to pairs of points in $S$.

Solution

Let $\mathcal{B}_1 = \{ S \cap B_M(x, r) \mid x \in M, r > 0 \}, \mathcal{B}_2 = \{ B_S(x, r) \mid x \in S, r > 0 \}$. Let $\mathcal{T}_1, \mathcal{T}_2$ be the topologies generated by $\mathcal{B}_1, \mathcal{B}_2$, respectively. Then $\mathcal{B}_1$ is a basis for the subspace topology on $S$, and $\mathcal{B}_2$ is a basis for the metric topology obtained by restricting the metric of $M$ to pairs of points in $S$. For any $x \in S$ and $r > 0$, $S \cap B_M(x, r) = B_S(x, r)$. Then $\mathcal{B}_2 \subset \mathcal{B}_1$, so $\mathcal{T}_2 \subset \mathcal{T}_1$. We will show that $\mathcal{T}_2$ is finer than $\mathcal{T}_1$. Let $U$ be an open set in $\mathcal{T}_1$. Let $p \in U$. Then $p \in (B_M(x, r) \cap S) \subset U$ for some $x \in M$ and $r > 0$. Let $r’ = r - d(x, p)$. Then $B_S(p, r’) \subset B_M(x, r)$ because

Therefore, $U$ is open in $\mathcal{T}_2$.

Therefore, $\mathcal{T}_1 = \mathcal{T}_2$.