Hi Math people. Can someone tell me what's wrong with my proof, below? My professor did not give me credit for it, but did not provide feedback, either. She posted the answers online and her proof took a completely different tack than mine.

I get that I did the proof differently. But I'm curious what is wrong with my proof, especially the first part which I was pretty confident of. But all feedback is appreciated.

My approach, for the 'left-to-right' proof, was to take the two possible cases (x=y and x not= y) and show that only the expected on was in R and R inverse. Seems pretty straightforward to me.

Thanks, everyone.

: A relation $\displaystyle R$ on a set $\displaystyle A$ is antisymmetric if and only if $\displaystyle R \cap R^{-1} \subseteq \{(a,a)\colon a \in A\}$.Prove

: ``$\displaystyle \Rightarrow$'': Assume $\displaystyle R$ is antisymmetric. That means $\displaystyle xRy$ and $\displaystyle yRx$ implies that $\displaystyle x=y$.Proof

Suppose $\displaystyle x,y \in A, x=y$, and $\displaystyle xRy$. That means $\displaystyle (x,y) \in R$ and $\displaystyle (y,x) \in R^{-1}$. Since $\displaystyle x=y$, we have that $\displaystyle (x,x) \in R, (x,x) \in R^{-1}$, therefore $\displaystyle (x,x) \in R \cap R^{-1}$.

Now, suppose $\displaystyle x,y \in A, x \ne y$. Since $\displaystyle R$ is antisymmetric, we can't have both $\displaystyle xRy$ and $\displaystyle yRx$. WLOG, say $\displaystyle xRy$, then $\displaystyle (x,y) \in R, (x,y) \notin R^{-1}$ and $\displaystyle (x,y) \notin R \cap R^{-1}$.

``$\displaystyle \Leftarrow$'': Assume $\displaystyle R \cap R^{-1} \subseteq \{(a,a)\colon a \in A\}$.

Suppose $\displaystyle x,y \in A, x \ne y, (x,y) \in R \cap R^{-1}$.\\

$\displaystyle (x,y) \in R \cap R^{-1}\; \Rightarrow\; xRy \text{ and } xR^{-1}y\; \Rightarrow\; yRx$.

So we have $\displaystyle xRy,\, yRx$, and $\displaystyle x \ne y$, therefore, $\displaystyle R$ is not antisymmetric. In other words, if $\displaystyle x \ne y$ and $\displaystyle (x,y) \in R$, and $\displaystyle (x,y) \in R^{-1}$, then $\displaystyle R$ is not antisymmetric.

But since we have that only ``twin'' elements may be in $\displaystyle R \cap R^{-1}$, $\displaystyle R$ must be antisymmetric (that's my argument and I'm sticking to it!).