# Topology proof: smallest topology and base for topology

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• Oct 18th 2012, 09:09 AM
Kiefer
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.
• Oct 18th 2012, 10:07 AM
Plato
Re: Topology proof: smallest topology and base for topology
Quote:

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*

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

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

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

Also $G_1\in\mathcal{T}_1~\&~G_2\in\mathcal{T}_2$ are in then $G_1\cap G_2$ must belong to $\mathcal{T}^*$ because typologies are closed under finite intersection.
• Oct 18th 2012, 02:42 PM
Kiefer
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
• Oct 18th 2012, 02:49 PM
Plato
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 $\mathcal{T}_0$ is smaller.
But haven't you already incuded $\mathcal{T}_0$ in $\mathcal{T}^*~?$