Finite Intersection of Open Sets of Normed Vector Space is Open

Theorem

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector 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 $M$.

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

For each $i \in \closedint 1 n$, we have $x \in U_i$.

Thus:

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

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

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

So:

$\epsilon > 0$.

Let $y \in \map {B_\epsilon} x$.

Then $\norm {x - y} < \epsilon$

Hence:

$\forall i \in \closedint 1 n : \norm {x - y} < \epsilon_i$

In other words:

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

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

So:

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

The result follows.

$\blacksquare$

2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.3$: Normed and Banach spaces. Topology of normed spaces