Intersection of Topologies is a Topology

Theorem

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

Then $\tau := \displaystyle \bigcap_{i \in I} {\tau_i}$ 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:

$\displaystyle \forall i \in I: \bigcup_{j \in J} {U_j} \in \tau_i$

Therefore we have:

$\displaystyle \bigcup_{j \in J} {U_j} \in \bigcap_{i \in I} {\tau_i} = \tau$

$\Box$

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:

$\displaystyle U_1 \cap U_2 \in \bigcap_{i \in I} {\tau_i} = \tau$

$\Box$

$X \in \tau$

By the definition of a topology:

$\forall i \in I: X \in \tau_i$

Thus by definition of set intersection we have that:

$\displaystyle X \in \bigcap_{i \in I} {\tau_i} = \tau$

$\Box$

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

$\blacksquare$