The Interior of Unions of Sets in a Metric Space

Recall from the Interior and Boundary Points of a Set in a Metric Space page that if $(M, d)$ is a metric space and $S \subseteq M$ then a point $a \in S$ is said to be an interior point of $S$ if there exists an $r > 0$ such that:

(1)

\begin{align} \quad B(a, r) \subseteq S \end{align}

In other words, a point $a$ is an interior point of $S$ if there exists an open ball centered at $x$ that is fully contained in $S$. Furthermore, the set of all interior points of $S$ is called the interior of $S$ and is denoted $\mathrm{int} (S)$.

We will now look at some very important theorems regarding the unions of interiors of sets.

Theorem 1: Let $(M, d)$ be a metric space and let $\mathcal F$ be a collection of subsets of $M$. Then $\displaystyle{\mathrm{int} \left ( \bigcup_{S \in \mathcal F} S\right ) \supseteq \bigcup_{S \in \mathcal F} \mathrm{int} (S) }$.

• Proof: Let $x \in \bigcup_{S \in \mathcal F} \mathrm{int}(S)$. Then $x \in \mathrm{int} (S)$ for some $S^* \in \mathcal F$. So there exists an $r > 0$ such that:

(2)

\begin{align} \quad B(x, r) \subseteq S^* \end{align}

• But since $S^* \in \mathcal F$ we have that also:

(3)

\begin{align} \quad S^* \subseteq \bigcup_{S \in \mathcal F} S \end{align}

• Thus $\displaystyle{ B(x, r) \subseteq S^* \subseteq \bigcup_{S \in \mathcal F} S}$ which shows that $\displaystyle{x \in \mathrm{int} \left ( \bigcup_{S \in \mathcal F} S\right )}$. Hence:

(4)

\begin{align} \quad \mathrm{int} \left ( \bigcup_{S \in \mathcal F} S\right ) \supseteq \bigcup_{S \in \mathcal F} \mathrm{int} (S) \quad \blacksquare \end{align}

Note that in general equality in Theorem 1 above does not hold for any arbitrary collection $\mathcal F$ of subsets of $M$.