Intersection of Topologies is Topology

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

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

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

$(O1)$: Union of Open Sets
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:
 * $\displaystyle \bigcup_{j \mathop \in J} {U_j} \in \bigcap_{i \mathop \in I} {\tau_i} = \tau$

$(O2)$: Intersection of Open Sets
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:
 * $\displaystyle U_1 \cap U_2 \in \bigcap_{i \mathop \in I} {\tau_i} = \tau$

$(O3)$: Set Itself
By the definition of a topology:
 * $\forall i \in I: S \in \tau_i$

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

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