Difference between revisions of "Notes:Basis for a topology/Attempt 2"

Revision as of 09:19, 8 August 2016

Overview

Last time I tried to merge the definitions, which got very confusing! This time I shall treat them as separate things and go from there.

Definitions

Here we will use GBasis for a generated topology (by a basis) and TBasis for a basis of an existing topology.

GBasis

Let [ilmath]X[/ilmath] be a set and [ilmath]\mathcal{B}\subseteq\mathcal{P}(X)[/ilmath] be a collection of subsets of [ilmath]X[/ilmath]. Then we say:

• [ilmath]\mathcal{B} [/ilmath] is a GBasis if it satisfies the following 2 conditions:
  1. [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath] - every element of [ilmath]X[/ilmath] is contained in some GBasis set.
  2. [ilmath]\forall B_1,B_2\in\mathcal{B}\forall x\ B_1\cap B_2\exists B_3\in\mathcal{B}[B_1\cap B_2\ne\emptyset\implies(x\in B_3\subseteq B_1\cap B_2)][/ilmath]^{[Note 1][Note 2]}

Then [ilmath]\mathcal{B} [/ilmath] induces a topology on [ilmath]X[/ilmath].

Let [ilmath]\mathcal{J}_\text{Induced} [/ilmath] denote this topology, then:

• [ilmath]\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}_\text{Induced}\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\big][/ilmath]

TBasis

Suppose [ilmath](X,\mathcal{ J })[/ilmath] is a topological space and [ilmath]\mathcal{B}\subseteq\mathcal{P}(X)[/ilmath] is some collection of subsets of [ilmath]X[/ilmath]. We say:

• [ilmath]\mathcal{B} [/ilmath] is a TBasis if it satisfies both of the following:
  1. [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{J}][/ilmath] - all the basis elements are themselves open.
  2. [ilmath]\forall U\in\mathcal{J}\exists\{B_\alpha\}_{\alpha\in I}[\bigcup_{\alpha\in I}B_\alpha=U][/ilmath]

If we have a TBasis for a topological space then we may talk about its open sets differently:

• [ilmath]\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\big][/ilmath]

Facts

1. [ilmath]\mathcal{B} [/ilmath] is a GBasis of [ilmath]X[/ilmath] inducing [ilmath](X,\mathcal{J}')[/ilmath] [ilmath]\implies[/ilmath] [ilmath]\mathcal{B} [/ilmath] is a TBasis for the [ilmath](X,\mathcal{J}')[/ilmath]
2. [ilmath](X,\mathcal{ J })[/ilmath] is a topological space with a TBasis [ilmath]\mathcal{B} [/ilmath] [ilmath]\implies[/ilmath] [ilmath]\mathcal{B} [/ilmath] is a GBasis and it generates [ilmath](X,\mathcal{ J })[/ilmath]

