Intersection of Topologies is Topology

Theorem
Let $\struct {\tau_i}_{i \mathop \in I}$ be an arbitrary indexed set of topologies on a set $S$.

Then $\tau := \ds \bigcap_{i \mathop \in I} \tau_i$ is also a topology on $S$.

Proof
Each of the open set axioms are examined in turn:

Let $\struct {U_j}_{j \mathop \in J}$ 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:
 * $\ds \forall i \in I: \bigcup_{j \mathop \in J} {U_j} \in \tau_i$

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

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:
 * $\ds U_1 \cap U_2 \in \bigcap_{i \mathop \in I} \tau_i = \tau$

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

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

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