# Homework Help: Does a refl/anti-symm relation on a set A have this property?

1. Dec 2, 2014

### pandaBee

1. The problem statement, all variables and given/known data
Let $R$ be an ordered relation on a set $A$ that is reflexive and anti-symmetric.
If there is a chain of elements in $R$ that begins and ends with the same element, say the element $x \in A$ is it true that all the elements of $R$ sandwiched in between the ones beginning and ending in $x$ are equal to each other?

2. Relevant equations
For clarification Let the 'chain' of elements in $R$ be the following:
$(x,b_0), (b_0,b_1), ..., (b_n, x)$
let $Z = \{b_0, b_1, ..., b_n\}$
If $R$ is a reflexive and anti-symmetric ordered relation are all elements of $Z$ equivalent?
3. The attempt at a solution
I came up with a few antisymmetric, reflexive ordered relations and tested the validity of this argument and it passed in each one so I am guessing that the theorem above may be true since I have not found a counter-example yet. I just have no idea how to express it in a proof. The examples I came up with were:

The set $G=\{(x,y) \in P\text{ x }P~ |~ x \ \text{genes come from}\ y\}$ where $P$ is the set of all people,
The set $D=\{(x,y) \in ℤ\text{ x }ℤ ~|~x \ \text{divides}\ y\}$
And also the standard less than or equal to ordered relation.

In every case it was implied that if a element of the set under which the relation is ordered under begins and ends a chain of ordered pairs in the ordered relation, then all of the elements are equivalent to one another, as a consequence of anti-symmetry and reflexivity.

I'm unsure of how to start this proof. I've tried defining a set containing all the intermediate elements (in this case being the $b_j$'s in the set $Z$ and assuming that there is at least two elements that are not equal to one another in the set and trying to find a contradiction, but I ran into a dead end there as well. Any suggestions? Or perhaps someone can think of a counter example ruling out the theorem completely?

Last edited: Dec 2, 2014
2. Dec 3, 2014

### RUber

I think in general, you only need to show that it is true for any element y such that x R y R x implies that x=y.
I do not have a clear mental image of a rigorous proof, but I would start with the basics:

reflexive means that x R x for all x.
anti-symmetric means that x R y implies y ~R x.
So, x ~R x. If y R x and y ~R x, does this mean y = x?

Also, I think ordering is useful since there should not be anything between x and x in an ordered relation other than more x.