Topological Basis

  • #1

Homework Statement


Let (χ,τ) be a topological space and β be a collection of subsets of χ. Then β is a basis for τ if and only if:
1. β ⊂ τ
2. for each set U in τ and point p in U there is a set V in β such that p ∈ V ⊂ U.

2. Relevant definitions

Let τ be a topology on a set χ and let β ⊂ τ. Then β is a basis for the topology τ if and only if every open set in τ is the union of elements of β.

The Attempt at a Solution


I'm proving the if and only if. So for one direction, I have the following:
Suppose β ⊂ τ and for each set U in τ and point p in U, there is a set, call it Vp in β such that p ∈ Vp ⊂ U.
Consider the union ∪ Vp. Every point p in U is also in the union ∪ Vp. So it follows that U ⊂ ∪ Vp.
Notice every point p in the union ∪ Vp is also in U. So it follows that ∪ Vp ⊂ U.
This shows U = ∪ Vp, hence U is a union of elements of β, and therefore β is a basis.

For the other direction, this is what I attempted, but I am not sure the logic follows:
Suppose β is a basis. Then every open set U in τ is the union of elements of β, call them Bi for i in some index. So U = ∪ Bi, with all Bi open.
I'm thinking this follows from the definition that all such collections of {Bi} ⊂ τ, which implies that β ⊂ τ. (satisfying point 1)
Now Suppose there is some point p in U. Then it follows that p is in at least one such Bi, call it V. Therefore there exists a set V in β such that p ∈ V ⊂ U, satisfying point 2.
 

Answers and Replies

  • #2
Dick
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Homework Statement


Let (χ,τ) be a topological space and β be a collection of subsets of χ. Then β is a basis for τ if and only if:
1. β ⊂ τ
2. for each set U in τ and point p in U there is a set V in β such that p ∈ V ⊂ U.

2. Relevant definitions

Let τ be a topology on a set χ and let β ⊂ τ. Then β is a basis for the topology τ if and only if every open set in τ is the union of elements of β.

The Attempt at a Solution


I'm proving the if and only if. So for one direction, I have the following:
Suppose β ⊂ τ and for each set U in τ and point p in U, there is a set, call it Vp in β such that p ∈ Vp ⊂ U.
Consider the union ∪ Vp. Every point p in U is also in the union ∪ Vp. So it follows that U ⊂ ∪ Vp.
Notice every point p in the union ∪ Vp is also in U. So it follows that ∪ Vp ⊂ U.
This shows U = ∪ Vp, hence U is a union of elements of β, and therefore β is a basis.

For the other direction, this is what I attempted, but I am not sure the logic follows:
Suppose β is a basis. Then every open set U in τ is the union of elements of β, call them Bi for i in some index. So U = ∪ Bi, with all Bi open.
I'm thinking this follows from the definition that all such collections of {Bi} ⊂ τ, which implies that β ⊂ τ. (satisfying point 1)
Now Suppose there is some point p in U. Then it follows that p is in at least one such Bi, call it V. Therefore there exists a set V in β such that p ∈ V ⊂ U, satisfying point 2.

That sounds good to me. I'm a little fuzzy on what you are saying about condition 1. But note both sides of your 'if and only if' state that ##\beta \subset \tau## - so there is not much to prove there.
 

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