conditions for a collection of subsets to be a basis for some topology
Not just any collection of subsets of X can be a basis for a topology
on X. For instance, if we took π to be all open intervals of length 1 in β, π isnβt the basis for any topology on β: (0,1) and (.5,1.5) are unions of elements of π, but their intersection
(.5,1) is not. The collection formed by arbitrary unions of members of π isnβt closed under finite intersections and isnβt a topology.
Weβd like to know which collections β¬ of subsets of X could be the basis for some topology on X. Hereβs the result:
Theorem.
Proof.
First, weβll show that if β¬ is the basis for some topology π― on X, then it satisfies the two conditions listed.
π― is a topology on X, so Xβπ―. Since β¬ is a basis for π―, that means X can be written as a union of members of β¬: since every xβX is in this union, every xβX is contained in some member of β¬. That takes care of the first condition.
For the second condition: if B1 and B2 are elements of β¬, theyβre also in π―. π― is closed under intersection, so B1β©B2 is open in π―. Then B1β©B2 can be written as a union of members of β¬, and any xβB1β©B2 is contained by some basis element in this union.
Second, weβll show that if a collection β¬ of subsets of X satisfies the two conditions, then the collection π― of unions of members of β¬ is a topology on X.
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β’
β βπ―: β is the null union of zero elements of β¬.
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β’
Xβπ―: by the first condition, every X is contained in some member of β¬. The union of all the members of β¬ is then all of X.
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β’
π― is closed under arbitrary unions: Say we have a union of sets TΞ±βπ―β¦
βΞ±βITΞ± =βΞ±βIβΞ²βJΞ±BΞ² (since each TΞ± is a union of sets in β¬) =βΞ²ββΞ±βIJΞ±BΞ² Since thatβs a union of elements of β¬, itβs also a member of π―.
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β’
π― is closed under finite intersections: since a collection of sets is closed under finite intersections if and only if it is closed under pairwise intersections, we need only check that the intersection of two members T1,T2 of π― is in π―.
Any xβT1β©T2 is contained in some B1xβT1 and B2xβT2. By the second condition, xβB1xβ©B2x gets us a B3x with xβB3xβB1xβ©B2xβT1β©T2. Then
T1β©T2=βxβT1β©T2B3x which is in π―.
β
Title | conditions for a collection of subsets to be a basis for some topology |
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Canonical name | ConditionsForACollectionOfSubsetsToBeABasisForSomeTopology |
Date of creation | 2013-03-22 14:21:49 |
Last modified on | 2013-03-22 14:21:49 |
Owner | waj (4416) |
Last modified by | waj (4416) |
Numerical id | 4 |
Author | waj (4416) |
Entry type | Proof |
Classification | msc 54A99 |
Classification | msc 54D70 |