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Let $(\mathbb R^2, \|\cdot\|)$ be two dimensional normed space and $S=\{x\in\mathbb R^2 : \|x\|=1\}$ the unit sphere of $(\mathbb R^2, \|\cdot\|)$. Since $S$ bounded and closed in $\mathbb R^2$ then it is compact. Does $S$ homeomorphic to the segment $[0,1]$ with usual topolgy? And if so then how to construct such an homeomorphism?

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    $\begingroup$ No, they are not homeomorphic. Homeomorphic means that they are "topologically the same" -- intuitively if not formally it should be clear that the sphere and the interval are very different. $\endgroup$ – Mees de Vries May 23 '18 at 11:11
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    $\begingroup$ There are not even homotopy equivalent (check the fundamental groups). $\endgroup$ – Watson May 23 '18 at 11:43
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No, they are hot homeomorphic. Suppose there exists a homeomorphism $f:\Bbb S^1\longrightarrow [0,1]$, let $p\in\Bbb S^1$ such that $f(p)\neq 0$ and $f(p)\neq 1$. Then $f:\Bbb S^1\setminus \{p\}\longrightarrow [0,1]\setminus\{f(p)\}$ should be still a homeomorphism. But the source is still connected whereas the target is not connected.

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  • $\begingroup$ You wrote $\backslash \{f(q)\}$ $\endgroup$ – John Cataldo May 23 '18 at 11:23
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    $\begingroup$ Thanks for the comment! $\endgroup$ – InsideOut May 23 '18 at 11:24
  • $\begingroup$ Thanks for answer. Could you more precise. $K=[0,1]\setminus {f(p)}$ is not connected thet it can be represented in the form $A\bigcup B$, where $A,B$ nonempty disjoint closed (or open) sets. But the sets $[0,f(p))$ and $(f(p),1]$ are not closed (or open)? $\endgroup$ – golomorfMath May 23 '18 at 12:31
  • $\begingroup$ They are not open or closed as subset of $\Bbb R$, however they are with respect to the induced topology. $\endgroup$ – InsideOut May 23 '18 at 12:39
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Another way of showing that $S^1$ is not homeomorphic to $[0,1]$ is by analyzing auto-homeomorphisms:

Definition. A topological space $X$ is called homogenous if for any $x,y\in X$ there exists a homeomorphism $f:X\to X$ such that $f(x)=y$.

The idea here is that in a way homogenous spaces have no special points.

Lemma. Let $X,Y$ be two homeomorphic spaces. Then $X$ is homogenous if and only if $Y$ is. In other words homogeneity is a topological invariant.

I leave the proof as an exercise.

With that you can easily see that $S^1$ is indeed homogenous: treat $S^1\subseteq\mathbb{C}$ and then for any $x,y\in S^1$ a homeomorphism that we are looking for can be for example $z\mapsto zx^{-1}y$ with complex multiplication on the right side.

Now $[0,1]$ is not homogenous. Indeed, if $f:[0,1]\to[0,1]$ is a homeomorphism then $f(0)=0$ or $f(0)=1$ as a consequence of the intermediate value theorem. So $0$ and $1$ are special points in $[0,1]$.

Note that this kind of argument can be extended to any dimension: an $n$-dimensional sphere is not homeomorphic to an $m$-dimensional cube because $n$-dimensional sphere is homogenous unlike $m$-dimensional cube (boundary points on a $m$-dimensional cube are special).

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Another way to see that $S^1$ and $[0,1]$ are not homeomorphic using standard topology (i.e. not using homology/homotopy) is by using the Brouwer Fixed Point Theorem, which asserts that any continuous function $$f \colon [0,1] \to [0,1]$$ has at least one fixed point $x_0$.

But any non-zero rotation map on $S^1$ is continuous without any fixed points, for instance $$r \colon S^1 \to S^1, e^{i\theta} \mapsto e^{i \theta + \varphi},$$ with $\varphi \neq 0$.

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