Difference between revisions of "Topology generated by a basis/Statement"

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(Created page with "<noinclude> {{Requires references|grade=A|msg=I could do this now but I can't be bothered!}} __TOC__ ==Statement== </noinclude>Let {{M|X}} be a set and let {{M|\mathcal{B}...")
 
(Removed wrong version, added warning and note explaining.)
 
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'''{{iff}}'''
 
'''{{iff}}'''
 
* we have both of the following conditions:
 
* we have both of the following conditions:
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*# {{M|1=\bigcup\mathcal{B}=X}} (or equivalently: {{M|1=\forall x\in X\exists B\in\mathcal{B}[x\in B]}}) ''and''
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*# {{M|1=\bigcup\mathcal{B}=X}} (or equivalently: {{M|1=\forall x\in X\exists B\in\mathcal{B}[x\in B]}}<ref group="Note">By the [[implies-subset relation]] {{M|1=\forall x\in X\exists B\in\mathcal{B}[x\in B]}} really means {{M|X\subseteq\bigcup\mathcal{B} }}, as we only require that all elements of {{M|X}} be in the union. Not that all elements of the union are in {{M|X}}. ''However:''
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*# {{M|1=\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V]}}<ref group="Note">We could of course write:
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* {{M|\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))}} by definition. So clearly (or after some thought) the reader should be happy that {{M|\mathcal{B} }} contains only subsets of {{M|X}} and he should see that we cannot as a result have an element in one of these subsets that is not in {{M|X}}.
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Thus {{M|\forall B\in\mathcal{B}[B\in\mathcal{P}(X)]}} which is the same as (by [[power-set]] and [[subset of|subset]] definitions) {{M|\forall B\in\mathcal{B}[B\subseteq X]}}.
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* We then use [[Union of subsets is a subset of the union]] (with {{M|B_\alpha:\eq X}}) to see that {{M|\bigcup\mathcal{B}\subseteq X}} - as required.</ref>) ''and''
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*# {{M|\forall U,V\in\mathcal{B}\big[U\cap V\neq\emptyset\implies \forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]\big]}}<ref group="Note">We could of course write:
 
* {{M|1=\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)]}}</ref>
 
* {{M|1=\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)]}}</ref>
−
Note that we could also say:
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*#* {{Caveat|{{M|1=\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V]}} is commonly said or written; however it is wrong}}, this is slightly ''beyond'' just abuse of notation.<ref group="Note">Suppose that {{M|U,V\in\mathcal{B} }} are given but disjoint, then there are no {{M|x\in U\cap V}} to speak of, and {{M|x\in W}} may be vacuously satisfied by the absence of an {{M|X}}, ''however'':
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* Let {{M|\mathcal{B} }} be a collection of [[sets]], then {{M|(\bigcup\mathcal{B},\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\})}} is a [[topological space]] {{iff}} {{M|1=\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V]}}
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* {{M|x\in W\subseteq U\cap V}} is taken to mean {{M|x\in W}} [[logical and|and]] {{M|W\subseteq U\cap V}}, so we must still show {{M|\exists W\in\mathcal{B}[W\subseteq U\cap V]}}
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** This is just condition {{M|2}} from above, clearly {{M|1}} isn't needed as {{M|1=\bigcup\mathcal{B}=\bigcup\mathcal{B} }} (obviously/trivially)<noinclude>
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** '''This is not always possible as {{M|W}} would have to be [[empty set|{{M|\emptyset}}]] for this to hold!''' We do not require {{M|\emptyset\in\mathcal{B} }} (as for example in the [[metric topology]])</ref>
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<noinclude>
 
==Notes==
 
==Notes==
 
<references group="Note"/>
 
<references group="Note"/>

Latest revision as of 21:59, 15 January 2017

Grade: A
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Statement

Let [ilmath]X[/ilmath] be a set and let [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] be any collection of subsets of [ilmath]X[/ilmath], then:

  • [ilmath](X,\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\})[/ilmath] is a topological space with [ilmath]\mathcal{B} [/ilmath] being a basis for the topology [ilmath]\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\}[/ilmath]

if and only if

  • we have both of the following conditions:
    1. [ilmath]\bigcup\mathcal{B}=X[/ilmath] (or equivalently: [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath][Note 1]) and
    2. [ilmath]\forall U,V\in\mathcal{B}\big[U\cap V\neq\emptyset\implies \forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]\big][/ilmath][Note 2]
      • Caveat:[ilmath]\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V][/ilmath] is commonly said or written; however it is wrong, this is slightly beyond just abuse of notation.[Note 3]

Notes

  1. ↑ By the implies-subset relation [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath] really means [ilmath]X\subseteq\bigcup\mathcal{B} [/ilmath], as we only require that all elements of [ilmath]X[/ilmath] be in the union. Not that all elements of the union are in [ilmath]X[/ilmath]. However:
    • [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] by definition. So clearly (or after some thought) the reader should be happy that [ilmath]\mathcal{B} [/ilmath] contains only subsets of [ilmath]X[/ilmath] and he should see that we cannot as a result have an element in one of these subsets that is not in [ilmath]X[/ilmath].
    Thus [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{P}(X)][/ilmath] which is the same as (by power-set and subset definitions) [ilmath]\forall B\in\mathcal{B}[B\subseteq X][/ilmath].
  2. ↑ We could of course write:
    • [ilmath]\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)][/ilmath]
  3. ↑ Suppose that [ilmath]U,V\in\mathcal{B} [/ilmath] are given but disjoint, then there are no [ilmath]x\in U\cap V[/ilmath] to speak of, and [ilmath]x\in W[/ilmath] may be vacuously satisfied by the absence of an [ilmath]X[/ilmath], however:
    • [ilmath]x\in W\subseteq U\cap V[/ilmath] is taken to mean [ilmath]x\in W[/ilmath] and [ilmath]W\subseteq U\cap V[/ilmath], so we must still show [ilmath]\exists W\in\mathcal{B}[W\subseteq U\cap V][/ilmath]
      • This is not always possible as [ilmath]W[/ilmath] would have to be [ilmath]\emptyset[/ilmath] for this to hold! We do not require [ilmath]\emptyset\in\mathcal{B} [/ilmath] (as for example in the metric topology)

References


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