Criterion for a Set to be Closed in a Metric Subspace

# Criterion for a Set to be Closed in a Metric Subspace

Recall from the Criterion for a Set to be Open in a Metric Subspace page that if $(M, d)$ is a metric space and $S \subseteq M$ such that $(S, d)$ is a metric subspace of $(M, d)$ then a set $X$ in the metric subspace, i.e., $X \subseteq S$ is open in $S$ if and only if there exists an open set $A$ in $M$ such that:

(1)\begin{align} \quad X = A \cap S \end{align}

We will now look at an analogous theorem which will show us that a set $Y \subseteq S$ is closed in $S$ if and only if there exists a closed set $B$ in $M$ such that:

(2)\begin{align} \quad Y = B \cap S \end{align}

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

**Proof:**$\Rightarrow$ Suppose that $Y$ is closed in $S$. Then $S \setminus Y$ is open in $S$. Hence there exists an open set $A$ in $M$ such that:

\begin{align} \quad (S \setminus Y) = A \cap S \end{align}

- Taking the complement of both sides gives us:

\begin{align} \quad Y = S \setminus (A \cap S) = S \setminus A = S \cap (A^c) = S \cap (M \setminus A) \end{align}

- Let $B = M \setminus A$. Then $B$ is a closed set in $M$ and $Y = B \cap S$.

- Suppose that $Y = B \cap S$ for some closed set $B$ in $M$. Then since $B$ is closed in $M$ there exists an open set $A$ in $M$ such that $B = M \setminus A$. So:

\begin{align} \quad Y = B \cap S = S \cap B = S \cap (M \setminus A) = (S \cap M) \setminus A = S \setminus A \end{align}

- But $A$ is open in $S$, so $S \setminus A$ is closed in $S$. Thus $Y$ is closed in $S$. $\blacksquare$