Difference between revisions of "Notes:Basis for a topology"

Revision as of 10:58, 26 June 2016

Problem

There's "basis for a topology" and "basis that generates a topology", the two are very similar constructs, and it is important to be able to move between them. This page is just to write down some concrete notes that prove any claims I may want to make.

Definitions

Basis for a topology

Given a topological space, [ilmath](X,\mathcal{J})[/ilmath] a basis for the space is a collection, [ilmath]\mathcal{B} [/ilmath] of subsets of [ilmath]X[/ilmath] such that:

1. [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{J}][/ilmath] - the elements of [ilmath]\mathcal{B} [/ilmath] are open in [ilmath]X[/ilmath].
2. [ilmath]\forall U\in\mathcal{J}\ \exists\{B_\alpha\}_{\alpha\in I}\subseteq\mathcal{B}[\bigcup_{\alpha\in I}B_\alpha=U][/ilmath]

Basis criterion

If we have a basis, [ilmath]\mathcal{B} [/ilmath] for a topological space [ilmath](X,\mathcal{ J })[/ilmath] then we can talk about open sets differently:

• A subset of [ilmath]X[/ilmath], [ilmath]U\in\mathcal{P}(X)[/ilmath], is open in [ilmath]X[/ilmath] if and only if [ilmath]\forall p\in U\exists B\in\mathcal{B}[p\in B\subseteq U][/ilmath]

Topology generated by a basis

Given a set, [ilmath]X[/ilmath] and a collection of subsets of [ilmath]X[/ilmath], [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] we call [ilmath]\mathcal{B} [/ilmath] a topological basis^{[Note 1]} or something!? if it satisfies:

1. [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath]^{[Note 2]} and
2. [ilmath]\forall B_1,B_2\in\mathcal{B}\ \forall x\in B_1\cap B_2\ \exists B_3\in\mathcal{B}[x\in B_3\subseteq B_1\cap B_2][/ilmath]^{[Note 3]}

Then [ilmath]\mathcal{B} [/ilmath] generates a unique topology on [ilmath]X[/ilmath]. This topology has [ilmath]\mathcal{B} [/ilmath] as a basis.

Alec's interpretation

We can use the Basis Criterion above to define the open sets:

Distilling the page

Let us make the following "artificial" definitions:[ilmath]\def\D#1{\mathrm{D} #1}[/ilmath]

• [ilmath]\mathrm{D}1[/ilmath] - Definition 1 - given a topological space [ilmath](X,\mathcal{ J })[/ilmath] we can define a basis, [ilmath]\mathcal{B} [/ilmath] as follows:
  1. [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{J}][/ilmath] - the elements of [ilmath]\mathcal{B} [/ilmath] are open in [ilmath]X[/ilmath].
  2. [ilmath]\forall U\in\mathcal{J}\ \exists\{B_\alpha\}_{\alpha\in I}\subseteq\mathcal{B}[\bigcup_{\alpha\in I}B_\alpha=U][/ilmath]
• [ilmath]\mathrm{D}2[/ilmath] - Definition 2 - so called "Basis Criterion"
  • Given a collection of subsets of a set, [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath], a subset, [ilmath]U[/ilmath], of the set [ilmath]X[/ilmath] is "[ilmath]\mathrm{D}2[/ilmath]" if and only if
    • [ilmath]\forall p\in U\exists B\in\mathcal{B}[p\in B\subseteq U][/ilmath]
• [ilmath]\mathrm{D}3[/ilmath] - Definition 3 - a system of subsets of a set [ilmath]X[/ilmath], [ilmath]\mathcal{D} [/ilmath] is called [ilmath]\mathrm{D}3[/ilmath] if:
  1. [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath]^{[Note 4]} and
  2. [ilmath]\forall B_1,B_2\in\mathcal{B}\ \forall x\in B_1\cap B_2\ \exists B_3\in\mathcal{B}[x\in B_3\subseteq B_1\cap B_2][/ilmath]

Then we can start saying "[ilmath]\D{1}\implies \D2[/ilmath] defines a topology" and such.

Workings

There are a few ways to go.

John M. Lee's path

1. Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and [ilmath]\mathcal{B} [/ilmath] as [ilmath]\D1[/ilmath] collection of sets, then:
   • [ilmath]U\in\mathcal{J} [/ilmath] if and only if [ilmath]U[/ilmath] is [ilmath]\D2[/ilmath]
2. Suppose [ilmath]\mathcal{B}'[/ilmath] is [ilmath]\D3[/ilmath] - then there is a unique topology on [ilmath]X[/ilmath] for which [ilmath]\mathcal{B}'[/ilmath] is [ilmath]\D1[/ilmath] on.
   • He sidesteps the proof of uniqueness.

Alec's first attempt

1. Suppose [ilmath]\mathcal{B} [/ilmath] is [ilmath]\D3[/ilmath], then the collection of all [ilmath]\D2[/ilmath] subsets is a topology on [ilmath]X[/ilmath], call this topology [ilmath]\mathcal{K} [/ilmath]
2. [ilmath]\mathcal{B} [/ilmath] is [ilmath]\D1[/ilmath] (wrt: [ilmath](X,\mathcal{ K })[/ilmath])
3. Corollary: A [ilmath]\D3[/ilmath] collection of subsets is a basis for the topology it generates
4. Suppose [ilmath]\mathcal{B}'[/ilmath] is [ilmath]\D1[/ilmath] (wrt: [ilmath](X,\mathcal{ J })[/ilmath]), then it is [ilmath]\D3[/ilmath].
5. The topology generated by [ilmath]\mathcal{B}'[/ilmath] is the same as [ilmath]\mathcal{J} [/ilmath].

These are all "easy", however to complete this we need:

• There is no other topology for which [ilmath]\mathcal{B}'[/ilmath] is a basis OR
• There is no other topology which can be generated by [ilmath]\mathcal{B} [/ilmath]

That is we still do not know uniqueness.

Notes

1. ↑ Not what Lee actually says, check this!
2. ↑ Lee actually says:
   • [ilmath]\bigcup_{B\in\mathcal{B} }B=X[/ilmath]
   Clearly I have [ilmath]\bigcup_{B\in\mathcal{B} }B\subseteq X[/ilmath] but I do not have equality.... hmm...
3. ↑ There are many abuses of notation here. Make sure they're clear and understood!
4. ↑ Dubious, actually require: [ilmath]\bigcup_{B\in\mathcal{B} }B=X[/ilmath]

Links

1. http://www.cmi.ac.in/~anirbit/topology.pdf - appears to deal with exactly my problem.

But strangely, appears to call a set "closed" whenever it is not open!