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

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) The intersection of any two elements of $$\vartheta$$ is an element of $$\vartheta$$;
 * 2) $$X$$ 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 $$\bigcap_{i=1}^n X_i \in \vartheta$$.

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 $$X \in \varnothing$$.

Hence $$P(0)$$ is true

From Intersection of One Set, 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 $$\bigcap_{i=1}^k X_i \in \vartheta$$.

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

Induction Step
This is our induction step:

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

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

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

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

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 $$\bigcap_{i=1}^n X_i \in \vartheta$$.

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.