Criterion for a Set to be Open in a Metric Subspace

Consider a metric space $(M, d)$ and a subset $S \subseteq M$ so that $(S, d)$ is a metric subspace of $(M, d)$. The ball centered at $x \in S \subseteq M$ with radius $r$ can refer to either the set of points $y \in S$ such that $d(x, y) < r$ or the set of points $y \in M$ such that $d(x, y) < r$. To make the distinction, we write:

(1)

\begin{align} \quad B_S (x, r) = \{ y \in S : d(x, y) < r \} \quad \mathrm{and} \quad B_M (x, r) = \{ y \in M : d(x, y) < r \} \end{align}

By the definitions above, we see that $B_S (x, r) \subseteq B_M(x, r)$ since $S \subseteq M$. In fact, from the diagram below, we more generally see that:

(2)

\begin{align} \quad B_S (x, r) = B_M(x, r) \cap S \end{align}

We will now look at an important theorem in determining whether a set in a metric subspace is open or not.

Theorem 1: Let $(M, d)$ be a metric space and let $(S, d)$ be a metric subspace of $(M, d)$. A set $X \subseteq S$ is open in $S$ if and only if there exists an open set $A$ in $M$ such that $X = A \cap S$.

• Proof: $\Rightarrow$ Let $X \subseteq S$ be open in $S$. Then for all $x \in X$ there exists a ball (in $S$) centered at $x$ with radius $r > 0$ such that:

(3)

\begin{align} \quad x \in B_S(x, r) \subseteq S \end{align}

• Since $X$ is open in $S$ we have that $X$ can be written as the union of all of these open balls, that is:

(4)

\begin{align} \quad X = \bigcup_{x \in X} B_S(x, r) \end{align}

• But we know that $B_S(x, r) = B_M(x, r) \cap S$, and so:

(5)

\begin{align} \quad X = \bigcup_{x \in X} [B_M(x, r) \cap S] \\ \quad X = \left ( \bigcup_{x \in X} B_M(x, r) \right ) \cap S \end{align}

• Let $A = \bigcup_{x \in X} B_M(x, r)$. Then $A$ is an open set in $M$ since every point in $A$ is an interior point of $A$. Therefore $X = A \cap S$ where $A$ is an open set in $M$.

• $\Leftarrow$ Suppose that $X = A \cap S$ for some open set $A$ in $M$. To show that $X$ is open, we need to prove that $X = \mathrm{int}(X)$. Since $\mathrm{int}(X) \subseteq X$ trivially, we will show that $\mathrm{int} (X) \supseteq X$.

• Let $x \in A \cap S$. Then $x \in A$ and $x \in S$. Since $x \in A$ and $A$ is open in $M$ we have that there exists a ball (in $M$) centered at $x$ with some radius $r > 0$ that is fully contained in $M$, that is:

(6)

\begin{align} \quad B_M(x, r) \subseteq A \end{align}

• But we know that $B_S (x, r) = B_M(x, r) \cap S$, and so:

(7)

\begin{align} \quad B_S(x, r) = B_M(x, r) \cap S = A \cap S = X \end{align}

• Therefore the ball (in $S$) centered at $x$ with radius $r > 0$ is fully contained in $X$, so $x \in \mathrm{int} (X)$ and we conclude that $X = \mathrm{int} (X)$. So for every $x \in X$ there exists a ball (in $S$) centered at $x$ with radius $r$ that is fully contained in $X$, so $X$ is open in $S$. $\blacksquare$