Is the closure of an open connected subset of $\mathbb{R}^{n}$ a topological manifold?

Question

If we remove the connectedness restriction, there are easy counter examples, as in: $\left(\frac{1}{2}, \frac{1}{1}\right) \cup \left(\frac{1}{4}, \frac{1}{3 }\right) \cup \left(\frac{1}{6}, \frac{1}{5}\right) \cup \cdots$ which has $0$ in the closure but is not locally euclidean there.

Answer 1

No, that's not true. Imagine three open triangles in $\mathbb{R}^2$ "meeting" at a vertex (they don't actually contain the point, but it's in their common closure). The triangles are also linked by an open annulus. See this picture:

Then the point at the center prevents this space from being a topological manifold (with boundary). It can be given the structure of a "stratified manifold" though.

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You can also adapt your own counterexample. Consider the space $X = \bigcup_{n \in \mathbb{N}^*} (\frac{1}{n+1}, \frac{1}{n})$. Then $(X \times \mathbb{R}) \cup (\mathbb{R} \times (0,1))$ is open and connected, but its closure isn't a topological manifold in any possible meaning of the word (look at the point $(0,2)$, that space isn't even locally connected).

Answer 2

No. Consider $X=\{|y|>|x|\}\cup\{|y|>1\}\cup\{|x|>1\}$. This is an open and connected set. Its closure contains $0$ which has no neighborhood in $X$ homeomorphic to any $\mathbb R^n$.

Answer 3

No: look at an open cube .

Edit (October 2, 2014)
To clarify matters (since the question mentioned just the word "manifold"):
$\bullet $ The closure of the open cube is not a "topological manifold" but a "topological manifold with boundary"

$\bullet \bullet $ The closure of the open cube is not a "differential manifold" nor even a " differential manifold with boundary", but a "differential manifold with corners", the corners being the eight vertices of the cube.
This last notion of "differential manifold with corners" is an advanced, quite specialized notion of differential topology which is never (as far as I know) discussed in textbooks.

As a consequence I think that there is no rigorous treatment of the Stokes theorem in textbooks that directly applies to a cube!
Onre must resort to ad hoc trickery that relies on the fact that the vertices of a cube have measure zero and that they may be deleted as far as integration is concerned.
My learned friends in analysis once told me that there are versions in the research literature (by De Giorgi maybe?) that apply in particular to cubes but which are stated in terms of currents, a generalization to manifolds of Schwartz's distributions on $\mathbb R^n$.

To end on a surrealistic note: Lang, at the very end of his book Real Analysis invokes Hironaka's resolution of singularities to state Stokes's theorem in great generality.
(For non algebraic geometers: Hironaka's result is one of the most difficult papers in the history of mathematics: 217 pages in the Annals of Mathematics ...)