I'm currently learning about equivalence relations. I understand that an equivalence relation is a relation that is reflexive, symmetric, and transitive. But I'm having trouble proving the transitive property.

For example:

Suppose we have the relation ~ on the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ by the rule f~g iff there is a $k \in \mathbb{R}$ such that $f(x) = g(x)$ for every $x \geq k$ is an equivalence relation.

$$$$ To prove that it is reflexive I said:

Suppose we have a real number k such that for every $x$, $x \geq k$. Then we can choose any k and f(x) = f(x) for all x. So f~f. Thus the relation is reflexive.

^It seems a little short.. and trivial so I'm wondering if this part is even right. $$$$ To prove that it is symmetric I said:

Suppose we have two functions such that f(x) = g(x). Then we can similarly say that g(x) = f(x). So it follows that f~g and g~f. Therefore the relation is symmetric

^Again, I'm a little confused about this. It feels too short.

$$$$ Now I'm having trouble proving the transitive property.

Could someone look over my proofs for the reflexive and symmetric properties of the relation and give me a hint on how to approach the transitive property?


In the argument for symmetry, you've shown that if $f = g$ then $f \sim g$ and $g \sim f$. But this does not guarantee symmetry, which requires that if $f \sim g$ then $g \sim f$. To do this, $f \sim g$ shows that there is a constant $k$ such that $f(x) = g(x)$ for all $x \geq k$, and you need to show that there is a constant $l$ such that $g(x) = f(x)$ for all $x \geq l$.

For transitivity, you must show that if $f \sim g$ and $g \sim h$ then $f \sim h$. In your case, this means there are constants $k$ such that $f(x) = g(x)$ for $x \geq k$ and $l$ such that $g(x) = h(x)$ for $x \geq l$, and you need to show that there is a constant $m$ such that $f(x) = h(x)$ for all $x \geq m$. How might you produce such an $m$ in terms of the other data you have available?

  • $\begingroup$ So was my reflexive proof ok? As for symmetry, should I be saying something like: Suppose f~g, then we can say that f(x) = g(x) for some $x \geq k$ but this is also equivalent to saying g(x) = f(x), so g~f. For the transitivity part can we consider k > l? Then we know that x is also greater than l, so f(x) = h(x)? $\endgroup$ – Pandamonium Oct 11 '14 at 2:40
  • $\begingroup$ Yes, it looks good to me. In a proof that a relation is an equivalence relation, reflexivity and symmetry tend to be relatively easy, and may feel almost tautological, whereas transitivity is often (but not always) somewhat more involved. $\endgroup$ – Travis Willse Oct 11 '14 at 2:43
  • $\begingroup$ So for symmetry can we just consider k = l? Also, for transitivity, I didn't really produce an m. I just considered k > l. Is this sufficient for proving transitivity? $\endgroup$ – Pandamonium Oct 11 '14 at 2:48
  • $\begingroup$ For symmetry, that's right---in this case the equivalence relation is defined in terms of equality, which is itself symmetric, so there's not much to prove. For transitivity, you must show there is some $m$ that satisfies the condition. This doesn't mean you have to produce such an $m$, but doing so is often the easiest way of doing this. I think you have the right idea, but you need to be careful about wording. A priori, you just know there are $k$ and $l$ that satisfy the criteria, and you need to make some comment about why you can assume, e.g., that $k > l$. $\endgroup$ – Travis Willse Oct 11 '14 at 3:04

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