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if we work in $\mathbb{R}^2\(0,0)$ with euclidean topology and we set following equivalence relation $P$ on this space: $(x,y)P(x',y')$ iff there exists $a$ in $\mathbb{R}^2\(0,0)$ such that $(x,y) = a(x',y').$ how we define the open sets in quotient space $\mathbb{R}^2-\{(0,0)\}/P.$
Let $f: \mathbb{R}^2\(0,0) \to \mathbb{R}^2\(0,0)/P$ the quotient function: then the open sets in $\mathbb{R}^2\(0,0)/P$ are defined like: $\{f(A) \mid A\text{ open and saturated }\}$ but I have troubles by doing it for this example. Can someone help me?

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Do you really mean that $R$ is the irrational numbers? Because if so, $0\notin R$, and there’s no reason to write $R\setminus\{0\}$. – Brian M. Scott Aug 22 '13 at 16:07
If $a$ has to be an irrational number, then $P$ is not an equivalence relation. – Dan Rust Aug 22 '13 at 16:07
Please check to make sure that Stefan didn't change the meaning of your question with his edit. Should it be $R^2\cup\{(0,0)\}$ instead of $R^2-\{(0,0)\}$? If the edit is correct, then you probably mean real or rational numbers, rather than irrational. – Cameron Buie Aug 22 '13 at 16:08
I think, $R$ stands for the field of real numbers, otherwise the question does not make sense. One can also consider the case of other fields, like rational numbers. – studiosus Aug 22 '13 at 16:10
I think the OP meant for $R$ to be the real numbers and so the quotient space would be the space $\mathbb{R}P^1\cong S^1$. – Dan Rust Aug 22 '13 at 16:11

I’m going to assume that your $R$ is really $\Bbb R$, the set of real numbers.

HINT: Note that $\langle x,y\rangle\mathbin{P}\langle x',y'\rangle$ if and only if $\langle x,y\rangle$ and $\langle x',y'\rangle$ lie on the same straight line through the origin of $\Bbb R^2$. For $0\le\theta<\pi$ let $L_\theta'$ be the line through the origin containing the point $\langle\cos\theta,\sin\theta\rangle$, and let $L_\theta=L_\theta'\setminus\{\langle 0,0\rangle\}$. The sets $L_\theta$ for $\theta\in[0,\pi)$ are the $P$-equivalence classes and therefore also the fibres of the quotient map $f$ and the points of the quotient space. This means that you can identify the points of the quotient space with points of the interval $[0,\pi)$: each $L_\theta$ is a point of the quotient space, and it corresponds to the point $\theta$ in the set $[0,\pi)$.

To figure out the quotient topology $\tau$ on $[0,\pi)$, remember that a subset $U$ of $[0,\pi)$ is in $\tau$ if and only if $\bigcup_{\theta\in U}L_\theta$ is open in $\Bbb R^2\setminus\{\langle 0,0\rangle\}$. (It may help to imagine bending the interval $[0,\pi)$ around into a circle, to get the space obtained by identifying $0$ and $\pi$ in the closed interval $[0,\pi]$.)

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what do you mean by $[0,\pi)$ ? – Koen Aug 23 '13 at 13:33
@Koen: The standard meaning: $[0,\pi)=\{x\in\Bbb R:0\le x<\pi$, the half-closed, half-open interval from $0$ to $\pi$. – Brian M. Scott Aug 23 '13 at 17:15

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