Show that $S^1$ is not homeomorphic to either $\mathbb{R}^1$ or $\mathbb{R}^2$

$\mathbf{My \ solution}$:

So first we will show that $S^1$ is not homeomorphic to $\mathbb{R}^1$. To show that they are not homeomorphic we need to find a property that holds in $S^1$ but does not hold in $\mathbb{R}^1$ or vice-versa. $S^1$ is compact however $\mathbb{R}^1$ is not compact.

The set $\{1\} $ is closed, and the map $$f: \Bbb R^2 \longrightarrow \Bbb R,$$ $$(x, y) \mapsto x^2 + y^2$$ is continuous. Therefore the circle $$\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\} = f^{-1}(\{1\})$$ is closed in $\Bbb R^2$.

Set $S^1$ is also bounded, since, for example, it is contained within the ball of radius $2$ centered at 0 of $\Bbb R^2$ (in the standard topology of $\Bbb R^2$).

Hence it is also compact.

However real line $\Bbb R^1$ is not because there is a cover of open intervals that does not have a finite subcover. For example, intervals (n−1, n+1) , where n takes all integer values in $\mathbb{Z}$, cover $\mathbb{R}$ but there is no finite subcover.

Hence $S^1$ can not be isomorphic to $\mathbb{R}^1$.

How to show now that $S^1$ is not homeomorphic to $\mathbb{R}^2$? Can i show it now in the same way? They can not be homeomorphic since $S^1$ is compact however $\mathbb{R}^2$ not. How to show that $\mathbb{R}^2$ is not compact?

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    $\begingroup$ $\Bbb R^2$ is covered by all open balls around the origin, but not by finitely many of these. -- For an alternative proof ide: $S 1$ minus two point sis disconnected, but $\Bbb R^2$ minus two points is connected. $\endgroup$ – Hagen von Eitzen May 12 '18 at 16:41
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    $\begingroup$ $\mathbb{R}^2$ is not compact for the same reason that $\mathbb{R}$ isn't. Consider growing balls of radius $n$, for example, to build an open cover without finite subcover. Another argument is that removing a single point from a circle leaves a connected space, while that is not true of $\mathbb{R}$. $\endgroup$ – user296602 May 12 '18 at 16:41
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    $\begingroup$ Open balls of radius 2 centered at every point with integer coordinates would work. No finite collection of those covers $\Bbb R^2$. $\endgroup$ – G Tony Jacobs May 12 '18 at 16:49
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    $\begingroup$ Not to mention, with the euclidean metric $\mathbb R^2$ is not closed and bounded. $\endgroup$ – fleablood May 12 '18 at 16:52
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    $\begingroup$ A closed subspace of a compact Hausdorff space is compact. So if $\Bbb R^2$ were compact then $\Bbb R\times \{0\},$ which is closed in $\Bbb R^2,$ and is homeomorphic to $\Bbb R$, would also be compact.... Also no metric space with an unbounded metric can be compact. $\endgroup$ – DanielWainfleet May 13 '18 at 3:48

To prove that $\mathbb R^2$ is not compact: Assume that it is. The image of a compact space under a continuous map is compact. The mapping $f:\mathbb R^2 \to \mathbb R, (x,y)\mapsto x$ is continuous and has image $\mathbb R$. Hence $\mathbb R$ is compact. But you yourself showed that $\mathbb R$ is not compact. Contradiction.

To show that $S^1$ is not homeomorphic to $\mathbb R^2$: Observe that $S^1$ is compact but $\mathbb R^2$ isn't. Done.


You can certainly show these using compactness. The following proofs, however, I find simpler:

The removal of any one point from $\Bbb R$ results in a disconnected space, but if you remove one point from $S^1$, you still have a connected space.

The removal of any two points from $S^1$ results in a disconnected space, but if you remove two points from $\Bbb R^2$, you still have a connected space.

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    $\begingroup$ Why do you say connectedness is "simpler" than compactness? $\endgroup$ – GEdgar May 12 '18 at 17:15
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    $\begingroup$ I find it simpler, conceptually. YMMV. $\endgroup$ – G Tony Jacobs May 12 '18 at 17:19

Assume that $\mathbb{R}^2$ is homeomorphic to $S^1$. You already showed that $S^1$ is compact. Let $\mathcal{O}=\{O_i\}_{i\in I}$ be an open cover of $\mathbb{R}^2$ and $f: \mathbb{R}^2\to S^1$ a homeomorphism. Then $f(\mathcal{O})=\bigcup\limits_{i\in I}f(O_i)$ is an open cover for $S^1$, by compactness of $S^1$ it must contain a finite open subcover $\{f(O_{i_j})\}_{j=1}^k$. This gives,

$$\{f(O_{i_j})\}_{j=1}^k \supset S^1 \implies \{O_{i_j}\}_{j=1}^k\}\supset \mathbb{R}^2$$

which means that $\mathbb{R}^2$ is compact, contradiction.


If $f:S^{1} \rightarrow \mathbb{R}^{1}$ was a homeomorphism then, since $S^{1}$ is compact, $f(S^{1}) = \mathbb{R}^{1}$ would have to be compact. A contradiction.

The same argument works with $\mathbb{R}^{2}$.

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    $\begingroup$ I think OP already knows this. Their question was how to show that $\mathbb R^2$ is not compact. $\endgroup$ – John Gowers May 12 '18 at 16:52
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    $\begingroup$ I can't see anything there that isn't verbatim contained in the OP. And the OP contains the question "How to show that R2 is not compact?" which this answer completely ignores. $\endgroup$ – fleablood May 12 '18 at 16:54

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