Symmetric and transitive but not reflexive I am trying to do an exercise in Munkres' Topology where I am supposed to show the mistake in the following proof about equivalence relations:
Since C is symmetric, aCb implies bCa. Since C is transitive, aCb and bCa imply aCa.

This has been asked here before, but I don't get how the answers show a flaw in the argument above; The top answer just make a comment about a specific relation. I couldn't find what the second answer was referring to.
I also saw a supposed counter-example for $x,y \in \mathbb{R}$. In this case $xCy$ if there is an integer $n$ such that $n<x,y<n+1$. This, of course, becomes a problem if one or both of the elements are integers. However, can $\mathbb{Z}$ even be considered part of the set since the relation is never true for integers?

The mistake is that the proof assumes that $a$ relates to anything at all. If $a$ does not relate to anything, then the relation can still be symmetric and transitive, but it is not reflexive. The example you gave is true for some integers (i.e., $6\,C\,1$), but does not appear to be symmetric.
