1. Relations prove

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2. Originally Posted by Cronus
2)Prove:
Every maximal chain in a finite poset contains a minimal element of the poset.
A chain is totally ordered subset. Because of the finite character every chain in a finite poset has a maximal term.
If that term is not maximal in the whole poset then the chain is not maximal.

Originally Posted by Cronus
1)Prove:
We necessarily get an equivalence relation when we form the transitive closure of the symmetric closure of the reflexive closure of a relation.
What is the definition of each of: transitive closure, symmetric closure and reflexive closure of a relation?

3. Originally Posted by Plato
What is the definition of each of: transitive closure, symmetric closure and reflexive closure of a relation?
I don't really have definitions... What I know is that if a relation R on a set S fails to have a certain property, you may be able to extend R to a relation R* on S that does have that property.

So for example for S = {0,1,2,3} and R={<0,1>,<0,2>,<1,1>,<1,3>,<2,2>,<3,0>} we know that R is not reflexive, symmetric or transitive. However, if we add the ordered pairs <0,0> and <3,3>, these pairs along with the original ones will give me the closure of R with respect to reflexivity.

4. Well I would assume that there must be some minimal additions to accomplish closure.
For example, just by uniting the diagonal with any relation makes it reflexive.

Also, I find this to be a trivial question for this reason.
Any relation that is reflexive, symmetric, and transitive is by definition an equivalence relation.

5. Yeah, but I think the order in which you form your closures matters... I.E. If the transitive closure adds something that messes up with the symmetric qualities, then we're screwed!

6. So got b... still working on a... any ideas?

This is awesome!

7. Originally Posted by Cronus
Yeah! THis forum rocks my socks
!!