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

Revision as of 17:29, 5 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]

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]!