Is any open subset of $\mathbb{R}^{n}$ homeomorphic to $\mathbb{R}^{n}$?

Question

I read somewhere that any open interval $(a,b)$ is homeomorphic to $\mathbb{R}$. Is that true? Also, is any open subset of $\mathbb{R}^{n}$ homeomorphic to $\mathbb{R}^{n}$?

Thanks!

Yes Can you find a homeomorphism from (a,b)? – M.B. Dec 17 '14 at 23:21 — M.B., Dec 17 '14 at 23:21

Simon S · Answer 1 · 2014-12-18T00:08:05.840

4

Yes. No; for example, $(1,2) \cup (3,4)$ is open in $\mathbb{R}$ but not homeomorphic to it.

edited Dec 18 '14 at 00:08

answered Dec 17 '14 at 23:21

Simon S

26,524

Is it true for a connected open set? – Mike Pierce Dec 17 '14 at 23:23
3

@mapierce271: For $n>1$, $\Bbb{R}^n$ with a point deleted is open, connected, and not homeomorphic to $\Bbb{R}^n$. You need some stronger condition... – Micah Dec 17 '14 at 23:28

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

The answer is no. Because not every open set has the same topological properties of $\mathbb{R}^n$. Simple examples can be given when considering an open disconnected set and sets with holes. For example, in $\mathbb{R}^2$ a set without holes is called simply connected and you can say that every connected, simply connected open set is homeomorphic to $\mathbb{R}^2$.

For higher dimensions, answering the question of wether a topological space has $n$-dimensional holes or not is an absolutely non-trivial question and there are different ways of giving partial answers to it, mainly (but not exclusively) the study of the homotopy and the homology groups of the space. These groups are invariant under homeomorphisms, which means that homeomorphic spaces have the same groups associated to them, but the converse is not always true!

A complete answer to the question is given here: Characterization of the Subsets of Euclidean Space which are Homeomorphic to the Space Itself

As a comment to this - the homotopy type is not good enough for $n>2$ anymore! The Whitehead manifold is a contractible open submanifold of $\Bbb R^3$ that's not homeomorphic to $\Bbb R^3$. — , Dec 18 '14 at 01:45

score 1 · Answer 3 · answered Dec 17 '14 at 23:25

It is true for an interval. For example, $\tan x$ is an homeomorphism between $(-\pi/2, \pi/2)$ and $\mathbb{R}$, and each interval $(a, b)$ is homeomorphic to that latter interval.

However, not for any open set. If two sets are homeomorphic, they share the same topological properties, in particular conectedness. So if you have any open set which isn't connected $(e.g. (0, 1) \cup (2, 3)$, you can't be homeomorphic because $\mathbb{R}^n$ is connected.

score 1 · Answer 4 · answered Dec 17 '14 at 23:32

1

Hint for the 2nd question: In $\Bbb{R}^2$ consider the open set $\{(x,y)\mid 1<x^2+y^2<2\}$. Could it be homeomorphic to the whole $\Bbb{R}^2$?

answered Dec 17 '14 at 23:32

janmarqz

10,538

Is any open subset of $\mathbb{R}^{n}$ homeomorphic to $\mathbb{R}^{n}$?

4 Answers4