Would also work?
Why or why not?
Good day to all,
We have just started relations and I came across the following question in my textbook:
Let X = {1, 2}. List all the partial orders that can be defined on X.
My solution:
I began by computing X x X (Cartesian product) which gave me:
X x X ={(1,1), (1,2), (2,1), (2,2)}
Since we are looking for partial orders, the relation has to be simultaneously reflexive, antisymmetric and transitive.
If the relation R on X is reflexive then it must contain (1,1) and (2,2)
The ordered pairs (1,2) and (2,1) cannot belong to R for if they did and R is also antisymmetric then that would imply 1=2, which is false.
Therefore I concluded that the list of partial orders is the set:
{(1,1), (2,2)}
I was wondering if my logic is flawed and if so what errors have I committed?
Finally, is it possible in this problem to determine the number of partial orders (cardinality)?
Any advice would be greatly appreciated.
Kindest regards
Thank you Plato for your quick response.
I am not sure. The definition of antisymmetric states:
for every a,b in X ((a,b) belongs to R and (b,a) belongs to R implies a=b)
If (1,2) belongs to R and (2,1) does not belong to R (based on the set you listed) then the hypothesis of the implication is false which means that the implication is true. If this is a valid reasoning could we not say the same for the set: {(1,1), (2,1), (2,2)}
Slightly confused!
There is something that I am obviously not understanding. (2,1) does not belong to the set you listed.
I believe the set is antisymmetric, since for it not to be antisymmetric we would have to have: and a different than b. Therefore the set you provided is a partial order as well (reflexive, transitive and antisymmetric).
I apologize as I realize this may be obvious to many.