Open and Closed Set Differences in Metric Spaces

# Open and Closed Set Differences in Metric Spaces

Suppose that $(M, d)$ is a metric space and that $A, B \subseteq M$. Suppose that we know that $A$ is an open subset and $B$ is a closed subset. What can we say about the differences $A \setminus B$ and $B \setminus A$? Are they necessarily open? Are they necessarily closed? The theorem below will tell us that $A \setminus B$ is always open and $B \setminus A$ is always closed. In the theorems below, we use the important fact that $A \setminus B = A \cap (M \setminus B)$ and $B \setminus A = B \cap (M \setminus A)$:

Theorem 1: Let $(M, d)$ be a metric space and let $A, B \subseteq M$. If $A$ is an open subset and $B$ is a closed subset then $A \setminus B$ is an open subset. |

**Proof:**Let $A, B \subseteq M$ and let $A$ be an open subset and let $B$ be a closed subset. Then $B^c = M \setminus B$ is an open subset and

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

- Then $A \setminus B$ is the intersection of two open sets. A finite intersection of open sets is open, so $A \setminus B$ is open. $\blacksquare$

Theorem 2: Let $(M, d)$ be a metric space and let $A, B \subseteq M$. If $A$ is an open subset and $B$ is a closed subset then $B \setminus A$ is a closed subset. |

**Proof:**Let $A, B \subseteq M$ and let $A$ be an open subset and let $B$ be a closed subset. Then $A^c = M \setminus A$ is a closed subset and:

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

- Then $B \setminus A$ is the intersection of two closed sets. Any union of closed sets is closed (in particular, a finite union of closed sets is closed), so $B \setminus A$ is closed. $\blacksquare$