0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

"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.

$\endgroup$
4
  • 2
    $\begingroup$ The OP states: "For $\lambda>0$". This answer seems to be for $\lambda \ne 0$. $\endgroup$ – paw88789 Oct 20 '20 at 12:13
  • $\begingroup$ It works also for $\lambda >0$. But then the transversal has ''twice'' as much elements. $\endgroup$ – Wuestenfux Oct 20 '20 at 15:34
  • $\begingroup$ Okay, thank you, but I'd need some further clarification. How so that for reflexivity $\lambda$ takes a certain value, but that value doesn't have to stay the same for the symmetry and transitivity? Also I thought that the definition of symmetry - if $x \sim y $ then $y \sim x$ - in this case means if $(a, b, c)= \lambda (d, e, f)$ then $(d, e, f)= \lambda (a, b, c)$. This question is similar to the above, but why can there be a certain value ($\lambda$) before $(d, e, f)$ in the first part of the if clause and a different one before $(a, b, c)$, specifically $\frac{1}{\lambda}$ in 2nd part? $\endgroup$ – Treex Oct 20 '20 at 19:50
  • $\begingroup$ The definition is as follows: $(a,b,c)\sim (x,y,z)$ if there exists $\lambda\ne 0$ (or $\lambda >0$) such that $(a,b,c) = \lambda(x,y,z)$. For reflexivity, you can choose $\lambda$ appropriately. $\endgroup$ – Wuestenfux Oct 21 '20 at 6:41
3
$\begingroup$

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.


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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.