Both $R_1$ and $R_2$ are equivalence relations on the set $A$.($A$ is finite)

Define the binary relation $R$ on $A\times A$as follows.

$$R = \{((a_1, a_2),(b_1, b_2)) | (a_1, b_1) \in R_1,(a_2, b_2) \in R_2\}$$

Prove that $R$ is an equivalence relation.

I know a binary relation including reflexivity, Symmetry Transitivity,and assuming $A=\{a_1,a_2,\ldots,a_n\}$ then $$A \times A=\{(a_1,a_1),(a_1,a_2), \ldots,(a_n,a_n)\}$$

But I have no idea how to prove it.


1 Answer 1


One way to unravel the definition is to use infix notation for relation $R.$ This is a relation on $A \times A,$ so it holds or not between two pairs $(a,b)$ and $(c,d)$ in the product $A \times A.$ in infix notation then: $$ (a,b)R(c,d) \leftrightarrow aR_1 c\ \rm{and}\ bR_2 d. \tag{1}$$

Reflexive property: This holds provided for all $(x,y) \in A \times A$ we have $(x,y)R(x,y)$ or using $(1),$ $$ xR_1 x \ \rm{and} \ yR_2 y,$$ which follows from reflexivity of $R_1$ and $R_2$ individually, since they are assumed equivalence relations.

Symmetric property: This holds provided for all $w,x,y,z$ in $A$ we have if $(w,x) R (y,z)$ then also $(y,z)R (w,x),$ and again unraveling by using $(1)$ we see this holds because $R_1$ and $R_2$ are each individually symmetric since they are asumed to be equivalence relations.

Transitive property: Here we take any six elements $u,v,w,x,y,z \in A$ and then assume that both $(u,v)R(w,x)$ and $(w,x)R(y,z)$ hold. We then wznt to show that then $(u,v)R(y,z)$ holds also. Again a careful application of the re-write in $(1)$ finishes things, this time since $R_1$ and $R_2$ are each individually transitive.

Added note: The above argument did not assume $A$ is finite. But of course it would thus cover that case. In my opinion, the notation would have been more cumbersome if we had to work with a specific listing of the elements of $A$ in the finite case.

  • $\begingroup$ Thanks any way, but according to my statement ,$R$ should be $$ (a,c)R(b,d) \leftrightarrow aR_1 b\ \rm{and}\ cR_2 d $$ , instand of $$ (a,b)R(c,d) \leftrightarrow aR_1 b\ \rm{and}\ cR_2 d $$. But I don't think this hurt the provement.Thanks again! $\endgroup$
    – Chen Jason
    Mar 29, 2021 at 5:15
  • 1
    $\begingroup$ @ChenJason You're right. In my setup I was defining $(a,b)R(c,d)$ and should have put $aR_1c \ \rm{and} \ bR_2d.$ If you interchange $b$ and $c$ in your correction it matches mine. $\endgroup$
    – coffeemath
    Mar 29, 2021 at 10:34

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