Difference between revisions of "Locally Euclidean topological space"
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*** Choose: {{M|U:\eq \varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)}}, by the discussion above we see {{M|U\in\mathcal{O}(p,X)}} | *** Choose: {{M|U:\eq \varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)}}, by the discussion above we see {{M|U\in\mathcal{O}(p,X)}} | ||
**** Choose: {{M|\epsilon\eq\delta}}, we found {{M|\delta}} above, recall {{M|\delta\in\mathbb{R}_{>0} }}, so obviously {{M|\epsilon\in\mathbb{R}_{>0} }} | **** Choose: {{M|\epsilon\eq\delta}}, we found {{M|\delta}} above, recall {{M|\delta\in\mathbb{R}_{>0} }}, so obviously {{M|\epsilon\in\mathbb{R}_{>0} }} | ||
| − | {{Requires work|grade=A|msg=The bulk of the proof is done here, the "not trivial part" is that {{M|\varphi'}} restricts to a homeomorphism onto {{M|B_\delta(\varphi'(p);\mathbb{R}^n)}}, then compose that with a translation, we use [[translations are homeomorphisms]] and we're basically done. [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 16:55, 19 February 2017 (UTC)}} | + | {{Requires work|grade=A|msg=The bulk of the proof is done here, the "not trivial part" is that {{M|\varphi'}} restricts to a homeomorphism onto {{M|B_\delta(\varphi'(p);\mathbb{R}^n)}}, then compose that with a translation, we use [[translations are homeomorphisms]] and we're basically done. [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 16:55, 19 February 2017 (UTC) |
| + | * "''[[Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself]]''" will be needed and is ready}} | ||
=References= | =References= | ||
<references/> | <references/> | ||
{{Definition|Topology|Manifolds|Topological Manifolds|Smooth Manifolds}} | {{Definition|Topology|Manifolds|Topological Manifolds|Smooth Manifolds}} | ||
Latest revision as of 14:32, 20 February 2017
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- Caveat:Need to do locally euclidean of dimension [ilmath]n[/ilmath]!
Contents
Definition
Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space, we say it is locally Euclidean if:
- [ilmath]\forall p\in X\exists n\in\mathbb{N}_0\exists U\in\mathcal{O}(p;X)\underbrace{\exists V\in\mathcal{O}(\mathbb{R}^n)\exists\varphi\in\mathcal{F}(U,V)[U\cong_\varphi V]}_{\exists\text{homeomorphism }\varphi:U\rightarrow\text{some open subset of }\mathbb{R}^n } [/ilmath]
- In words: for all points, [ilmath]p\in X[/ilmath], there is an [ilmath]n[/ilmath] such that there is an open neighbourhood of [ilmath]p[/ilmath] which is homeomorphic to some open set in [ilmath]\mathbb{R}^n[/ilmath] for the given {{M|n}].
Equivalent definitions
Some open ball at the origin
Claim:
- [ilmath]\Big(\forall p\in X\exists n\in\mathbb{N}_0\exists U\in\mathcal{O}(p;X)\exists V\in\mathcal{O}(\mathbb{R}^n)\exists\varphi\in\mathcal{F}(U,V)[U\cong_\varphi V]\Big)[/ilmath] [ilmath]\iff[/ilmath] [ilmath]\Big(\forall p\in X\exists n\in\mathbb{N}_0\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists\varphi\in\mathcal{F}(U,B_\epsilon(0;\mathbb{R}^n)[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)]\Big)[/ilmath]
Proof:
[ilmath]\implies[/ilmath]
- Let [ilmath]p\in X[/ilmath] be given
- Choose [ilmath]n:\eq n'[/ilmath] where [ilmath]n'[/ilmath] is the [ilmath]n\in\mathbb{N}_0[/ilmath] posited to exist by the LHS of the [ilmath]\iff[/ilmath]
- We now obtain:
- [ilmath]U'\in\mathcal{O}(p;X)[/ilmath] posited to exist by the LHS,
- [ilmath]V'\in\mathcal{O}(\mathbb{R}^n)[/ilmath] posited to exist by the LHS, such that:
- [ilmath]\varphi':U'\rightarrow V'[/ilmath] posited to exist by the LHS is a homeomorphism.
- [ilmath]V'\in\mathcal{O}(\mathbb{R}^n)[/ilmath] posited to exist by the LHS, such that:
- [ilmath]U'\in\mathcal{O}(p;X)[/ilmath] posited to exist by the LHS,
- Recall that "an open set in a metric space contains an open ball about all of its points", this means:
- [ilmath]\exists\delta\in\mathbb{R}_{>0}[B_\delta(\varphi'(p);\mathbb{R}^n)\subseteq V'][/ilmath]
- As open balls are open sets and "the pre-image of an open set under a homeomorphism is open" we see:
- [ilmath]\varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)[/ilmath] is open in [ilmath]X[/ilmath]
- We must now show that [ilmath]p\in\varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)[/ilmath] (so we can say [ilmath]\varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)\in\mathcal{O}(p,X)[/ilmath] shortly)
- Note that [ilmath]f(p)\in B_\delta(\varphi'(p);\mathbb{R}^n)[/ilmath] as [ilmath]d(\varphi'(p),\varphi'(p)):\eq 0[/ilmath] regardless of the metric used
- As [ilmath]0<\delta[/ilmath] we see [ilmath]\varphi'(p)\in B_\delta(\varphi'(p);\mathbb{R}^n)[/ilmath]
- Thus [ilmath]p\in\varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)[/ilmath] - by the definition of pre-image
- As [ilmath]0<\delta[/ilmath] we see [ilmath]\varphi'(p)\in B_\delta(\varphi'(p);\mathbb{R}^n)[/ilmath]
- Note that [ilmath]f(p)\in B_\delta(\varphi'(p);\mathbb{R}^n)[/ilmath] as [ilmath]d(\varphi'(p),\varphi'(p)):\eq 0[/ilmath] regardless of the metric used
- Choose: [ilmath]U:\eq \varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)[/ilmath], by the discussion above we see [ilmath]U\in\mathcal{O}(p,X)[/ilmath]
- Choose: [ilmath]\epsilon\eq\delta[/ilmath], we found [ilmath]\delta[/ilmath] above, recall [ilmath]\delta\in\mathbb{R}_{>0} [/ilmath], so obviously [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath]
- We now obtain:
- Choose [ilmath]n:\eq n'[/ilmath] where [ilmath]n'[/ilmath] is the [ilmath]n\in\mathbb{N}_0[/ilmath] posited to exist by the LHS of the [ilmath]\iff[/ilmath]
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The bulk of the proof is done here, the "not trivial part" is that [ilmath]\varphi'[/ilmath] restricts to a homeomorphism onto [ilmath]B_\delta(\varphi'(p);\mathbb{R}^n)[/ilmath], then compose that with a translation, we use translations are homeomorphisms and we're basically done. Alec (talk) 16:55, 19 February 2017 (UTC)
- "Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself" will be needed and is ready
