$x-y> 1$. How is this relation neither symmetric nor anti symmetric?

Question

I have a assigned worksheet that claims so.

"This is irreflexive, as $x – x = 0$, which is never $> 1$.

This is neither symmetric nor antisymmetric, example: $x=5, y=1$ works, $y=5, x=1$ doesn’t (would imply antisymmetric), but $x=2, y=1$ doesn’t work and neither does $x=1, y=2$, so can’t be antisymmetric.

This is transitive. If the difference between $x$ and $y$ is greater than $1$ and between $y$ and $z$ is greater than $1$, then the difference between $x$ and $z$ must be greater than $2$."

I can appreciate that the relation is not symmetrical but am having confusions about the anti-symmetry.

The way I understand anti-symmetry in relations is that if $(x,y)$ values exist such that $xRy$ and $yRx$, then $x=y$. Now, since there are no values $(x,y)$ that hold the relation, we cannot dismiss this anti-symmetry. I have stumbled upon similar cases in discrete mathematics where if you cannot dismiss a hypothesis, it is accepted as true. Wouldn't it be true here too?

Why does $x=2, y=1$ doesn’t work and neither does $x=1, y=2$ dismiss anti-symmetry? Which condition of anti-symmetry is this breaking?

It would be helpful if you could also state its transitivity and reflexiveness.

Slide from my lecture regarding anti-symmetry.

Answer 1

To clarify the relation R, xRy when x - y > 1.
R is not reflexive because not (x - x > 1).
R is not symmetric because xRy and yRx implies 0 > 2.
R is transitive. If xRy, yRz, then x - y and y - z > 1.
Thus x - z > 2.

Answer 2

Let's recall the definitions. A relation $\sim$ is $\textit{reflexive}$ if $x\sim x$, $\textit{transitive}$ if $x\sim y$ and $y\sim z$ imply $x\sim z$, $\textit{symmetric}$ if $x\sim y$ implies $y\sim x$, and, $\textit{antisymmetric}$ if $x\sim y$ implies $¬(y\sim x)$ instead.

The relation used above is "$x\sim y$ if $x-y>1$". Let's check the properties.

1. $x\sim x$ ? No, because the set $\{x\in\mathbb{R}: x-x>1\}$ is empty.
2. $x\sim y \text{ and } y\sim z\implies x\sim z$ ? Yes, because if $x-y>1$ and $y-z>1$, addition leads to $x-z>2 \implies x-z>1\iff x\sim z$.
3. $x\sim y \implies ¬(y\sim x)$? This is true, because $$ x\sim y\implies x-y>1 \implies y-x<1\implies ¬(y\sim x) $$

So this relation is transitive and antisymmetric, but not reflexive. Antisymmetry excludes symmetry.