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?

  • $\begingroup$ A basic theorem says that a continuous image of a compact space is necessarily compact. $\endgroup$ Oct 21 '20 at 8:11
  • $\begingroup$ The image of a compact set under a continuous function is also compact. Therefore, $\mathbb{R}$ and the unit circle can't be homeomorphic. $\endgroup$
    – TwoStones
    Oct 21 '20 at 8:11
  • $\begingroup$ Another way to say it is that compactness is a topological property, so as they differ in that property they cannot have the same topological structure. $\endgroup$
    – Garmekain
    Oct 21 '20 at 8:15
  • $\begingroup$ A homeomorphism preserves all topological properties. So if two spaces differ in some respect (compactness, etc.) then they cannot be homeomorphic. $\endgroup$ Oct 21 '20 at 8:16
  • $\begingroup$ So with this, I can show that the mapping $g$ from the circle to the real line is not continuous. Since its image will not be compact? $\endgroup$
    – user140878
    Oct 21 '20 at 8:24

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.


Topological properties

The property "compactness" of a topological space is a topological property, i.e. it is expressed purely in terms of the elements of a topological space and of its associated set of the "open" subsets.

In particular, for compactness, if you look at Heine-Borel definition: the topological space $(X,\mathcal U)$ is compact iff, for every subset $\mathcal U'\subseteq\mathcal U$ such that $\bigcup\mathcal U'=X$ there is a finite subset $\mathcal U''\subseteq\mathcal U'$ such that $\bigcup\mathcal U''=X$.

Now, the important fact is that:

Homeomorphisms preserve all topological properties

In other words, if you have a homeomorphism $\phi:(X,\mathcal U)\to(Y,\mathcal V)$, any topological property that holds for $(X,\mathcal U)$ must also hold for $(Y,\mathcal V)$ and vice versa.

To see that, first note that homeomorphisms are bijections and both them ($\phi$) and their inverses ($\phi^{-1}$) are continuous on the whole domains ($X$ and $Y$, respectively). Continuity of $\phi$ means that $\phi^{-1}(V)\in\mathcal U$ for every $V\in\mathcal V$, and continuity of $\phi^{-1}$ means that $\phi(U)\in\mathcal V$ for every $U\in\mathcal U$.

Compactness example

How does this translate into preserving topological properties (e.g. compactness)? We can easily mount a proof which might go "back and forth" from $(X,\mathcal U)$ to $(Y,\mathcal V)$, and vice versa, a few times. For example, let $(X,\mathcal U)$ be compact, and our goal is to prove that $(Y,\mathcal V)$ is compact:

  • Take any subcollection $\mathcal V'\subseteq V$ such that $\bigcup\mathcal V'=Y$.
  • Map it back using $\phi^{-1}$ - you get a subcollection $\mathcal U'=\{\phi^{-1}(V)\mid V\in\mathcal V'\}$. Due to continuity of $\phi$, you have $\mathcal U'\subseteq\mathcal U$. Also, due to $\phi$ being bijection, you can easily show that $\bigcup\mathcal U'=X$
  • Now use compactness of $(X,\mathcal U)$: there exists a finite subcollection $\mathcal U''\subseteq\mathcal U'$ such that $\bigcup\mathcal U''=X$.
  • Go back and map $\mathcal U''$: let $\mathcal V''=\{\phi(U)\mid U\in\mathcal U''\}$
  • To finish the proof, conclude that $\mathcal V''\subseteq \mathcal V'$ and that $\mathcal V''$ is finite (both easy) and also that $\bigcup\mathcal V''=Y$ (follows from $\phi$ being a bijection).

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