Difference between revisions of "Locally Euclidean topological space of dimension n"
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==Definition== | ==Definition== | ||
| + | : {{Caveat|I think this {{M|n}} might have to be unique}} as later (see [[topological manifold]]) we'll talk about the "well-defined-ness" of {{M|n}}!<ref group="Note">As usual, [[well-defined-ness]] means we have an [[equivalence relation]] in play, and we're [[quotient by an equivalence relation|quotienting something]]. I'm not quite sure what yet though! | ||
| + | * I would have thought that a "locally euclidean of dimension n" space is really just something such that there exists an {{N}} for all points... | ||
| + | {{XXX|Solve this}}</ref> | ||
Let {{Top.|X|J}} be a [[topological space]] and let {{M|n\in\mathbb{N}_0}} be given. We say that {{M|X}} is ''locally Euclidean of dimension {{M|n}}'' if: | Let {{Top.|X|J}} be a [[topological space]] and let {{M|n\in\mathbb{N}_0}} be given. We say that {{M|X}} is ''locally Euclidean of dimension {{M|n}}'' if: | ||
* {{M|\forall p\in X\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists \varphi\in\mathcal{F}\big(U,B_\epsilon(0;\mathbb{R}^n)\big)\big[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)\big]}} | * {{M|\forall p\in X\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists \varphi\in\mathcal{F}\big(U,B_\epsilon(0;\mathbb{R}^n)\big)\big[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)\big]}} | ||
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# We require there be an open set containing {{M|p}} homeomorphic to the [[open unit ball]], {{M|\mathbb{B}^n}} | # We require there be an open set containing {{M|p}} homeomorphic to the [[open unit ball]], {{M|\mathbb{B}^n}} | ||
See the [[Locally euclidean]] page for more information. | See the [[Locally euclidean]] page for more information. | ||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Definition|Topology|Manifolds|Smooth Manifolds|Topological Manifolds}} | {{Definition|Topology|Manifolds|Smooth Manifolds|Topological Manifolds}} | ||
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See Locally euclidean, it's the same thing but with [ilmath]n[/ilmath] fixed before the [ilmath]\forall p\in x[/ilmath] part. Alec (talk) 17:07, 19 February 2017 (UTC)
Definition
- Caveat:I think this [ilmath]n[/ilmath] might have to be unique as later (see topological manifold) we'll talk about the "well-defined-ness" of [ilmath]n[/ilmath]![Note 1]
Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]n\in\mathbb{N}_0[/ilmath] be given. We say that [ilmath]X[/ilmath] is locally Euclidean of dimension [ilmath]n[/ilmath] if:
- [ilmath]\forall p\in X\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists \varphi\in\mathcal{F}\big(U,B_\epsilon(0;\mathbb{R}^n)\big)\big[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)\big][/ilmath]
- Caveat:Or perhaps...
- [ilmath]\exists n\in\mathbb{N}_0\forall p\in X\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists \varphi\in\mathcal{F}\big(U,B_\epsilon(0;\mathbb{R}^n)\big)\big[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)\big][/ilmath]
- Where the dimension, [ilmath]n[/ilmath], is the [ilmath]n[/ilmath] that must exist in the first quantifying clause.
TODO: Verdict needed after investigation
Equivalent definitions
We posit that there must be an open ball of radius [ilmath]\epsilon[/ilmath] about [ilmath]0\in\mathbb{R}^n[/ilmath], it actually works if:
- We require there to be any open set containing [ilmath]p[/ilmath] to be homeomorphic to any open set of [ilmath]\mathbb{R}^n[/ilmath]
- We require there be an open set containing [ilmath]p[/ilmath] homeomorphic to all of [ilmath]\mathbb{R}^n[/ilmath]
- We require there be an open set containing [ilmath]p[/ilmath] homeomorphic to the open unit ball, [ilmath]\mathbb{B}^n[/ilmath]
See the Locally euclidean page for more information.
Notes
- ↑ As usual, well-defined-ness means we have an equivalence relation in play, and we're quotienting something. I'm not quite sure what yet though!
- I would have thought that a "locally euclidean of dimension n" space is really just something such that there exists an [ilmath]n[/ilmath] for all points...
TODO: Solve this
