View source for Basis for a topology

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==Definition==
Let {{Top.|X|J}} be a [[topological space]] and let {{M|\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))}} be any collection of subsets of {{M|X}}<ref group="Note">We could say something else instead of {{M|\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))}}:
* Let {{M|\mathcal{B}\in\mathcal{P}(\mathcal{J})}} - so {{M|\mathcal{B} }} is explicitly a collection of [[open sets]], then we could drop condition {{M|1}}. Or!
* Let {{M|\mathcal{B}\subseteq\mathcal{J} }}. But it is our convention to ''not'' say "let {{M|A\subseteq B}}" but "let {{M|A\in\mathcal{P}(B)}}" instead. To emphasise that the [[power-set]] is possibly in play.
We do not do these because it (sort of) violates the [[Doctrine of Least Surprise]], we usually deal with subsets of the ''space'' not subsets of the ''[[set system]]'' on that space.<br/>
That is a weird way of saying if we have a structure (eg [[topological space]], [[measurable space]], so forth) say {{M|(A,\mathcal{B})}} we usually deal with (collections of) subsets of {{M|A}} and specify they must be in {{M|\mathcal{B} }}.</ref>. We say {{M|\mathcal{B} }} is ''a basis for the [[topology]] {{M|\mathcal{J} }}'' if both of the following are satisfied:
# {{M|1=\forall B\in\mathcal{B}[B\in\mathcal{J}]}} 
# {{M|1=\forall U\in\mathcal{J}\exists\{B_\alpha\}_{\alpha\in I}\subseteq\mathcal{B}[\bigcup_{\alpha\in I}B_\alpha=U]}}
=={{link|Basis criterion|topology}}==
{{:The basis criterion (topology)/Statement}}
==[[Topology generated by a basis]]==
{{:Topology generated by a basis/Statement}}
==See also==
* [[Sub-basis for a topology]]

==Notes==
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==References==
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{{Definition|Topology}}
=OLD PAGE=
==Definition==
Let {{M|X}} be a set. A ''basis'' for a topology on {{M|X}} is a collection of subsets of {{M|X}}, {{M|\mathcal{B}\subseteq\mathcal{P}(X)}} such that<ref name="Top">Topology - Second Edition - James R. Munkres</ref>:
# {{M|\forall x\in X\exists B\in\mathcal{B}[x\in B]}} - every element of {{M|X}} belongs to at least one basis element.
# {{M|\forall B_1,B_2\in\mathcal{B},x\in X\ \exists B_3\in\mathcal{B}[x\in B_1\cap B_2\implies(x\in B_3\wedge B_3\subseteq B_1\cap B_2)]}}<ref group="Note">This is a great example of a hiding if-and-only-if, note that:
* {{M|(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2}} (by the [[implies-subset relation]]) so we have:
** {{M|(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2\implies(x\in B_3\wedge B_3\subseteq B_1\cap B_2)}}
* Thus {{M|(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\iff x\in B_1\cap B_2}}
This pattern occurs a lot, like with the axiom of extensionality in set theory. </ref> - if any 2 basis elements have non empty intersection, there is a basis element within that intersection containing each point in it.
Note that:
* The elements of {{M|\mathcal{B} }} are called ''basis elements''<ref name="Top"/>
==Topology generated by {{M|\mathcal{B} }}==
If {{M|\mathcal{B} }} is such a basis for {{M|X}}, we define the ''topology {{M|\mathcal{J} }} generated by {{M|\mathcal{B} }}''<ref name="Top"/> as follows:
* A subset of {{M|X}}, {{M|U\subseteq X}} is considered open (equivalently, {{M|U\in\mathcal{J} }}) if:
** {{M|1=\forall x\in U\exists B\in\mathcal{B}[x\in B\wedge B\subseteq U]}}<ref group="Note">Note that each basis element is itself is open. This is because {{M|U}} is considered open if forall x, there is a basis element containing {{M|x}} with that basis element {{M|\subseteq U}}, if {{M|U}} is itself a basis element, it clearly satisfies this as {{M|B\subseteq B}} {{Todo|Make this into a claim}}</ref>
{{Begin Theorem}}
Claim: This {{M|\mathcal{(J)} }} is indeed a topology
{{Begin Proof}}
{{Todo|Do this, see page 81 in Munkres - shouldn't be hard!}}
{{End Proof}}{{End Theorem}}
==See also==
* [[Topological space]]
* [[Sub-basis for a topology]]
* [[Common topological constructs]]
==Notes==
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==References==
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{{Definition|Topology}}
{{Theorem Of|Topology}}