Finite Intersection of Open Sets is Open

Theorem

Let $\struct {S, \tau}$ be a topological space.

Let $S_1, S_2, \ldots, S_n$ be open sets of $\struct {S, \tau}$.

Then:

$\ds \bigcap_{i \mathop = 1}^n S_i$

is also an open set of $\struct {S, \tau}$.

That is, the intersection of any finite number of open sets of a topology is also in $\tau$.

Conversely, if the intersection of any finite number of open sets of a topology is also in $\tau$, then:

$(1): \quad$ The intersection of any two elements of $\tau$ is an element of $\tau$

$(2): \quad S$ is itself an element of $\tau$.

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 $M$.

Let $\struct {S, \NN}$ be a neighborhood space.

Let $n \in \N_{>0}$ be a natural number.

Let $\ds \bigcap_{i \mathop = 1}^n U_i$ be a finite intersection of open sets of $\struct {S, \NN}$.

Then $\ds \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\struct {S, \NN}$.

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 $M$.