Let $R^2$ be a binary relation on some set X. I need to prove or disprove the following claim:
"If $R^2$ is antisymmetric, then R is antisymmetric."
I know the opposite claim (if R is ANS then $R^2$ is ANS) is false, but I have struggled to find a counterexample for the former claim. Therefore I am leaning towards it being true, and wanted to prove it using the "set" definiton of antisymmetry, which says:
"R is antisymmetric ⇐⇒ R ∩ R^(-1) ⊆ ∆(X)." (∆(X) = {(x, x) ∈ X × X; x ∈ X} being a diagonal relation and R^(-1) being the inverse relation to R)
The claim would then look like this: "If $R^2$ ∩ ($R^2$)^(-1) ⊆ ∆(X), then R ∩ R^(-1) ⊆ ∆(X)."
Alternatively I could also use the more common definition of antisymmetry:
"R is antisymmetric ⇐⇒ (∀x, y ∈ X : xRy ∧ yRx ⇒ x = y)."
I have gotten stuck and can't seem to get to any conclusion. Thanks in advance.
You can't prove it, since it is false. On $\{1,-1\}$, consider the binary relation $R$ defined by $a\mathrel Rb$ if $ab=-1$. Then $a\mathrel{R^2}b\iff a=b$. So, $R^2$ is antisymmetric. But $R$ isn't (you have $1\mathrel R-1$ and $-1\mathrel R1$, but you don't have $1\mathrel R1$).