Finite Intersection of Open Sets of Metric Space is Open

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Theorem

Let $M = \left({A, d}\right)$ be a metric space.

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


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


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


Proof

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

For each $i \in \left[{1 \,.\,.\, n}\right]$, we have $x \in U_i$.

Thus:

$\exists \epsilon_i > 0: B_{\epsilon_i} \left({x}\right) \subseteq U_i$

where $B_{\epsilon_i} \left({x}\right)$ is the open $\epsilon_i$-ball of $x$.


Let $\displaystyle \epsilon = \min_{i \mathop = 1}^n \left\{{\epsilon_i}\right\}$.

Then:

$B_\epsilon \left({x}\right) \subseteq B_{\epsilon_i} \left({x}\right) \subseteq U_i$

for all $i \in \left[{1 \,.\,.\, n}\right]$.


So:

$\displaystyle B_\epsilon \left({x}\right) \subseteq \bigcap_{i \mathop = 1}^n U_i$

The result follows.

$\blacksquare$


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