Proof of facts

1. [ilmath]\mathcal{B} [/ilmath] is a GBasis of [ilmath]X[/ilmath] inducing [ilmath](X,\mathcal{J}')[/ilmath] [ilmath]\implies[/ilmath] [ilmath]\mathcal{B} [/ilmath] is a TBasis for the [ilmath](X,\mathcal{J}')[/ilmath]
   • Let [ilmath]\mathcal{B} [/ilmath] be a GBasis, suppose it generates the topological space [ilmath](X,\mathcal{J}')[/ilmath], we will show it's also a TBasis of [ilmath](X,\mathcal{J}')[/ilmath]
     1. [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{J}'][/ilmath] must be shown
        • Let [ilmath]B\in\mathcal{B} [/ilmath] be given.
          • Recall [ilmath]U\in\mathcal{J}'\iff\forall x\in U\exists B'\in\mathcal{B}[x\in B'\wedge B'\subseteq U][/ilmath]
          • Let [ilmath]x\in B[/ilmath] be given.
            • Choose [ilmath]B':=B[/ilmath]
              • [ilmath]x\in B'[/ilmath] by definition, as [ilmath]x\in B'=B[/ilmath] and
              • we have [ilmath]B\subseteq B[/ilmath], by the implies-subset relation, if and only if [ilmath]\forall p\in B[p\in B][/ilmath] which is trivial.
          • Thus [ilmath]B\in\mathcal{J}'[/ilmath]
        • Since [ilmath]B\in\mathcal{B} [/ilmath] was arbitrary we have shown [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{J}'][/ilmath]
     2. [ilmath]\forall U\in\mathcal{J}'\exists\{B_\alpha\}_{\alpha\in I}[\bigcup_{\alpha\in I}B_\alpha=U][/ilmath] must be shown
        • Let [ilmath]U\in\mathcal{J}'[/ilmath] be given.
          • Then, as [ilmath]\mathcal{B} [/ilmath] is a GBasis, by definition of the open sets generated: [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B\wedge B\subseteq U][/ilmath]
          • Let [ilmath]p\in U[/ilmath] be given
            • Define [ilmath]B_p[/ilmath] to be the [ilmath]B[/ilmath] that exists such that [ilmath]p\in B_p[/ilmath] and [ilmath]B_p\subseteq U[/ilmath]
          • We now have an .... thing, like a sequence but arbitrary, [ilmath](B_p)_{p\in U} [/ilmath] or [ilmath]\{B_p\}_{p\in U} [/ilmath] - I need to come down on a notation for this - such that:
            • [ilmath]\forall B_p\in(B_p)_{p\in U}[p\in B_p\wedge B_p\subseteq U][/ilmath]
          • We must now show [ilmath]\bigcup_{p\in U}B_p=U[/ilmath], we can do this by showing [ilmath]\bigcup_{p\in U}B_p\subseteq U[/ilmath] and [ilmath]\bigcup_{p\in U}B_p\supseteq U[/ilmath]
            1. [ilmath]\bigcup_{p\in U}B_p\subseteq U[/ilmath]
               • Using the union of subsets is a subset of the union we see:
                 • [ilmath]\bigcup_{p\in U}B_p\subseteq U][/ilmath]
            2. [ilmath]U\subseteq\bigcup_{p\in U}B_p[/ilmath]
               • Using the implies-subset relation we see: [ilmath]U\subseteq\bigcup_{p\in U}B_p\iff\forall x\in U[x\in\bigcup_{p\in U}B_p][/ilmath], we will show the RHS instead.
                 • Let [ilmath]x\in U[/ilmath] be given
                   • Recall, by definition of union, [ilmath]x\in\bigcup_{p\in U}B_p\iff\exists q\in U[x\in B_q][/ilmath]
                     • Choose [ilmath]q:=x[/ilmath] then we have [ilmath]x\in B_x[/ilmath] (as [ilmath]p\in B_p[/ilmath] is one of the defining conditions of choosing each [ilmath]B_p[/ilmath]!)
               • Thus [ilmath]U\subseteq\bigcup_{p\in U}B_p[/ilmath]
          • We have shown [ilmath]\bigcup_{p\in U}B_p=U[/ilmath]
        • Since [ilmath]U\in\mathcal{J}'[/ilmath] was arbitrary we have shown this for all [ilmath]U\in\mathcal{J}'[/ilmath]
     3. We have now shown [ilmath]\mathcal{B} [/ilmath] is a TBasis but not of what topology! Warning:Not finished!
   • This completes the proof.

Notes

1. ↑ Note that [ilmath]x\in B_3\subseteq B_1\cap B_2[/ilmath] is short for:
   • [ilmath]x\in B_3\wedge B_3\subseteq B_1\cap B_2[/ilmath]
2. ↑ Note that if [ilmath]B_1\cap B_2[/ilmath] is empty (they do not intersect) then the logical implication is true regardless of the RHS of the [ilmath]\implies[/ilmath]} sign, so we do not care if we have [ilmath]x\in B_3\wedge B_3\subseteq B_1\cap B_2[/ilmath]! Pick any [ilmath]x\in X[/ilmath] and aany [ilmath]B_3\in\mathcal{B} [/ilmath]!