Set Theory

[odumath]

I am taking a class that covers "set theory". Please see URLs below for the problem and my answer.

Problem:
http://img33.imageshack.us/img33/2795/problemx.jpg

My Answer:
http://img30.imageshack.us/img30/3598/answerx.jpg


I have no trouble following the function "R" (first function). For this one (I think) the answer is "yes" for all... reflexive, symmetry, and transitive.

On the 2nd function (S), I am not totally sure about the "transitive" relation though.

Based on the 2nd diagram, all relations are drawn: S = {(2, 3), (1, 2), (2, 1), (3, 1), (1, 3)}

What I don't understand is the following:

1. While (3, 1) is given, should it or should it NOT be drawn? Why? (3,2) doesn't exist, so logically (3,1) shouldn't exist, right?

2. Based on the previous statement, I could argue though that it does not matter whether (3,2) exist. I still should be able to go from C to A.


Again, given that (3,2) does not exist, I'm not sure if the provided (transitive) relationships (1,3) and (3,1) should be included or excluded.

Finally, assuming that I should leave (1,3) in the graph, is this "one-way" relationship TRANSITIVE OR NOT?

Thanks,
odumath

 

[daon]

I've never seen these diagrams, but if 3~1 then yeah, draw it. I'm not sure why you think (3,1) shouldn't exist (they gave you a specific relation in which is DOES). This is an example of a non-symmetric and non-transitive relation.

\(\displaystyle (3\sim 1) \,\, and \,\, (1\sim 2) \,\, but \,\, 3 \not \sim 2\)

 

[odumath]

daon,

could you pls expand as to why it's NOT transitive?

Thanks,
odumath

 

[daon]

I believe I showed exactly why it isn't. Transitivity means the following must be true: if aRb and bRc then aRc I picked a=3, b=1, c=2. I don't think it could be made any clearer...

 

[odumath]

Thanks, Daon.

I appreciate your help.

odumath