Finite Intersection of Open Sets of Metric Space is Open

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Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $U_1, U_2, \ldots, U_n$ be open sets in $M$.


Then $\ds U = \bigcap_{i \mathop = 1}^n U_i$ is open in $M$.


That is, a finite intersection of open subsets is open.


Proof 1

Let $x \in U$.

By the definition of intersection:

$\forall i \in \closedint 1 n: x \in U_i$.

Thus, since $U_i$ is open in $M$:

$\forall i \in \closedint 1 n: \exists \epsilon_i \in \R_{>0}: \map {B_{\epsilon_i} } x \subseteq U_i$

where $\map {B_{\epsilon_i} } x$ is the open $\epsilon_i$-ball of $x$.


Let $\ds \epsilon = \min_{i \mathop = 1}^n \set {\epsilon_i}$.

Then, by Open Ball contains Smaller Open Ball:

$\forall i \in \closedint 1 n: \map {B_\epsilon} x \subseteq \map {B_{\epsilon_i} } x$


By definition of intersection:

$\map {B_\epsilon} x \subseteq U$

The result follows.

$\blacksquare$


Proof 2

Aiming for a contradiction, suppose $U$ is not open in $M$.

By definition of open set, there exists a $y \in U$ such that:

$\ds \forall \epsilon \in \R_{>0}: \map {B_\epsilon} y \setminus \bigcap_{i \mathop = 1}^n U_i \ne \O$.

By De Morgan's law, we have

$\ds \bigcup_{i \mathop = 1}^n \paren {\map {B_\epsilon} y \setminus U_i} \ne \O$.

Thus, there exists $i \in \closedint 1 n$ such that $\map {B_\epsilon} y \setminus U_i \ne \O$.

As $U_i$ is open and $\epsilon$ is arbitrarily given, by Open Ball of Point Inside Open Ball, $y \notin U_i$. This is a contradiction.

Thus, $U$ is open in $M$.

$\blacksquare$


Also see


Sources

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