Can R be a subset of S and still not have the same reflexivity as S?

[tn0c]

Hi, this is my first time posting here, and I am trying to prove the following proofs and I do not know how to start:

Suppose R and S are relations on set A

1. If R is reflexive and R is a subset of S, then S is reflexive.
2. If R is symmetric and R is a subset of S, then S is symmetric.
3. If R is transitive and R is a subset of S, then S is transitive.

I understand the definitions well enough, but I do not know how to specify these conditions.
I am not certain that any of these are true. Any help would certainly be appreciated.

Attempt on #1:

Assume R is reflexive
Assume R is a subset of S.

Since R is reflexive, for all x in R, xRx ---> xRx
and since R is a subset of S, all x in R is in S as well.

Thus x in S, xSx ---> xSx (this is the step I am not sure of, nor do I know if I have anything else to work off of.)

Therefore S is reflexive. The other two questions I set up the same way, but I use the other definitions. It definitely is not enough, and in the original problem it states (prove or show counterexample). So I am wondering if this is suitable.

1. let R, S be relations on A
R= { (1,1) (2,2) (1,2) (2,1)}
S= { (1,1) (2,2) (1,2) (2,1) (1,3) (2,3) (3,2) (3,1)}

Clearly R is reflexive and R is in S. But S is not reflexive as it does not contain (3,3).

So for reflexivity at least this does not seem to be true. I am however trying to figure out if that is the case for symmetric and transitive.

 

[Dick]

Hi, this is my first time posting here, and I am trying to prove the following proofs and I do not know how to start:

Suppose R and S are relations on set A

1. If R is reflexive and R is a subset of S, then S is reflexive.
2. If R is symmetric and R is a subset of S, then S is symmetric.
3. If R is transitive and R is a subset of S, then S is transitive.

I understand the definitions well enough, but I do not know how to specify these conditions.
I am not certain that any of these are true. Any help would certainly be appreciated.

Attempt on #1:

Assume R is reflexive
Assume R is a subset of S.

Since R is reflexive, for all x in R, xRx ---> xRx
and since R is a subset of S, all x in R is in S as well.

Thus x in S, xSx ---> xSx (this is the step I am not sure of, nor do I know if I have anything else to work off of.)

Therefore S is reflexive.


The other two questions I set up the same way, but I use the other definitions. It definitely is not enough, and in the original problem it states (prove or show counterexample). So I am wondering if this is suitable.

1. let R, S be relations on A
R= { (1,1) (2,2) (1,2) (2,1)}
S= { (1,1) (2,2) (1,2) (2,1) (1,3) (2,3) (3,2) (3,1)}

Clearly R is reflexive and R is in S. But S is not reflexive as it does not contain (3,3).

So for reflexivity at least this does not seem to be true. I am however trying to figure out if that is the case for symmetric and transitive.

You are missing the sense of the 'subset'. R is a subset of AxA, i.e. ordered pairs (x,y) where x is in A and y is in A. If A={1,2,3} then your example for R isn't reflexive either. It's not reflexive on A={1,2,3}. It's only reflexive on the subset of A, {1,2}. That's not A. Make R reflexive on {1,2,3} and then say why if R is a subset of S as a set of ordered pairs, then S is reflexive.