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pandaBee
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Homework Statement
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?
Homework 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?
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?
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