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I found this exercise I'm not able to solve.

EXERCISE : Let's call $X$ the topological space obtained quotienting $\mathbb{R}^{n}$ by the equivalence relation $\sim$ :

$x\sim y \Leftrightarrow x=y$ or $||x||=||y||$ or $||x||\cdot ||y||=1$.

Is $X$ an Hausdorff space ($T2$)?? Is $X$ compact?? Can anyone help me?

The only idea i had is this one but I had plenty of doubt about its correctness.. Using hyperspherical coordinates ( can I say that $\mathbb{R}^{n}$ is omeomorph to $\mathbb{R}\times \left[ {0,\pi } \right] \times \cdots \times \left[ {0,\pi } \right] \times \left[ {0,2\pi } \right[$ ??? And if that was true, using the fact that $\sim$ act only on the norm and not on the angles, can I say that $\mathbb{R}^{n}/\sim {\rm{ }} = \left( {\mathbb{R}\times \left[ {0,\pi } \right] \times \cdots \times \left[ {0,\pi } \right] \times \left[ {0,2\pi } \right[} \right)/\sim {\rm{ }} = \mathbb{R}/\sim$ ??? Thanks in advance.

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Hint: there is one equivalence class for every norm $\leq 1$. – Grumpy Parsnip Sep 7 '11 at 14:27
$\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}\times[0,\pi)...$ because the one on the right has boundary/corners/etc while the one on the left is a smooth manifold. (Just look at regular polar coordinates). – Jason DeVito Sep 7 '11 at 14:35
up vote 2 down vote accepted

I imagine this space like this: you identify all elements with the same norm, so the orbits are spheres. Furthermore, you identify elements of norm $k$ with elements of norm $1/k$, therefore you can consider only spheres with radius $\leq 1$.

Denote by $\pi$ the canonical projection $x \mapsto [x]$ where $[x]$ is the equivalence class of $x$.

Pick $x,y$ two different points in $X$. Then $\|x\|\neq \|y\|$ and $\|x\|\|y\|\neq 1$. Now it is enough to find two neighborhoods $U,V$ for $x,y$ respectively such that $\pi(U) \cap \pi(V)=\emptyset$. You can consider $U,V$ to be open balls of carefully chosen radii. There are two cases to study: when both $\|x\|,\|y\|$ are on the same side of $1$, and for example when $\|x\|<1<\|y\|$.

As for compactness, pick $\{O_n\}$ an open cover for $X$. Define $X_n=\{x \in [0,1] : \exists y \in O_n \text{ with }\|y\|=x \text{ or }\|y\|=1/x\}$. Then $\{X_n\}$ is an open cover for $[0,1]$ and there is a finite subcover (eventually renumber) $\{X_1,...,X_n\}$. Then the corresponding $O_1,O_2,...,O_n$ cover $X$. This proves that $X$ is compact.

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You need only prove that $X$ is Hausdorff. Once you know that, it is clear that the map $[0,1]\to X: t \mapsto \text {class of } (t,0,\ldots ,0)$ is a homeomorphism because: a) it is bijective b) $[0,1]$ is compact and c) $X$ is Hausdorff . As a consequence $X$ is compact. – Georges Elencwajg Sep 7 '11 at 15:01
@Georges, isn't the compactness of $X$ quite clear even without Hausdorffness, because the canonical projection restricted to the closed unit ball of $\mathbb{R}^n$ (which is compact) is surjective. – LostInMath Sep 7 '11 at 15:05
@LostInMath: Sure, but Georges's argument has the virtue of identifying the quotient space with something rather familiar :) – t.b. Sep 7 '11 at 15:08
Thank you very much for you answers!!! Really really usefull! I have only a little last question... In the proof of the Hausdorfness. When dealing with this kind of exercises I have to choose not simply two open sets (containing $x$ and $y$) but two saturated sets. I'm pretty sure that's not a problem cause, seems to me, that in $\mathbb{R}^{n}$ every open set is saturated but I'm not able to proof that... Thank you again!! – Lorenzo Rossi Sep 7 '11 at 15:39
Caro Lorenzo, your display of gratitude is quite refreshing: bravo! Be careful that a saturated open set here is necessarily a union of spheres (with of course zero added in if it was in the original open set), so: no, not every open set in $\mathbb R^n$ is saturated. And it is good news that you could not prove such a false statement! – Georges Elencwajg Sep 7 '11 at 15:53

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