Symmetric and transitive but not reflexive

Question

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?

See Equivalence relation on a set $X$ : simmetry and transitivity are expressed as "for all $a,b,c \in X$ if ..." while reflexivity is stated as : "for all $a, \ aRa$ ". There is no "if" : it must hold for all elements of $X$. — Mauro ALLEGRANZA, Sep 10 '15 at 18:48

Plutoro · Accepted Answer · 2015-09-10T17:55:46.953

3

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.

edited Sep 10 '15 at 17:55

answered Sep 10 '15 at 17:48

Plutoro

23,378

How can 6C1? The comma is not a binary operation. So, e.g., 5.5C5.6 because 5<5.5,5.6<6. – Avatrin Sep 10 '15 at 18:15
$6C1$ Since 4 is an integer such that $4<6$ and $4+1=5>1$. – Plutoro Sep 10 '15 at 18:18
Oh, I see. I interpreted that as meaning that both x and y had to be between n and n+1. So n<x<n+1 and n<y<n+1. – Avatrin Sep 10 '15 at 19:49
I think you are actually right. Funny how that can be interpreted in different ways. – Plutoro Sep 10 '15 at 19:52

Symmetric and transitive but not reflexive

1 Answers1