Difference between revisions of "Topology generated by a basis/Statement"
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'''{{iff}}''' | '''{{iff}}''' | ||
* we have both of the following conditions: | * we have both of the following conditions: | ||
| − | *# {{M|1=\bigcup\mathcal{B}=X}} (or equivalently: {{M|1=\forall x\in X\exists B\in\mathcal{B}[x\in B]}}) ''and'' | + | *# {{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:'' |
| − | *# {{M| | + | * {{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}}. |
| + | 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]}}. | ||
| + | * 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'' | ||
| + | *# {{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> | ||
| + | *#* {{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'': | ||
| + | * {{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]}} | ||
| + | ** '''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> | ||
<noinclude> | <noinclude> | ||
==Notes== | ==Notes== | ||
Latest revision as of 21:59, 15 January 2017
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Contents
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]
- we have both of the following conditions:
- [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
- [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
- ↑ 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].
- We then use Union of subsets is a subset of the union (with [ilmath]B_\alpha:\eq X[/ilmath]) to see that [ilmath]\bigcup\mathcal{B}\subseteq X[/ilmath] - as required.
- ↑ 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]
- ↑ 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)
- [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]
References
