Definition of a manifold NOT A DUPLICATE: Homeomorphism in the definition of a manifold for example is slightly different.
A manifold according to Wikipedia, a book by Spivak, and serveral other books has the following in common:

$\forall x\in M\exists n\in\mathbb{N}_{\ge 0}\exists U$ neighbourhood of $x$ such that $U$ is homeomorphic to (an open subset) of $\mathbb{R}^n$

I don't like the "neighbourhood" part as:

A set $U\subseteq M$ is a neighbourhood to $x$ if $\exists V$ open in $M$ with $[x\in V\wedge V\subseteq U]$

There is no requirement for $U$ to be open. It could be closed!
Problem:

Suppose that $U$ is homeomorphic to $\mathbb{R}^n$ by a function $f:U\rightarrow\mathbb{R}^n$, we know that $f$ is bijective and continuous by definition. This means it is surjective. Thus $f^{-1}(\mathbb{R}^n)=U$
By continuity of $f$ this means that $U$ is open in $M$

This is a contradiction, as $U$ need not be open.
I would be much happier if the definition was "there exists an open set containing $x$ that is homeomorphic to $\mathbb{R}^n$

The open subset of $\mathbb{R}^n$ part
The definition requires there exists a neighbourhood (not all neighbourhoods) homeomorphic to $\mathbb{R}^n$ is this the same as requiring the neighbourhood be homeomorphic to an open subset of $\mathbb{R}^n$?
Thoughts:

I understand that any open interval (in $\mathbb{R}$) is homeomorphic to all of $\mathbb{R}$ however the union of two distinct intervals is open but not homeomorphic to all of $\mathbb{R}$, using this sort of logic it suggests that:
I require (probably through the Hausdorff property) the ability to find a small enough connected (I suspect) open set. Then the two would be equivalent.

I could prove this if I assume the manifold has a countable topological basis (because then it is metricisable and I can use open balls) but I'd like to prove it for all manifolds.
 A: The problem in the box you've labelled "Problem" is not a problem: the chart $f$ is required to be continuous on $U$, but not on $M$, so $f^{-1}(\mathbb{R}^n)$ could be closed in $M$ (although it will always be open too). The topological subspace $\{0, 1\} \times (0, 1)$ of $\mathbb{R}^2$ is a $1$-manifold in which every point has a neighbourhood that is homeomorphic with $\mathbb{R}$ and is both open and closed.
It doesn't make any essential difference in the definition of an $n$-manifold whether you require the domains of the charts to be neighbourhoods or open neighbourhoods or whether you require the ranges of the charts to be open subsets of $\mathbb{R}^n$ or the whole of $\mathbb{R}^n$: any point in an open subset of $\mathbb{R}^n$ has a neigbourhood that is homeomorphic with the whole of $\mathbb{R}^n$; this means that if $M$ is a topological space and $x$ is a point of $M$, then $x$ has a neighbourhood homeomorphic with $\mathbb{R}^n$ iff for some open $V \subseteq \mathbb{R}^n$, $x$ has a neighbourhood homeomorphic with $V$.
A: For a topological manifold, $M$ with topology $\mathcal{J}$
I've done half of the question. I have shown that:

$\forall x\in M\exists V\text{ neighbourhood of }x\exists n\in\mathbb{N}_{\ge0}[x\in V\wedge V\cong\mathbb{R}^n]$
  $$\iff$$
  $\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}[x\in U\wedge U\cong\mathbb{R}^n]$

Which is the first half done.
I will write the proof in later, I'm doing other things right now, but this is proved, no handwaving - formally proved!

The second part is that:

$\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}\exists V[V\in\mathcal{O}(\mathbb{R}^n)\wedge x\in U\wedge U\cong V]$
  $$\iff$$
  $\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}[x\in U\wedge U\cong\mathbb{R}^n]$

I've outlined this but not completed it. 

With these two facts a manifold is a manifold regardless of:


*

*which notion of neighbourhood you use

*What you demand a neighbourhood is homeomorphic to


I will write the proof in later, I've done one on paper
