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Let $S^2$ be the unit sphere in $\mathbb{R}^3$. That is $S^2=\{(x,y,z); x^2+y^2+z^2=1\}$. Prove that $\sim$ defined by $(x,y,z) \sim (a,b,c)$ if and only if $(x,y,z)=-(a,b,c)$ is an equivalence relation on $S^2$. Find five distinct equivalence classes given by $\sim$.

My attempt $x^2+y^2+z^2=1$ and multiplying both sides by $-1$ gives $-x^2-y^2-z^2=-1$ so $(x,y,z)=-(x,y,z)$ thus $(x,y,z)\sim -(x,y,z)$ so the relation is reflexive. Now assume $(x,y,z) \sim (a,b,c)$ thus $x^2+y^2+z^2=-a^2-b^2-c^2$ multiplying both sides by $-1$ gives $a^2+b^2+c^2=-x^2-y^2-z^2$ thus $(a,b,c)=-(x,y,z)$ so $(a,b,c) $\sim$ (x,y,z)$ so the relation is symmetric. Now assume $(x,y,z) \sim (a,b,c)$ and $(a,b,c) \sim (d,e,f)$ now since the relation is symmetric $(a,b,c) \sim (x,y,z)$ and $(d,e,f) \sim (a,b,c)$ then $(x,y,z) \sim (d,e,f)$

I'm even less unsure about the classes and I know the above proof is not correct.

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    Is that a copy of the problem statement or did you reword it? – 2017-01-25
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    This is how it was worded exactly to me – 2017-01-25
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    The equivalence relation given makes no sense to me. – 2017-01-25
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    Can you help explain why it doesn't make sense? I mean I'm looking at it also thinking now that it can't be reflexive either just in the case of $(1,0,0) \not=-(1,0,0)$ so maybe it was a mistake. – 2017-01-25
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    You have just pointed out one example. Note that the relation is reflexive if every element of $S^2$ is related to itself. – 2017-01-25
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    You might want to contact your instructor/tutor for a better version of the problem. – 2017-01-25
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    Maybe the relation was meant to be: $(x,y,z)\sim(a,b,c)$ iff $(x,y,z)=(a,b,c)$ or $(x,y,z)=-(a,b,c)$. – 2017-01-25

1 Answers 1

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A relation $R$ over a set $X$ needs to satisfy three properties in order to be an equivalence relation:

  • It must be reflexive, that is, every element in $X$ is related to itself. $\forall x\in X: xRx$;

  • It must be symmetric, that is, if $x$ is related to $y$, $y$ must be related to $x$. $xRy \implies yRx$;

  • It must be transitive, that is, if $x$ is related to $y$ and $y$ is related to $z$, $x$ must be related to $z$.

Let me stress that you have defined $\sim$ as:

$(x,y,z) \sim (a,b,c) \iff (x,y,z) = -(a,b,c)$

Unfortunately, that violates the reflexive property, that is, we don't have that $(x,y,z)$ is equivalent to itself for all triples $(x,y,z) \in S^2$. Take $(1,0,0)$ for example. If $\sim$ is reflexive, then we must have $(1,0,0) \sim (1,0,0) \iff (1,0,0) = -(1,0,0)$ by definition. But $1 \neq -1$ so it is not reflexive.

I believe this problem arises because of the way the problem was worded to you. In English, one could say, for example, "show that the relation we get by identifying all points $(x,y,z)$ with $-(x,y,z)$ is an equivalence relation." Although it looks the same, it is not the same as you wrote.

Or you could see written, "prove that $\sim$ is an equivalence relation if $(x,y,z) \sim -(x,y,z)$"; which would read as I wrote just above, being implicit that every element $(x,y,z)$ is related to itself, in addition to being related to $-(x,y,z)$.

If this was what was meant when the problem was handed to you, then yes we have an equivalence relation.

If that is the case, if we are to take implicitly that $(x,y,z)$ is related to itself, then the reflexive property is true by definition of the relation.

The symmetric property is shown by:

$$(x,y,z) \sim (a,b,c) \iff \begin{cases}(a,b,c) = -(x,y,z)\iff -(a,b,c) = (x,y,z)\ \text{or}\\(a,b,c) = (x,y,z)\end{cases}\tag{1}$$

And we would have symmetry if it is true that $(a,b,c) \sim (x,y,z)$. But

$$(a,b,c) \sim (x,y,z) \iff \begin{cases}(x,y,z) = (a,b,c)\ \text{or}\\(x,y,z) = -(a,b,c)\end{cases}\tag{2}$$

But either case we need in $(2)$ happens in $(1)$ and every case that can happen in $(1)$ is needed in $(2)$, thus we have symmetry.

Can you do something similar to show the transitivity?

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    Could you show how it would go withe the way you think it is worded? – 2017-01-25
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    @HighSchool15 you may want to check my answer now – 2017-01-25
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    Your terminology is incorrect. Where you have "transitive" you should have "reflexive". Transitive means $a\sim b \land b\sim c\implies a\sim c$. ... As per the comment by drhab, I think the OP mis-stated the Q. – 2017-01-26
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    @user254665 You are most right, going to change it. – 2017-01-26