0
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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).
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.