Proofs of symmetry and transitivity

[stiffy]

Let ? be a partition of set A. Define a relation R on A by: xRy iff (?X??)(x, y?X).
We can define R as R={(x,y)?X×X|?X??(xRy)}

Symmetry: Let x and y be arbitrary elements of A. Since x, y?A, then x, y???. This is because, A=?? (given in another proof). So, ?X in the partition ? such that x,y?X. This implies x?X and y?X. Knowing, X=[x] (given in another proof) we can say x?[x] and y?[x]. We can see that y?[x] gives us the relation yRx by the definition of equivalence classes. But then, (y, x)?X×X, so (y, x)??_(X??) X×X=R. Therefore, R is symmetric.

Transitivity: Let x, y, and z be arbitrary elements of A.
.
.
Same steps as above
.
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So, if x, y, z?X we can conclude X=[x] such that x?[x], y?[x], and z?[x]. We see that yRx and zRx. We have proven symmetry, so we can conclude xRz also. Using quantifiers, ?x??y?A?z?A ((yRx ? xRz)==>yRz). But then, (y,z)?X×X, so (y,z)??_(X??) X×X=R. Therefore, R is transitive.


This is new notation for a concept we already learned, so I am just wont to know if these proofs are correct and if I am using the notation properly and that my assumptions are correct. I am still pretty confused by proofs pertaining to relations.

Thanks,

--Dan

 

[pka]

Let ? be a partition of set A. Define a relation R on A by: xRy iff (?X??)(x, y?X).
We can define R as R={(x,y)?X×X|?X??(xRy)}

Symmetry: Let x and y be arbitrary elements of A. Since x, y?A, then x, y???. This is because, A=?? (given in another proof). So, ?X in the partition ? such that x,y?X. This implies x?X and y?X. Knowing, X=[x] (given in another proof) we can say x?[x] and y?[x]. We can see that y?[x] gives us the relation yRx by the definition of equivalence classes. But then, (y, x)?X×X, so (y, x)??_(X??) X×X=R. Therefore, R is symmetric.

You cannot say that. All you know is x, y?A they may be is different cells of the partition.
The point is if x & y are in the same cell then y & x must be in the same cell. Thus symmetry is automatic.

Transitivity results from the simple fact the different cells are disjoint.

 

[stiffy]

So, for the proof of transitivity all I have to say is...
Let x, y, and z be arbitrary elements of A. Assume xRy and yRz. Then there are sets X and Y in ? such that x,y?X and y,z?Y. Since X and Y are sets in the partition ? of A, they must either be disjoint or equal. It is clear that y is in both set, so the sets must be equal rather than disjoint. Thus, there is a set X=Y that contains elements x and z such that xRz. So, we can conclude that R is transitive.
Correct?

Symmetry...
Let x and y be arbitrary elements in A. Assume xRy. Then x,y?X, where X is some set in the partition ? of A. Thus, x?X and y?X; moreover, y?X suggests yRx. Therefore, R is symmetric.
???

I don't understand when I should use the other notation; namely, R={(x,y)?X×X|?X??(xRy)} and ?X?? X×X=R. I can see that it was not neccesary for this example, because I was already given "Let ? be a partition of set A. Define a relation R on A by: xRy iff (?X??)(x, y?X)." in the claim.

Thanks again for your help.

 

[pka]

Here is a piece of advice. Do not get hung up in notation; get the concept.
I have taught the foundations course for years. Here is one question that I tell classes that will be on their final exam: “There is a two way relationship between equivalence relations on a set and partitions of that set. Give a complete discussion, in Standard English, of that relationship. Be sure you discuss both ‘ways’.”
Could you answer my question?