Locally Euclidean topological space of dimension $n$

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See Locally euclidean, it's the same thing but with $n$ fixed before the $\forall p\in x$ part. Alec (talk) 17:07, 19 February 2017 (UTC)

• Fleshed out a little bit but more work is needed Alec (talk) 12:47, 20 February 2017 (UTC)

Definition

Caveat:I think this $n$ might have to be unique as later (see topological manifold) we'll talk about the "well-defined-ness" of $n$!

Let $(X,\mathcal{ J })$ be a topological space and let $n\in\mathbb{N}_0$ be given. We say that $X$ is locally Euclidean of dimension $n$ if:

• $\forall p\in X\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists \varphi\in\mathcal{F}\big(U,B_\epsilon(0;\mathbb{R}^n)\big)\big[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)\big]$
• Caveat:Or perhaps...
  • $\exists n\in\mathbb{N}_0\forall p\in X\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists \varphi\in\mathcal{F}\big(U,B_\epsilon(0;\mathbb{R}^n)\big)\big[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)\big]$
• Where the dimension, $n$, is the $n$ 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 $\epsilon$ about $0\in\mathbb{R}^n$, it actually works if:

1. We require there to be any open set containing $p$ to be homeomorphic to any open set of $\mathbb{R}^n$
2. We require there be an open set containing $p$ homeomorphic to all of $\mathbb{R}^n$
3. We require there be an open set containing $p$ homeomorphic to the open unit ball, $\mathbb{B}^n$

See the Locally euclidean page for more information.