You are mostly correct.
Any one element relation is transitive by default. So is.
is not anti symmetrical,
The first part of my assignment was creating 9 relations based on the set M := {A,B,C} itself with ascending cardinality
I ended up having this, I was assured that this is correct:
R1= {(A,A)}
R2= {(A,A),(A,B)}
R3= {(A,A),(A,B),(A,C)}
R4= {(A,A),(A,B),(A,C),(B,A)}
R5= {(A,A),(A,B),(A,C),(B,A),(B,B)}
R6= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C)}
R7= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A)}
R8= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A),(C,B)}
R9= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A),(C,B),( C,C)}
Ok now I need to check R1, R2, R4 for properties.
R1= {(A,A)}
R1 is not reflexive, because it does not contain (B,B) and (C,C)
R1 not sure about transitivity. Can I use (A,A) for the formal definiton aRb ^ bRc => aRc? I dont know how far I can reach out with the definition.
Do I need 3 different pairs or is one pair also enough to prove it? Like: I have (A,A) and (A,A). Therefore I also have (A,A)?
R1 is symmetrical (A,A) is the same as (A,A)
R1 is also anti symmetrical (even though I struggle with the definition, but I know that in this case it is because A and A are identical.)
R2= {(A,A),(A,B)}
R2 is not reflexive, because it does not contain (B,B) and (C,C)
R2 is transitive? Almost same case like above. Can I prove like that? (A,A) and (A,B), therefore (A,B)
R2 is not symmetrical?
R2 is anti symmetrical because (A,A) is also a part of (A,B)
R4= {(A,A),(A,B),(A,C),(B,A)}
R4 is not reflexive, because it does not contain (B,B) and (C,C)
R4 totally no clue at all if it is transitive.
R4 is not symmetrical ?
R4 is anti symmetrical because (A,A) is also in (A,B),(A,C),(B,A)?
This took some time to wrote down, I hope most of it is correct.
Thank you very much!
Could you please be so kind and explain anti-symmetry for a dumbo like me? Is this arguement for example "(A,A) is part of (B,A)" correct and a solid proof?
And why is R2/R4 not symmetrical? I know it makes sense that it is not, but how would I argue?
Is R4 transitive or not?
transitive means, as you say "if aRb and bRc then aRc". Here, where you have "ARB" symbolized by (A, B) that is "if (X, Y) and (Y, Z) then (X, Z)".
For 1) {(A, A)} that is only "if (A, A) and (A, A) then (A, A)"
Of course "if (X, X) and (X, X) then (X, X)" is always true as long as (X, X) is in the relation so we don't often worry about that and can require that "X" is not the same as "Y" and that "Y" is not the same as "Z". But in a case where there are no such pairs, we can use the basic logical fact that "if X is false then the statement 'if X then Y' is TRUE for all Y.
Either way we get that the relation is transitive.
I hate to contradict Plato but R4= {(A,A),(A,B),(A,C),(B,A)} is not transitive because
We have (B, A) and (A, C) but do NOT have (B, C).
Thank you both!
One (maybe stupid question) about the last thing you wrote:
In R2 I said (A,A) ^ (A,B) => (A,B)
Cant I say the same here?
(A,A) ^ (A,B) => (A,B)
and
(A,A) ^ (A,C) => (A,C)
To show transitivity, you must show all pairs that satisfy the pattern yield the correct outcome. So, it is not a question of "can you say the same". You must say them to prove it is transitive.
Okay thank you very much guys! To make it clear and sum it up:
R1= {(A,A)}
R1 is not reflexive
R1 is transitive
R1 is symmetrical
R1 is anti symmetrical
R2= {(A,A),(A,B)}
R2 is not reflexive
R2 is transitive
R2 is not symmetrical
R2 is anti symmetrical
R4= {(A,A),(A,B),(A,C),(B,A)}
R4 is not reflexive
R4 is not transitive
R4 is not symmetrical
R4 is not anti symmetrical
Is this correct?