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 n!
Contents
[hide]Definition
Let (X,J) be a topological space, we say it is locally Euclidean if:
- ∀p∈X∃n∈N0∃U∈O(p;X)∃V∈O(Rn)∃φ∈F(U,V)[U≅φV]⏟∃homeomorphism φ:U→some open subset of Rn
- In words: for all points, p∈X, there is an n such that there is an open neighbourhood of p which is homeomorphic to some open set in Rn for the given {{M|n}].
Equivalent definitions
Some open ball at the origin
Claim:
- (∀p∈X∃n∈N0∃U∈O(p;X)∃V∈O(Rn)∃φ∈F(U,V)[U≅φV]) ⟺ (∀p∈X∃n∈N0∃U∈O(p;X)∃ϵ∈R>0∃φ∈F(U,Bϵ(0;Rn)[U≅φBϵ(0;Rn)])
Proof:
⟹
- Let p∈X be given
- Choose n:=n′ where n′ is the n∈N0 posited to exist by the LHS of the ⟺
- We now obtain:
- U′∈O(p;X) posited to exist by the LHS,
- V′∈O(Rn) posited to exist by the LHS, such that:
- φ′:U′→V′ posited to exist by the LHS is a homeomorphism.
- V′∈O(Rn) posited to exist by the LHS, such that:
- U′∈O(p;X) 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:
- ∃δ∈R>0[Bδ(φ′(p);Rn)⊆V′]
- As open balls are open sets and "the pre-image of an open set under a homeomorphism is open" we see:
- φ′−1(Bδ(φ′(p);Rn)) is open in X
- We must now show that p∈φ′−1(Bδ(φ′(p);Rn)) (so we can say φ′−1(Bδ(φ′(p);Rn))∈O(p,X) shortly)
- Choose: U:=φ′−1(Bδ(φ′(p);Rn)), by the discussion above we see U∈O(p,X)
- Choose: ϵ=δ, we found δ above, recall δ∈R>0, so obviously ϵ∈R>0
- We now obtain:
- Choose n:=n′ where n′ is the n∈N0 posited to exist by the LHS of the ⟺
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The bulk of the proof is done here, the "not trivial part" is that φ′ restricts to a homeomorphism onto Bδ(φ′(p);Rn), 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
