# Proving it doesn't exist a homeomorphism between $\mathbb R$ and $\mathbb R^n$, $n>1$.

I have to prove that for $n \ge 2$, there doesn't exist a homeomorphism between $\mathbb R$ and $\mathbb R^n$. Could anyone give me a hint on how could I prove this?

## 2 Answers

Hint: removing a point from $\mathbb{R}$ produces a disconnected space.

The notion of connectdness is a topological property, meaning that it is invariant under homeomorphisms (this is trivial since the definition involves only the open sets and does not refer to any other property of the space)

Now, assume for the sake of contradiction that there exists a homeomorphism between $\mathbb{R^n}$ and $\mathbb{R}$, say $h \colon \mathbb{R}^n \to \mathbb{R}$. It is very easy to check that $h \colon \mathbb{R}^n \setminus \{h^{-1}(0)\} \to \mathbb{R}\setminus \{0\}$ is still a homeomorphism (it is bijective and the restriction of continuous functions is continuous hence both this and its inverse are continuous). It is also known that continuous functions map connected spaces into connected spaces. Since $\mathbb{R}\setminus \{0\}$ is clearly not connected (I am sure that you can find a separation at first glance :D ) then all we need to show is that $\mathbb{R}^n \setminus \{h^{-1}(0)\}$ is connected. It is very intuitive but to make it a little bit more rigorous we can note that it is path connected: take two points, say $x$ and $y$: if the line between them does not pass through $0$ then we are done; otherwise pick any point $z$ that is not on that line and consider the path that goes from $x$ to $z$ with a line and from $z$ to $y$ with another line. This concludes :D

• Huh. I thought OP wanted just a hint! :-) Commented Oct 11, 2013 at 13:40
• well, what can I say... oh yes, it is not done since I only proved that there is no homeomorphism between $\mathbb{R}^n$ and $\mathbb{R}$... now it's up to him to conclude that there is no homeomorphism between $\mathbb{R}$ and $\mathbb{R}^n$ :D
– user67133
Commented Oct 11, 2013 at 20:03