General Intersection Property of Topological Space

Theorem
Let $\left({X, \vartheta}\right)$ be a topological space.

Let $X_1, X_2, \ldots, X_n$ be open sets of $\left({X, \vartheta}\right)$.

Then:
 * $\displaystyle \bigcap_{i=1}^n X_i$

is also an open set of $\left({X, \vartheta}\right)$.

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

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


 * $(1): \quad$ The intersection of any two elements of $\vartheta$ is an element of $\vartheta$;
 * $(2): \quad X$ is itself an element of $\vartheta$.

Proof
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * For any sets $X_1, X_2, \ldots, X_n \in \vartheta$, it follows that $\displaystyle \bigcap_{i=1}^n X_i \in \vartheta$.

Let $\mathbb S$ be any finite subset of $\vartheta$.

From Intersection of Empty Set, we have that:
 * $\mathbb S = \varnothing \implies \bigcap \mathbb S = X$

From the definition of a topology, we have that $X \in \vartheta$.

Hence $P(0)$ is true.

From Intersection of Singleton, we have that:
 * $\mathbb S = X_1 \implies \bigcap \mathbb S = X_1$

Thus $P(1)$ is trivially true.

Basis for the Induction
$P(2)$ is the case $X_1 \cap X_2 \in \varnothing$, which is our axiom:
 * The intersection of any two elements of $\vartheta$ is an element of $\vartheta$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * For any sets $X_1, X_2, \ldots, X_k \in \vartheta$, it follows that $\displaystyle \bigcap_{i=1}^k X_i \in \vartheta$.

Then we need to show:
 * For any sets $X_1, X_2, \ldots, X_{k+1} \in \vartheta$, it follows that $\displaystyle \bigcap_{i=1}^{k+1} X_i \in \vartheta$.

Induction Step
This is our induction step:

Consider the set $\displaystyle \bigcap_{i=1}^{k+1} X_i$.

This is $\displaystyle \left({\bigcap_{i=1}^k X_i}\right) \cap X_{k+1}$.

But from the induction hypothesis, we know that:
 * $\displaystyle \left({\bigcap_{i=1}^k X_i}\right) \in \vartheta$

So from the base case, it follows that:
 * $\displaystyle \bigcap_{i=1}^{k+1} X_i \in \vartheta$

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore, for any sets $X_1, X_2, \ldots, X_n \in \vartheta$, it follows that $\displaystyle \bigcap_{i=1}^n X_i \in \vartheta$.

The converse follows directly from the above.

In particular, the fact that $X$ is itself an element of $\vartheta$ follows from the definition of Intersection of Empty Set.