# The Interior of Intersections 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)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 interiors of both finite intersections and arbitrary intersections.

Theorem 1: Let $(M, d)$ be a metric space and let $S_1, S_2, ..., S_n \subseteq M$ be a finite collection of subsets of $M$. Then $\displaystyle{\mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right )= \bigcap_{i=1}^{n} \mathrm{int} (S_i)}$. |

**Proof:**Let $x \in \mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right )$. Then there exists an $r > 0$ such that:

- But $\displaystyle{\bigcap_{i=1}^{n} S_i \subseteq S_i}$ for each $i \in \{ 1, 2, ..., n \}$ and so $B(x, r) \subseteq S_i$ for each $i$. But this implies that $x \in \mathrm{int} (S_i)$ for each $i$ and thus $\displaystyle{x \in \bigcap_{i=1}^{n} \mathrm{int} (S_i)}$. This shows that:

- Now let $x \in \bigcap_{i=1}^{n} \mathrm{int}(S_i)$. Then $x \in \mathrm{int} (S_i)$ for all $i \in \{1, 2, ..., n \}$. So for each $i$ there exists an $r_i > 0$ such that:

- Let $r = \min \{ r_1, r_2, ..., r_n \}$. Then $B(x, r) \subseteq S_i$ for all $i \in \{ 1, 2, ..., n \}$ which shows that $\displaystyle{B(x, r) \in \bigcap_{i=1}^{n} S_i}$. Thus $\displaystyle{x \in \mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right )}$. This shows that:

- We conclude that then $\displaystyle{\mathrm{int} \left ( \bigcap_{i=1}^{n} S_i \right )= \bigcap_{i=1}^{n} \mathrm{int} (S_i)}$. $\blacksquare$

Theorem 2: Let $(M, d)$ be a metric space and let $\mathcal F$ be a finite collection of subsets of $M$. Then $\displaystyle{\mathrm{int} \left ( \bigcap_{S \in \mathcal F} S \right ) \subseteq \bigcap_{S \in \mathcal F} \mathrm{int} (S) }$. |

**Proof:**Let $\displaystyle{x \in \mathrm{int} \left ( \bigcap_{S \in \mathcal F} S \right )}$. Then $x$ is an interior point of $\displaystyle{\mathrm{int} \left ( \bigcap_{S \in \mathcal F} S \right )}$ and so there exists an $r > 0$ such that:

- Therefore $B(x, r) \subseteq S$ for all $S \in \mathcal F$, so $x \in \mathrm{int} (S)$ for all $S \in \mathcal F$. This implies that $\displaystyle{x \in \bigcap_{S \in \mathcal F} \mathrm{int} (S)}$ and so:

Note that in Theorem 1 we relied on the fact that were looking at a finite intersection to show equality. Equality in Theorem 2 does not hold in general though.