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 in $M$.


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


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


Proof

Let $\ds x \in \bigcap_{i \mathop = 1}^n U_i$.

By the definition of intersection, for each $i \in \closedint 1 n$, we have $x \in U_i$.

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

$\exists \epsilon_i > 0: \map {B_{\epsilon_i} } x \subseteq U_i$

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:

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

for all $i \in \closedint 1 n$.


So:

$\ds \map {B_\epsilon} x \subseteq \bigcap_{i \mathop = 1}^n U_i$

The result follows.

$\blacksquare$


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