Homeomorphic images of $\mathbb{R}^n$ Let $f : \mathbb{R}^n \to \Omega  \subset \mathbb{R}^n $ be a homeomorphism. If $\Omega$ is a proper subset of $\mathbb{R}^n$, then how do we show that $\mathbb{R}^n  \backslash \Omega$ is an unbounded set? 
Suppose, to the contrary, $\mathbb{R}^n  \backslash \Omega$ is bounded. It is a closed set of $\mathbb{R}^n$, and hence is compact. $\Omega$ is open in $\mathbb{R}^n$. Further, is we show $\Omega$ is closed, using our assumption, we would arrive at a contradiction that $\Omega$ is the empty set. 
Any help regarding how to do this? 
Other methods would be appreciated too. 
 A: Suppose on the contrary that $\mathbb{R}^n \setminus \Omega$ is non-empty and bounded. So there exist $p \in \mathbb{R}^n \setminus \Omega$ and a closed round ball $B$ centred at $p$ which contains $\mathbb{R}^n \setminus \Omega$. In other words, $\partial B \subset \Omega$. Let $\iota : S^{n-1} \cong \partial B \subset \mathbb{R}^n$ denote the 'inclusion'; This is a continuous map. So $\iota$ is a $(n-1)$-singular cycle. By construction, it is not a boundary in $\mathbb{R}^n \setminus \{ p \}$ and a fortiori it is not a boundary in $\Omega \subset \mathbb{R}^n \setminus \{p\}$.
Since $f : \mathbb{R}^n \to \Omega$ is a homeomorphism, $f^{-1} : \Omega \to \mathbb{R}^n$ is defined and continuous. Hence, the map $f^{-1} \circ \iota : S^{n-1} \to \mathbb{R}^n$ is a $(n-1)$-singular cycle. Because $\mathbb{R}^n$ is contractible, this cycle is a boundary in $\mathbb{R}^n$. Let $g : B^n \to \mathbb{R}^n$ be a $n$-chain such that $\partial g = f^{-1} \circ \iota$. Then $f \circ g : B^n \to \Omega$ is such that $\partial (f \circ g) = \iota$. This shows that $\iota$ is a boundary in $\Omega$, which yields the contradiction.
