# Topology Proof

• October 22nd 2012, 10:44 AM
Kiefer
Topology Proof
If T1 and T2 are topologies for X and B={(G1∩G2)|G1∈T1, G2∈T2},
Then T*={G|For all p, if p∈G, then A∈B such that p∈A⊆G} is a topology.
So I need to show
(1) T*⊆P(X)
(2) ∅∈T* and X∈T*
(3) if A,B∈T*, then A∩B∈T*
(4) if A⊆T*, then ∪A∈T*

This is part of a problem proving T~=∩{T|T is a topology for X such that T1 ⊆ T and T2 ⊆ T}=>{(G1∩G2)|G1∈T1, G2∈T2} is a base for T*...So is G an element of T~ (not sure what else it would be an element of)?

(I know the first two are kind of trivial, but if you are able to help me prove them anyways I would appreciate it)
• October 22nd 2012, 01:03 PM
Plato
Re: Topology Proof
Quote:

Originally Posted by Kiefer
If T1 and T2 are topologies for X and B={(G1∩G2)|G1∈T1, G2∈T2},
Then T*={G|For all p, if p∈G, then A∈B such that p∈A⊆G} is a topology.
So I need to show
(1) T*⊆P(X)
(2) ∅∈T* and X∈T*
(3) if A,B∈T*, then A∩B∈T*
(4) if A⊆T*, then ∪A∈T*

I find your notation so hard to follow.
(3) If $A\in\mathcal{B}~\&~B\in\mathcal{B}$ then $\exists G_1\in\mathcal{T}_1~\&~\exists G_2\in\mathcal{T}_2$ such that $A=G_1\cap G_2$;
$\exists H_1\in\mathcal{T}_1~\&~\exists H_2\in\mathcal{T}_2$ such that $B=H_1\cap H_2$.

$A\cap B=(G_1\cap G_2)\cap(H_1\cap H_2)=(G_1\cap H_1)\cap(G_2\cap H_2)$ but $(G_1\cap H_1)\in\mathcal{T}_1~\&~(G_2\cap H_2)\in\mathcal{T}_2$