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For a function $f: A\to B$, I have a relation $@$ on $A$ described by

$(\forall x,y \in A)\quad x @ y \Leftrightarrow f(x) + f(y)$

Is there any way to show that $@$ is an equivalence relation?

user3467433
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  • The answer of Rustyn is fine. Good for you to know: conversely if you start with an equivalence relation @ on set $A$ and $B$ denotes the set of equivalenceclasses then you can construct a function $f:A\rightarrow B$ sending each element of $A$ to the equivalenceclass from wich it is an element. This function $f$ will satisfy $x@y\iff f(x)=f(y)$. In that sense functions, equivalencerelations (and partitions) are closely related. – drhab Apr 03 '14 at 07:39
  • What does it mean $f(x)+f(y)$? – Asaf Karagila Apr 03 '14 at 13:28

1 Answers1

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To check if a relation @ is an equivalence relation, you need to verify three things:

  • @ is reflexive
  • @ is symmetric
  • @ is transitive

For your particular relation, reflexivity is apparent as $f(x) = f(x)$, (so $x@x$). Symmetry is also not very difficult. If $x@y$, do we have $y@x$? Of course, $x@y$ means $f(x)=f(y)\leftrightarrow f(y)=f(x) \leftrightarrow y@x$. Transitivity is also very easy to see, because $=$ itself is a transitive relation. Explicitly if $x@y \land y@z$, this means that $f(x)=f(y) \land f(y)=f(z)$. So, in particular, by transitivity of $=$, we have that $f(x) = f(z) \leftrightarrow x@z$.

Rustyn
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