Intersection of Topologies is Topology

Theorem
Let $\left({\tau_i}\right)_{i \in I}$ be an arbitrary indexed set of topologies for a set $X$.

Then $\tau := \displaystyle \bigcap_{i \mathop \in I} {\tau_i}$ is also a topology for $X$.

Proof
Proceed by verifying $\tau$ satisfies the conditions for a topology (on $X$).

Union in $\tau$
Let $\left({U_j}\right)_{j \in J}$ be an arbitrary indexed set, such that:
 * $\forall j \in J: U_j \in \tau$

Thus for all $i \in I$, we have by definition of set intersection that:
 * $\forall j \in J: U_j \in \tau_i$

Since $\tau_i$ is a topology for every $i \in I$, by definition we have:
 * $\displaystyle \forall i \in I: \bigcup_{j \mathop \in J} {U_j} \in \tau_i$

Therefore we have:
 * $\displaystyle \bigcup_{j \mathop \in J} {U_j} \in \bigcap_{i \mathop \in I} {\tau_i} = \tau$

Intersection in $\tau$
Let $U_1, U_2 \in \tau$.

Then by definition of set intersection:
 * $\forall i \in I: U_1, U_2 \in \tau_i$

Since $\tau_i$ is a topology for each $i \in I$, we obtain that:
 * $\forall i \in I: U_1 \cap U_2 \in \tau_i$

Therefore we have:
 * $\displaystyle U_1 \cap U_2 \in \bigcap_{i \mathop \in I} {\tau_i} = \tau$

$X \in \tau$
By the definition of a topology:
 * $\forall i \in I: X \in \tau_i$

Thus by definition of set intersection we have that:
 * $\displaystyle X \in \bigcap_{i \mathop \in I} {\tau_i} = \tau$

Thus, by definition, $\tau = \displaystyle \bigcap_{i \mathop \in I} {\tau_i}$ is a topology.