Here's the definition of a topology τ on X (see link):
Let X be a set and let τ be a family of subsets of X. Then τ is called a topology on X if:
1. Both the empty set and X are elements of τ.
2. Any union of sets in τ is an element of τ.
3. Any intersection of finitely many elements of τ is an element of τ.
So the question amounts to verifying 1, 2, and 3 for the intersection ⋂[i∈I]T_i
of the topologies T_i.
1: Since each T_i is a topology on X, ∅ ∈ T_i and X ∈ T_i for all i∈I.
Conclude ∅ ∈ ⋂[i∈I]T_i and X ∈ ⋂[i∈I]T_i.
2: Let V = {v_j : j∈J} be any collection of sets, each v_j of which is in ⋂[i∈I]T_i.
(In other words, let V ⊆ ⋂[i∈I]T_i).
Then V ⊆ T_i for each i ∈ I. So, because T_i is a topology, the union ⋃[j∈J]v_j is in T_i.
Conclude ⋃[j∈J]v_j ∈ ⋂[i∈I]T_i.
3: Similar to 2. Now assume V = {v_j : j∈J} is a finite collection (i.e., J is finite),
with each v_j in ⋂[i∈I]T_i. Again, V ⊆ T_i for each i ∈ I. So, because T_i is a topology,
and J is finite, the intersection ⋂[j∈ j]v_j is in T_i.
Conclude ⋂[j∈J]v_j ∈ ⋂[i∈ I]T_i.
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Verified answer
Here's the definition of a topology τ on X (see link):
Let X be a set and let τ be a family of subsets of X. Then τ is called a topology on X if:
1. Both the empty set and X are elements of τ.
2. Any union of sets in τ is an element of τ.
3. Any intersection of finitely many elements of τ is an element of τ.
So the question amounts to verifying 1, 2, and 3 for the intersection ⋂[i∈I]T_i
of the topologies T_i.
1: Since each T_i is a topology on X, ∅ ∈ T_i and X ∈ T_i for all i∈I.
Conclude ∅ ∈ ⋂[i∈I]T_i and X ∈ ⋂[i∈I]T_i.
2: Let V = {v_j : j∈J} be any collection of sets, each v_j of which is in ⋂[i∈I]T_i.
(In other words, let V ⊆ ⋂[i∈I]T_i).
Then V ⊆ T_i for each i ∈ I. So, because T_i is a topology, the union ⋃[j∈J]v_j is in T_i.
Conclude ⋃[j∈J]v_j ∈ ⋂[i∈I]T_i.
3: Similar to 2. Now assume V = {v_j : j∈J} is a finite collection (i.e., J is finite),
with each v_j in ⋂[i∈I]T_i. Again, V ⊆ T_i for each i ∈ I. So, because T_i is a topology,
and J is finite, the intersection ⋂[j∈ j]v_j is in T_i.
Conclude ⋂[j∈J]v_j ∈ ⋂[i∈ I]T_i.