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.