Topology proof: smallest topology and base for topology

Thm: If Τ_{1 }and Τ_{2} are topologies for X and Τ*=∩{ Τ | Τ is a topology for X such that Τ_{1} ⊆ Τ, Τ_{2} ⊆ Τ},

then

(1) Τ* is the smallest topology for X containing both Τ_{1 }and Τ_{2}; and

(2) {(G_{1}∩G_{2})| G_{1} ∈ Τ_{1}, G_{2} ∈ Τ_{2}} 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.

Re: Topology proof: smallest topology and base for topology

Quote:

Originally Posted by

**Kiefer** Thm: If Τ_{1 }and Τ_{2} are topologies for X and Τ*=∩{ Τ | Τ is a topology for X such that Τ_{1} ⊆ Τ, Τ_{2} ⊆ Τ}, then

(1) Τ* is the smallest topology for X containing both Τ_{1 }and Τ_{2}; and

(2) {(G_{1}∩G_{2})| G_{1} ∈ Τ_{1}, G_{2} ∈ Τ_{2}} is a base for Τ*

As I read this $\displaystyle \mathcal{T}^*$ is the intersection of all typologies on $\displaystyle X$ which contain $\displaystyle \mathcal{T}_1~\&~\mathcal{T}_2$

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

For part (2), note that if $\displaystyle G\in\mathcal{T}_1$ then $\displaystyle G\in\mathcal{T}^*$. WHY?

Also $\displaystyle G_1\in\mathcal{T}_1~\&~G_2\in\mathcal{T}_2$ are in then $\displaystyle G_1\cap G_2$ must belong to $\displaystyle \mathcal{T}^*$ because typologies are closed under finite intersection.

Re: Topology proof: smallest topology and base for topology

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

Re: Topology proof: smallest topology and base for topology

Quote:

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

Why is that?

Say $\displaystyle \mathcal{T}_0$ is smaller.

But haven't you already incuded $\displaystyle \mathcal{T}_0$ in $\displaystyle \mathcal{T}^*~?$