proving finite operation

Question

I often see that proof is only done in two operation, rather than finite.

For example, when we prove $(\mathbb{R},\tau)$ is a topological space, we only show that $O_1,O_2\in \tau \Rightarrow (O_1\cap O_2)\in \tau$, rather than $\bigcap_{i\in I}O_i \in \tau, O_i\in \tau$. How does proving two operation is equivalent to proving finite operation?

Answer 1

Suppose that $O_1, O_2 \in \tau \implies (O_1 \cap O_2) \in \tau$. We will show that the any finite intersection of open sets is also open by induction on the number of open sets we are intersecting.

For the base case, we consider the intersection of $1$ open set,

$$\cap_{i=1}^1 O_i = O_1 \in \tau,$$

so the intersection of $1$ open set is open.

Now let us suppose the intersection of $k$ open sets is open. We will show that the intersection of $k+1$ open sets is open:

$$\cap_{i=1}^{k+1} O_i = \underbrace{\cap_{i=1}^{k} O_i}_{\in \tau} \cap \underbrace{O_{k+1}}_{\in \tau} \in \tau.$$

By mathematical induction, the finite intersection of open sets is always open.