# 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?

• A basic theorem says that a continuous image of a compact space is necessarily compact. Oct 21 '20 at 8:11
• The image of a compact set under a continuous function is also compact. Therefore, $\mathbb{R}$ and the unit circle can't be homeomorphic. Oct 21 '20 at 8:11
• 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. Oct 21 '20 at 8:15
• A homeomorphism preserves all topological properties. So if two spaces differ in some respect (compactness, etc.) then they cannot be homeomorphic. Oct 21 '20 at 8:16
• 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? 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 $$n_1. This means that $$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).