Finite Intersection of Open Sets is Open

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

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

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

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

Proof
We prove by induction of $n$.

$(0): \quad \displaystyle \bigcap_{i \mathop = 1}^0 U_i = S$ which by definition of topological space is open.

$(n \implies n+1):$ Suppose $U_1, \ldots, U_n$ are open $\implies \displaystyle \bigcap_{i \mathop = 1}^n U_i$ is open.

Let $U_1, \ldots, U_{n+1}$ be open.


 * $\displaystyle \bigcap_{i \mathop = 1}^{n+1} U_i = \left( {\bigcap_{i \mathop = 1}^n U_i} \right)\cap U_{n+1}$ which as intersection two open sets by definition of topological space is open.