Homeomorphism and compact spaces I can't really pinpoint the idea behind the following example ( from wikipedia https://en.wikipedia.org/wiki/Homeomorphism )
"The Euclidean real line is not homeomorphic to the unit circle as a subspace of $\Bbb{R^2}$, since the unit circle is compact as a subspace of Euclidean $\Bbb{R^2}$ but the real line is not compact.
Why can't we have a homeomorphism between these spaces? The real line is considered open ( or closed and unbounded ), I believe, and thus is not compact. So we would have a mapping from a non-compact set onto a compact set ( either the mapping or its inverse, both need to be continuous ). Does this imply we can't construct a continuous mapping between these spaces?

Assume that $f:C\to\mathbb R$ is a homeomorphism, where $C$ is the unit circle. Clearly
the open intervals
$$
U_n=(-n,n), \quad n\in\mathbb N,
$$
form an open cover of $\mathbb R$, i.e. $\bigcup_{n\in\mathbb N}U_n=\mathbb R$. Set
$$
V_n=f^{-1}(U_n), \quad n\in\mathbb N.
$$
Then $V_n$'s are open subsets of $C$, in its relative topology, as inverses of open sets, and they form an open cover of $C$. Since $C$ is compact, the open cover $\{V_n\}_{n\in\mathbb N}$ of $C$ contains a finite subcover $V_{n_1},\ldots,V_{n_k}$. We may assume that $n_1<n_2<\cdots<n_k$. This means that $U_{n_1}\subset\cdots\subset U_{n_k}$, and hence $V_{n_1}\subset\cdots\subset V_{n_k}$. Thus
$$
C=\bigcup _{j=1}^k V_{n_j}=V_{n_k}=f^{-1}(-n_k,n_k).
$$
Contradiction.
