R is symmetric iff R = R^-1

[Testify]

Homework Statement

In Velleman's "How to Prove it", he gives a proof that "R is symmetric iff R = R^-1, which I find to be confusing when he is proving that $$R^{-1}\subseteq{R}$$:

Now suppose $$(x,y)\in R^{-1}$$. Then $$(y,x)\in R$$, so since R is symmetric, $$(x,y)\in R$$. Thus, $$R^{-1}\subseteq R$$ so R=R^-1

It seems to me that he is saying that since $$xRy\rightarrow yRx$$ and yRx, xRy, which makes no sense.

Basically my question is this: how this part of his proof could be correct?

 

[HallsofIvy]

It would help if you would tell us what "R" is! A relation?

It seems to me that he is saying that since xRy$\rightarrow$ yRx and yRx, xRy, which makes no sense.

If R is a relation, then it is a set of ordered pairs. $R^{-1}$ is defined as the set of pairs $\{(x, y)| (y, x)\in R\}$.

What he is saying is that if (x,y) is in R^-1, then (y, x) is in R. Since R is symmetric, (x, y) is in R and so $R^{-1}\subset R$.

 

[Testify]

Ah, okay. I guess I was stuck thinking that (x,y) was in R and didn't consider that R being symmetric could mean that if yRx then xRy.

And yes, R was a relation haha.

Thanks!