# Characterization of the Subsets of Euclidean Space which are Homeomorphic to the Space Itself

I have no real experience in topology (although I have done a course in metric spaces) but in the course of a project I am doing it has become useful to produce (if possible) a characterization of the subsets of arbitrary dimensional Euclidean Space (with the usual metric and topology) that are homeomorphic to the whole space.

I started by looking at the sorts of properties which are conserved under homeomorphism and found that such a subset is open and connected. I have also shown that convex open sets are homeomorphic to R^n.

However, what I am really looking for is an equivalence between subsets homeomorphic to R^n and subsets with a list of specific properties (e.g. open, convex). That I can use to identify any possible homeomorphic subset.

Hints and statements of characterization would be appreciated as starting points. However, I would like to work through the necessary proofs on my own if possible.

Thank You

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Can a closed set be homeomorphic to $\mathbb{R}^n$? What about a compact set? What about a non-simply connected set (e.g. an open annulus $\{ x\in \mathbb{R}^2 | 1 < |x| < 2\}$)? Must all sets homeomorphic to $\mathbb{R}^n$ be convex? – Billy Jun 23 '13 at 19:54
You're definitely on the right track by trying to come up with properties that need to be satisfied. Properties like that that do not change under homeomorphism are often referred to as "topological invariants". You might find some inspiration on this Wikipedia list. – fuglede Jun 23 '13 at 19:57
Thanks for your replies. I'll definitely be checking out the list. – James Wilsenach Jun 23 '13 at 20:15
It is. Thank you. – James Wilsenach Jun 24 '13 at 0:49

According to this previous question, a subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ if and only if it is open, contractible, and simply connected at infinity.
Note that the last condition is necessary. For example, the Whitehead manifold is a contractible open subset of $\mathbb{R}^3$, but it is not homeomorphic to $\mathbb{R}^3$.