Prove that relation $R\subseteq X\times X$, where $X= \mathbb{R}\times\mathbb{R}$, is an equivalence relation and construct its equivalence class. $R$ is defined as: $$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \Longleftrightarrow x_1^2+y_1^2 = x_2^2+y_2^2$$

The way I did prove that it is an equivalence relation was based on equality relationship between two or more quantities. To make it easier to read, we can also substitute $a = x_1^2+y_1^2$, $b = x_2^2+y_2^2$, $c = x_3^2+y_3^2$, at least I think we can.

The first property is reflexivity:
$\langle x_1, y_1\rangle R \langle x_1, y_1\rangle \Longleftrightarrow x_1^2+y_1^2 = x_1^2+y_1^2$, which is always true, as $a=a$.

Then we check for symmetry:
$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \Longrightarrow \langle x_2, y_2\rangle R \langle x_1, y_1\rangle$, so if $x_1^2+y_1^2 = x_2^2+y_2^2$, then $x_2^2+y_2^2 = x_1^2+y_1^2$. Using the substitution we can say that $(a=b) \Rightarrow (b=a)$, which is also true.

The last one is transitivity:
$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \wedge \langle x_2, y_2\rangle R \langle x_3, y_3\rangle \Longrightarrow \langle x_1, y_1\rangle R \langle x_3, y_3\rangle$, which means $x_1^2+y_1^2 = x_2^2+y_2^2 \wedge x_2^2+y_2^2 = x_3^2+y_3^2 \Longrightarrow x_1^2+y_1^2 = x_3^2+y_3^2$, using substitutions it is $(a=b \wedge b=c) \Longrightarrow (a=c)$.

Up to this moment I think that my line of thinking is quite correct, but I am not sure whether that proof is good enough and it is what I would like to know.

When it comes to equivalence class of $R$, I would say that the relation tells us about points on the common circle, but I have to write it up with the definition, which is:

Equivalence class of element $a\in A$ in regard to equivalence relation $R\subseteq A\times A$ is a set $[a]_R = \{b\in A\: |\: a R b\}$.

Sadly I can not get it completely, but I thought about
$$[\langle a,b \rangle]_R = \{\langle a, b\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$$ I would like you to tell me whether my proof (also usage of substitutions) and construction of equivalence class is right, also how could I get grasp of it, as lectures I attend are not good enough.

  • $\begingroup$ You first part(the relation part) looks good, the second part it should be $[\langle a,b \rangle]_R = \{\langle x, y\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$: this is the set of all elements of $A\times A$ that are in relation with $\langle a,b\rangle$(here $A$ is indeed $\Bbb R\times \Bbb R$) $\endgroup$ – Holo Dec 16 '18 at 21:14

I like your proof, but I'd write the equivalence classes as follows:

$$ [(a,b)]_R=\{(x,y)\in \mathbb{R}^2|a^2+b^2=x^2+y^2\} $$

which translates to: "The equivalence class of $(a,b)$ is the set of all points $(x,y)$ having the same distance from 0 as $(a,b)$.


Your idea is quite good and can be better formalized.

Consider a set $X$ and a map $f\colon X\to Z$, $Z$ any set. Then you can define a relation $\sim_f$ on $X$ by decreeing that $$ x\sim_f y\quad\text{if and only if}\quad f(x)=f(y) $$ Then $\sim_f$ is clearly an equivalence relation: the proof is easily based on the properties of equality, which is essentially what you did just with more complicated symbols.

For $a\in X$, its equivalence class is $[a]_{\sim_f}=\{x\in X:f(x)=f(a)\}$.

In your case, $X=\mathbb{R}\times\mathbb{R}$, $Z=\mathbb{R}$ and $f(x,y)=x^2+y^2$ (square of the distance of $(x,y)$ from the origin).

Thus the equivalence class of $(a,b)\in X$ is the set of all points in $X=\mathbb{R}\times\mathbb{R}$ that share the same distance from the origin as $(a,b)$.

  • $\begingroup$ Is it more correct to say "and a map $f\colon X\to Z$, where $Z$ is any set"? $\endgroup$ – manooooh May 23 at 19:14
  • 1
    $\begingroup$ @manooooh As correct. $\endgroup$ – egreg May 23 at 19:32

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.