Can a strict order relation imply an equality relation?

Question

Let's assume we have a partial or total order relation $R$ defined on a set $S$.

If $R$ was not strict (i.e. it denoted $\leq$ instead of $<$), an equality relation $E$ could be defined as such:

$$ xEy:=xRy\land yRx $$

In other words:

$$ x=y:=x\leq y\land y\leq x $$

For all $x,y\in S$.

But if $R$ was strict (i.e. it denoted $<$ instead of $\leq$), I can't think of a way that $E$ could be defined using $R$. Is it possible to do so?

Answer 1

For a total order, yes. $xEy \iff (\lnot (x R y) \land \lnot (y R x))$. I don't see any way to use a strict partial order to define equality.

Answer 2

Not quite. You are confusing $\vee$ with $\&$. What you want is that:

$$ x = y := x\leq y \ \& \ y \leq x$$.

But this only works if the set with the order relation on it is already known to be connected. That is:

$$x,y \in S \implies x \leq y \vee y \leq x $$

Also, there is no way that we can define equality simply in terms of strict-inequalities (by definition).

Answer 3

This is the closest I can get:

Define $x_\lt =\{y | y \lt x\}$ and $x_\gt =\{y | y \gt x\}$, then the relation defined by

$$xRy \text{ iff } x_\lt = y_\lt \text{ and } x_\gt = y_\gt$$

is definitely an equivalence, but I think you'd have to make other assumptions about your ordered set to make it coincide with equality.

Answer 4

For general partial order, no, you can't.

Let $S=\{a,b\}$ and $R$ be the empty relation, i.e. we have that it is strict ($¬aRa$ and $¬bRb$) as well as $¬ aRb$ and $¬bRa$.

Now you can't distinguish between $a$ and $b$ using only without explicitly using the symbol $=$, that is: if $φ(x,y)$ is a formula that doesn't use $=$ then we have $φ(a,a)⇔φ(a,b)⇔φ(b,a)⇔φ(b,b)$