Topology proof: smallest topology and base for topology

Thm: If Τ₁ and Τ₂ are topologies for X and Τ*=∩{ Τ | Τ is a topology for X such that Τ₁ ⊆ Τ, Τ₂ ⊆ Τ},
then
(1) Τ* is the smallest topology for X containing both Τ₁ and Τ₂; and
(2) {(G₁∩G₂)| G₁ ∈ Τ₁, G₂ ∈ Τ₂} is a base for Τ*

We know:
B is a base for Τ iff
(1) B⊆Τ and,
(2) For every G∈Τ, if p∈G, then there is B∈B such that p∈B⊆G.

Quote:

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Originally Posted by Kiefer

Thm: If Τ₁ and Τ₂ are topologies for X and Τ*=∩{ Τ | Τ is a topology for X such that Τ₁ ⊆ Τ, Τ₂ ⊆ Τ}, then
(1) Τ* is the smallest topology for X containing both Τ₁ and Τ₂; and
(2) {(G₁∩G₂)| G₁ ∈ Τ₁, G₂ ∈ Τ₂} is a base for Τ*

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As I read this is the intersection of all typologies on which contain

Exactly what difficulty are you having with part (1)? Please be complete in answering ?

For part (2), note that if then . WHY?

Also are in then must belong to because typologies are closed under finite intersection.

For part one, I can prove that T* is a topology but am not sure how to prove it is the smallest one in X containing T1 & T2

Quote:

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Originally Posted by Kiefer

For part one, I can prove that T* is a topology but am not sure how to prove it is the smallest one in X containing T1 & T2

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Why is that?
Say is smaller.
But haven't you already incuded in