The equivalence classes and their representatives

Question

A relation on $\mathbb{R}^3 \backslash \{(0,0,0)\} $ is defined by $(a, b, c)\sim(d, e, f)$ if and only if there is $\lambda \in \mathbb{R}$ with $\lambda > 0$, and $a=\lambda d$, $b=\lambda e$ and $c=\lambda f$. Prove that $\sim$ is an equivalence relation and find a transversal for the relation that is geometrically pleasant.

I can show that $\sim$ is an equivalence relation. For the reflexivity, simmetry and transitivity to hold simultaneously, $\lambda$ should be 1, except if there's a mistake in my reasoning.

Now if $\lambda$ is indeed 1 that also implies that if $(a, b, c)\sim(d, e, f)$ then $a=d$, $b=e$ and $c=f$, which means that every $(a, b, c)$ is related only to itself.

Now the transversal is apparently a sphere, but I don't see how that could be if every point represents its own equivalence class?

Answer 1

"I can show that ∼ is an equivalence relation. For the reflexivity, simmetry and transitivity to hold simultaneously, λ should be 1, except if there's a mistake in my reasoning."

No, $\lambda$ is arbitrary. For reflexivity, $\lambda=1$. But to prove symmetry, let $(a,b,c)=\lambda(x,y,z)$ for some $\lambda\ne 0$. Then $(x,y,z)=\frac{1}{\lambda}(a,b,c)$. Thus $(a,b,c)\sim (x,y,z)$ implies $(x,y,z)\sim(a,b,c)$. Similar for transitivity.

The quotient set is the projective plane over the reals. A transversal is given by $$\{(1,b,c)\mid b,c\in\Bbb R\}\cup \{(0,1,c)\mid c\in\Bbb R\}\cup \{(0,0,1)\}.$$ The points with first component 0 are the points at infinity.

Answer 2

Reflexive: $(a,b,c)=1\cdot(a,b,c)$, so $(a,b,c)\sim (a,b,c)$.

Symmetric: If $(a,b,c)\sim (d,e,f)$, then $\exists \lambda>0$ with $(a,b,c)=\lambda\cdot(d,e,f)$. Then$(d,e,f)=\frac{1}{\lambda}\cdot (a,b,c)$. So $(d,e,f)\sim (a,b,c).$.

Transitive: To you. (But it amounts to noting that the product of two positive values of $\lambda$ is again positive.)

Equivalence classes are rays (half lines) with endpoint at the origin. So the unit sphere (centered at the origin) indeed forms a transversal.

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Edit: To be accurate: The origin referred to above as the endpoint of each ray is not included in each ray. (I.e., they are open rays.)