Reflexive property for equivalence relations

Question

From Munkres, Section #3: Relations:

Define two points $(x_0, y_0)$ and $(x_1, y_1)$ of the plane to be equivalent if $y_0 - x_0^2$ = $y_1 - x_1^2$. Check that this is an equivalence relation and describe the equivalence classes.

I reckon if I can show this relation satisfies the reflexive, symmetric, and transitive properties then I will have solved the problem. But how do I apply the reflexive property to this problem?

This is what I understand:

-A relation on a set $A$ is a subset $C$ of the cartesian product $A \times A$.

-A relation $xCy$ is read as $(x, y) \in C$

-A relation is reflexive if $xCx$ for every $x$ in A.

Now I can think of the plane that the two points are in as the cartesian product $\Bbb{R} \times \Bbb{R}$. So I will have shown that the relation is reflexive if I can show $xCx$ for every $x$ in $\Bbb{R}$.

But how do I show this? Doesn't $xCx$ imply that $(x,x) \in C$? How does this $x$ relate to $(x_0, y_0)$ and $(x_1, y_1)$ in the stated problem? I don't think we are saying that all points where the $x$ and $y$ coordinates are equal are equivalent? I get that $y_0 = x_0^2$ is a parabola, and that we are describing parabolas in some way, but somehow it still seems unclear. Suppose the problem had said our points were equivalent if $y_0 - x_0^2$ = $2y_1 - x_1^2$. What would change?

Answer 1

For your given relation, obviously we have $y_0-x_0^2=y_0-x_0^2$ and therefore it is reflexive.
(Any two points $P_1$ and $P_2$ on the plane are equivalent if they are on the same parabola $y=x^2+k$ for some $k.$)

On the other hand, if you had something like $(x_0,y_o)\sim(x_1,y_1)$ if $y_0 - x_0^2=2y_1- x_1^2,$ the relation is reflexive for all the point such that $\{(x,0) : x\in\Bbb{R}\}.$

Answer 2

Your mistake is in saying that the relation is reflexive if $x\,C\,x$ for every $x$ in $\Bbb R$. But $C$ is a relation on $\Bbb R^2$, not on $\Bbb R$, so it should be: the relation is reflexive if $x\,C\,x$ for every $x$ in $\Bbb R^2$. Or to put it more clearly, if

$(x,y)\,C\,(x,y)$ for all $(x,y)\in\Bbb R^2$.

This means $y-x^2=y-x^2$, which is obviously true.

Answer 3

Your mistake has been revealed allready in the other answers. This is some convenient information concerning equivalence relations.

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If $f:X\to Y$ is a function then relation $R$ on $X$ prescribed by: $$aRb\iff f(a)=f(b)$$ is an equivalence relation.

This is quite easy to prove:

• $f(a)=f(a)$ (reflexive)
• $f(a)=f(b)\implies f(b)=f(a)$ (symmetric)
• $f(a)=f(b)\wedge f(b)=f(c)\implies f(a)=f(c)$ (transitive)

Further the equivalence classes are the non-empty fibers of $f$.

Under the extra condition that $f$ is surjective every fiber is non-empty and a one-to-one relation between $Y$ and the collection of equivalence classes exists. It links $y\in Y$ with equivalence class $f^{-1}(\{y\})=\{x\in X\mid f(x)=y\}$.

You can apply this on function $f:\mathbb R^2\to\mathbb R$ prescribed by $\langle x,y\rangle\mapsto y-x^2$.