Exercise 2.1.1 (Intersection of topologies is topology)

Proof. Let ${(T_i)}_{i\in I}$ be a family of topologies on $X$ and consider

We demonstrate that $T$ is a topology.

1.

Since $\varnothing \in T_i$ and $X\in T_i$ by the axioms of topology for all $i\in I$, we have $\varnothing ,X\in T$ as well.

2.

Let ${(A_j)}_{j\in J}$ be a family of sets in $T$. By construction, we have $A_j\in T_i$ for all $j\in J$ and for all $i\in I$. Since $T_i$, $i\in I$ are all topologies, we have ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j\in T_i$ for all $i\in I$. Finally, we conclude that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j\in T$ by construction.

3.

Let ${(A_n)}_{n=1}^N$ be a finite family of sets in $T$. By construction, we have $A_n\in T_i$ for all $1\le n\le N$ and for all $i\in I$. Since $T_i$, $i\in I$ are all topologies, we have ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{1\le n\le N}{A}_n\in T_i$ for all $i\in I$. This, we conclude that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{1\le n\le N}{A}_n\in T$ by construction.