Intersection of Topologies, I would like to know if the proof is well.

Question

Let {$\tau_i:i\in I$} be any collection of topologies on set $X$.Then the intersection $\bigcap_{i\in I}\tau _i$ is also a topology on $X$.

Proof: since it is a family $\left \{ \tau_i \right \}_{i\in I}$ of topologies, it has to $X,\emptyset \in \left \{ \tau_i \right \}_{i\in I}$. Hence for all $i\in I$, it has that $X,\emptyset \in \tau_i$, hence $X,\emptyset\in\bigcap_{i\in I}\tau _i$.

Now, let $\left\{U_j \right \}_{j \in J} \subset \bigcap_{i\in I}\tau _i $, hence $\forall_{i\in I}, \left\{U_j \right\}_{j\in J}\in \tau_i$, as $\tau_i$ is a topology $\forall_{i\in I}$, it has that $\bigcap_{j\in J}U _j\in \tau_i, \forall_j\in J$, hence $\bigcap_{j\in J}U _j\in \bigcap_{i\in I}\tau _i, \forall i$.

Let $\left\{U_j \right \}_{j\in J} \in \bigcap_{i\in I}\tau _i$, hence:

$\forall_{i\in I}, \left\{U_j \right \}_{j\in J} \in \tau_i$, since $\tau_i$ is a topology then, $\bigcup_{j\in J}U_j\in\bigcap_{i\in I}\tau_i$.

Hence $\left \{ \tau_i \right \}_{i\in I}\subset Top(X)$.

Answer 1

Your proof is essentially correct, but you should be more precise.

(1) You should not write $\left\{U_j \right \}_{j \in J} \in \bigcap_{i\in I}\tau _i$, but $\left\{U_j \right \}_{j \in J} \subset \bigcap_{i\in I}\tau _i$.

(2) When you consider intersections, you should mention that $J$ is finite.