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|\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|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: | *# {{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: | ||
* {{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> | ||
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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:
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]
References
