Can two elements of a set be equal but not identical I have some confusion regarding the equality produced by ordering of sets. My question is this: is it possible for a set to contain two elements which (when compared with some ordering relation) are equal, and nonetheless are distinct elements of the set.
An example of this might be a set of polynomials $\{x^2, x+2, x^3\}$ where they are ordered by evaluating them at $x=2$. Clearly the first two polynomials are "equal" on this ordering, but not the same element. Is this simply a forbidden ordering?
 A: $\newcommand{\p}{\preccurlyeq}$Definitions: A partial order relation $\p$ on a set $S$ mandates the following properties:

*

*Reflexivity: $s \p s$ for all $s \in S$

*Antisymmetry: Whenever $s \p t$ and $t \p s$, it holds that $t=s$

*Transitivity: If $s \p t$ and $t \p u$, then $s \p u$
A poset $(S,\p)$ becomes a total (or linear) order with the following property:

*

*Comparability/Connexity: For every $s,t \in S$, either $s \p t$ or $t \p s$

Let's formalize your order relation a bit more clearly. Let $S := \{x^2,x+2,x^3\}$. Define $\p$ on $S$ by
$$f \p g \iff f(2) \le g(2)$$
where $\le$ is the usual "less than or equal to" order relation on $\Bbb R$.
In this case, antisymmetry does not hold. This is because while
\begin{align*}
x+2 \p x^2 &\text{ because } 4 = x+2 \bigg|_{x=2} \le x^2 \bigg|_{x=2} = 4\\
x^2 \p x+2 &\text{ because } 4 = x^2 \bigg|_{x=2} \le x+2 \bigg|_{x=2} = 4\\
\end{align*}
it is obvious that $x^2 \ne x+2$ (at least in $\Bbb R[x]$, the space of polynomials over $\Bbb R$) despite $x^2 \p x+2$ and $x+2 \p x^2$ both holding.
Thus, we wouldn't even say $\p$ is an order relation on $S$.

Suppose for the sake of contradiction that there exist distinct $a,b$ in $S$, a set with partial order $\preccurlyeq$. Suppose, further, $a \approx b$, where $\approx$ means that $a \p b$ and $b \p a$ under $\p$. If $\p$ is indeed an order relation, however, antisymmetry requires $a=b$ in the usual sense of equality, i.e. they are identical elements.
