thehagman:
Ok, I'm sure it's me... I just want to understand the conceptual approach. Let me start by explaining my understanding of "symmetry".
If I have (1,2) and (2,1)... I have symmetry.
If I have (1,2), and (2,1) and (2,3) but not (3,2), I don't have "big picture" symmetry (balance), right.
Applying that "big picture balance), I tried to understand if I need to apply this also for transitivity.
Yes, as I can go from (1,2) and (2,4), I have transitivity from (1,3).
On the other hand, as (3,2) is missing, I don't have transitity for (3,1).
Now, giving that I don't have transitivity on the "bottom" do I still have transitivity on the "top"?
My view is not "balanced", right?
Is that's why you indicated there the S function does NOT have eitehr symmetry, reflexivity, and transitivity?
EEH
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by: thehagmanPosted on 2009-09-18 at 07:47:35ID: 25366520
S not symmetric, not reflexive, not transitive
However your argument needs clarification (esp. the blue arrow makes no sense to me):
To show that S is not transitive sou need to find just one triplet a,b,c such that
(a,b) in S
(b,c) in S
(a,c) not in S
We have (3,1) and (1,2), but not (3,2)
I think You try to argue: We don't have (3,2) and we have (1,2), hence we should not have (3,1) - this is ok, but somewhat confusing.