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Topological Basis Homework: If-Then Conditions
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[QUOTE="Numberphile, post: 5845191, member: 610206"] [h2]Homework Statement [/h2] 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. [B] 2. Relevant definitions[/B] 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. [h2]The Attempt at a Solution[/h2] 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 V[SUB]p[/SUB] in β such that p ∈ V[SUB]p[/SUB] ⊂ U. Consider the union ∪ V[SUB]p[/SUB]. Every point p in U is also in the union ∪ V[SUB]p[/SUB]. So it follows that U ⊂ ∪ V[SUB]p[/SUB]. Notice every point p in the union ∪ V[SUB]p[/SUB] is also in U. So it follows that ∪ V[SUB]p[/SUB] ⊂ U. This shows U = ∪ V[SUB]p[/SUB], 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 B[SUB]i[/SUB] for i in some index. So U = ∪ B[SUB]i[/SUB], with all B[SUB]i[/SUB] open. I'm thinking this follows from the definition that all such collections of {B[SUB]i[/SUB]} ⊂ 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 B[SUB]i[/SUB], call it V. Therefore there exists a set V in β such that p ∈ V ⊂ U, satisfying point 2. [/QUOTE]
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Topological Basis Homework: If-Then Conditions
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