Intersection of Topologies is Topology

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Theorem

Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary indexed family 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:


Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets

Let $\family {U_j}_{j \mathop \in J}$ be an arbitrary indexed family, such that:

$\forall j \in J: U_j \in \tau$

Thus we have by definition of set intersection that:

$\forall i \in I: \forall j \in J: U_j \in \tau_i$

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:

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

$\Box$


Open Set Axiom $\paren {\text O 2 }$: Pairwise 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:

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

$\Box$


Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology

By the definition of a topology:

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

Thus by definition of set intersection we have that:

$\ds S \in \bigcap_{i \mathop \in I} \tau_i = \tau$

$\Box$


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

$\blacksquare$


Sources

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