Topology generated by a basis
From Maths
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]}
Note that we could also say:
- Let [ilmath]\mathcal{B} [/ilmath] be a collection of sets, then [ilmath](\bigcup\mathcal{B},\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\})[/ilmath] is a topological space if and only if [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]
- This is just condition [ilmath]2[/ilmath] from above, clearly [ilmath]1[/ilmath] isn't needed as [ilmath]\bigcup\mathcal{B}=\bigcup\mathcal{B}[/ilmath] (obviously/trivially)
Proof
Grade: A
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
The message provided is:
This is actually an important proof
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