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 (http://en.wikipedia.org/wiki/N-sphere#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.
Best Answer
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.