Prove: R∩R−1 is symmetric.

Question

The problem that I'm having is proving it - obviously. The only context that I am provided with is: "Prove: R∩R⁻¹ is symmetric."

If (x,y) ∈ R then (y,x) ∈ R⁻¹, and since it's the intersection, whatever elements are in the intersection must have both (x,y) and (y,x); making it symmetric, but how do I go about formally proving it?

How? Just as you did it there in your second parraph, first two lines up exactly until ",but how...": that is a formal proof. +1 — Timbuc, Apr 09 '15 at 01:20
I think you've basically proven it. All that need be shown is that whenever $(x, y)$ is in the set, $(y, x)$ is also. You've done that. — Brian Tung, Apr 09 '15 at 01:22

score 0 · Answer 1 · answered Apr 09 '15 at 01:22

0

Choose any $(x, y) \in R \cap R^{-1}$. To show that $R \cap R^{-1}$ is symmetric, it suffices to show that $(y, x) \in R \cap R^{-1}$. Indeed, since $(x, y) \in R \cap R^{-1}$, we know that $(x, y) \in R$ and $(x, y) \in R^{-1}$. The former implies that $(y, x) \in R^{-1}$ and the latter implies that $(y, x) \in R$. Hence, we conclude that $(y, x) \in R \cap R^{-1}$, as desired. $~~\blacksquare$

answered Apr 09 '15 at 01:22

Adriano

41,576

Thank you, I just came to this conclusion; I just wasn't sure if it was enough to assume that x and y could be an elements of R. – Justin Apr 09 '15 at 01:25

Prove: R∩R−1 is symmetric.

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