Let $\sim$ be the equivalence relation on $\mathbb{R}^2$ generated by $A=\{ (v,2v)\ |\ v \in \mathbb{R}^2\}$. Explicitly: $$\forall x,y \in \mathbb{R}^2 \hspace{5mm} x\sim y \Longleftrightarrow \exists n \in \mathbb{Z}:\ y = 2^n x $$ I want to study the quotient space $X=\mathbb{R}^2/\sim$
I don't know how to prove or disprove that:
- X is second (or first) countable
- X is Hausdorff
- X is compact
More generally, I'd like to have some easier to visualize homeomorphic space, if such a space exists.
I managed to prove the properties of X by myself, but I'd still appreciate a double check on the proof and, if available, a nicer description of the space:
Edit 1. X is not Hausdorff.
Take the sequence $(2^{-n}e_1)_{n \in \mathbb{N}}$ where $e_1 = (1,0)$.
Clearly, for all $n \in \mathbb{N}$, $2^{-n}e_1 \sim e_1$. So $[2^{-n}e_1]=[e_1]\rightarrow [e_1]$
On the other hand $2^{-n}e_1 \rightarrow 0$. So, by continuity of the projection, $[2^{-n}e_1]\rightarrow [0]$
But $0 \nsim e_1$, so the limit is not unique and X is not $T_2$.
Edit 2. X is second countable.
Since $\mathbb{R}^2$ is second countable, it is sufficient to prove that the projection $\pi: \mathbb{R}^2 \rightarrow X$ is open. Let $A$ be an open set of $\mathbb{R}^2$. Then:
$$\pi^{-1}(\pi(A))= \{\ 2^nx \ |\ x \in A, \ n \in \mathbb{Z}\} = \bigcup_{n \in \mathbb{Z}}\{\ 2^nx\ |\ x\in A\} $$
Given $n \in \mathbb{Z}$, let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2,\ x \mapsto 2^{-n}x$.
$f$ is continuous, hence $\{\ 2^nx\ |\ x\in A\}=f^{-1}(A)$ is open in $\mathbb{R}^2$.
$\pi^{-1}(\pi(A))$ is open in $\mathbb{R}^2$, since it is a union of open sets. This proves that $\pi(A)$ is open in X.
X is compact.
Let $\mathcal{U}$ be an open cover of X. Then there is $U \in \mathcal{U}$ such that $[0] \in U$.
Then $\pi^{-1}(U)$ is open and $0 \in \pi^{-1}(U)$, so there is $\varepsilon > 0$ such that $B(0,\varepsilon)\subseteq \pi^{-1}(U)$
Let $y \in \mathbb{R}^2,\ y \neq 0$. Since $2^{-n}\rightarrow 0$, there is $m \in \mathbb{N}$ such that $2^{-m}<\frac{\varepsilon}{||y||}$
So we have $||2^{-m}y|| = 2^{-m} ||y|| < \varepsilon$, from which $2^{-m}y \in B(0,\varepsilon) \subseteq \pi^{-1}(U)$
Since $y \sim 2^{-m}y$, clearly $y \in \pi^{-1}(U)$. So $\pi^{-1}(U)=\mathbb{R}^2$ and $U = \pi(\pi^{-1}(U)) = \pi(\mathbb{R}^2) = X$
$\{U\}$ is a finite subcover of $\mathcal{U}$. This proves X is compact and works as an alternative proof of the fact that X is Hausdorff.
It’s convenient to work in polar coordinates. Let $R=\{\langle r,\theta\rangle\in\Bbb R^2:1\le r<2\}$; each $\sim$-class has a unique representative in $R$. Roll $R$ into a torus $T$ by starting with $R^+=\{\langle r,\theta\rangle\in\Bbb R^2:1\le r\le 2\}$ and identifying $\langle 1,\theta\rangle$ and $\langle 2,\theta\rangle$ for $0\le\theta<2\pi$, and give this its usual topology; then $T$ is homeomorphic to $X\setminus\{[\mathbf{O}]\}$, where $\mathbf{O}$ is the origin in $\Bbb R^2$. Finally, $X$ itself is the only open set containing $\mathbf{O}$. (Note that not only is $X$ compact, but so is $X\setminus\{[\mathbf{O}]\}$!)