Show that unit circle is not homeomorphic to the real line 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?
 A: 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.
A: 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.
A: 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.
