Difference between revisions of "Basis for a topology"
From Maths
(Created page with "==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="...") |
m (Missed something silly) |
||
| Line 12: | Line 12: | ||
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: | 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: | * 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.{{ | + | ** {{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}} | {{Begin Theorem}} | ||
Claim: This {{M|\mathcal{(J)} }} is indeed a topology | Claim: This {{M|\mathcal{(J)} }} is indeed a topology | ||
Revision as of 16:43, 20 March 2016
Contents
Definition
Let [ilmath]X[/ilmath] be a set. A basis for a topology on [ilmath]X[/ilmath] is a collection of subsets of [ilmath]X[/ilmath], [ilmath]\mathcal{B}\subseteq\mathcal{P}(X)[/ilmath] such that[1]:
- [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath] - every element of [ilmath]X[/ilmath] belongs to at least one basis element.
- [ilmath]\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)][/ilmath][Note 1] - 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 [ilmath]\mathcal{B} [/ilmath] are called basis elements[1]
Topology generated by [ilmath]\mathcal{B} [/ilmath]
If [ilmath]\mathcal{B} [/ilmath] is such a basis for [ilmath]X[/ilmath], we define the topology [ilmath]\mathcal{J} [/ilmath] generated by [ilmath]\mathcal{B} [/ilmath][1] as follows:
- A subset of [ilmath]X[/ilmath], [ilmath]U\subseteq X[/ilmath] is considered open (equivalently, [ilmath]U\in\mathcal{J} [/ilmath]) if:
- [ilmath]\forall x\in U\exists B\in\mathcal{B}[x\in B\wedge B\subseteq U][/ilmath][Note 2]
Claim: This [ilmath]\mathcal{(J)} [/ilmath] is indeed a topology
TODO: Do this, see page 81 in Munkres - shouldn't be hard!
See also
Notes
- ↑ This is a great example of a hiding if-and-only-if, note that:
- [ilmath](x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2[/ilmath] (by the implies-subset relation) so we have:
- [ilmath](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)[/ilmath]
- Thus [ilmath](x\in B_3\wedge B_3\subseteq B_1\cap B_2)\iff x\in B_1\cap B_2[/ilmath]
- [ilmath](x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2[/ilmath] (by the implies-subset relation) so we have:
- ↑ Note that each basis element is itself is open. This is because [ilmath]U[/ilmath] is considered open if forall x, there is a basis element containing [ilmath]x[/ilmath] with that basis element [ilmath]\subseteq U[/ilmath], if [ilmath]U[/ilmath] is itself a basis element, it clearly satisfies this as [ilmath]B\subseteq B[/ilmath]
TODO: Make this into a claim
References
