How to show that $ \succ acyclic \implies \succ asymmetric $

Question

For a preference relation defined as

$$ \succ := \{ (x,y) \in X\times X : x\ is\ better\ than\ y \}$$

one has to show that

$$ \succ acyclic \implies \succ asymmetric $$

whereas

$$ acyclic := \forall\ n \ge 2,\ x_1 \succ x_2,\ x_2 \succ x_3,\ ... \ x_{n-1} \succ x_n \implies x_1 \ne x_n$$

and

$$ asymmetric := x \succ y \implies y \nsucc x$$

Now as much as I believe the solution should be obvious, unfortunately for me it's not.

My approach whould be to show that this is true for $n =2$ first of all, and then proof via induction.

However, even for $n = 2$, I do not manage to establish that the proposition holds, because the acyclic property is only

$$ x_1 \succ x_2 \implies x_1 \ne x_2 \implies x_1 \succ x_2 \lor x_2 \succ x_1$$

which is far from implying that $\succ$ is also asymmetric, i.e.,

$$ x_1\succ x_2 \implies x_2 \nsucc x_1$$

Which key point do I miss here?

Is it even smart to go for proof by induction here, or is there a better approach?

Thanks for all hints!

Answer 1

Assume that it is not asymmetric. There there are $x_1, x_2$ such that $x_1>x_2$ and $x_2>x_1$. Applying the acyclicity for $n=3, x_3=x_1$, you get $x_1\ne x_1$, a contradiction.

Answer 2

Assume that a relation $R$ is acyclic. Let $x$ and $y$ such that $xRy$ and $yRx$. Then $(x,y,x)$ is a $2$-cycle if $y\ne x$. This would contradict the acyclicity of $R$, hence $x=y$. Thus, $[xRy\land yRx]\implies[x=y]$, that is, $R$ is asymmetric.

Answer 3

Asymmetry is simply the $n=3$ case of acyclicity. In particular, the following are equivalent: $$(x\succ y)\wedge(y\succ z)\Longrightarrow(x\ne z)\\\neg\bigl((x\succ y)\wedge(y\succ z)\bigr)\vee(x\neq z)\\\bigl(\neg(x\succ y)\vee\neg(y\succ z)\bigr)\vee\neg(x=z)\\\neg(x\succ y)\vee\bigl(\neg(y\succ z)\vee\neg(x=z)\bigr)\\\neg(x\succ y)\vee\neg\bigl((y\succ z)\wedge(x=z)\bigr)\\\neg(x\succ y)\vee\neg(y\succ x)\\\neg(x\succ y)\vee(y\not\succ x)\\(x\succ y)\Longrightarrow(y\not\succ x)$$