Proving Limits Are Unique With Respect To A Pre-Ordered Set (Or Proset) Hello again guys and gals! I'm stuck yet again, and need some guidance. The problem I'm doing can be found in C. Pugh's, Real Mathematical Analysis, on PG. 191 - Problem 26 (I believe it is the 2nd Edition of the book - the ISBN is 978-1-4419-2941-9). Let me get right to the problem-statement which is stated exactly as it is given in the text; note, I didn't include parts ($b$)-($d$) in order to save space (sorry that this is long).
Problem 26-PG. 191 (C. Pugh's, Real Mathematical Analysis):  Let $\Omega$ be a set with a transitive relation $\preceq$. It satisfies the conditions that for all $\omega_{1},\omega_{2},\omega_{3}\in\Omega$, $\omega_{1}\preceq\omega_{1}$ and if $\omega_{1}\preceq\omega_{2}\preceq\omega_{3}$ then $\omega_{1}\preceq\omega_{3}$. A function $f:\Omega\rightarrow\mathbb{R}$ converges to a limit L with respect to $\Omega$ if, given any $\varepsilon>0$ there is an $\omega_{0}\in\Omega$ such that $\omega_{0}\preceq\omega$ implies $\big|f(\omega)-L\big|<\varepsilon$. We write $\lim_{\Omega}f(\omega)=L$ to indicate this convergence. Observe that:
$~~\bullet~$When $f(n)=a_{n}$ and $\mathbb{N}$ is given its standard order relation $\leq$, $\!\lim\limits_{n\rightarrow+\infty}\!\!a_{n}$ means the same thing as $\lim_{\mathbb{N}}f(n)$.
$~~\bullet~$When $\mathbb{R}$ is given with its standard order relation $\leq$, $\!\lim\limits_{t\rightarrow+\infty}\!\!f(t)$ means the same thing as $\lim_{\mathbb{R}}f(t)$
$~~\bullet~$Fix an $x\in\mathbb{R}$ and give $\mathbb{R}$ the new relation $t_{1}\preceq t_{2}$ when $|t_{2}-x|\leq|t_{1}-x|$. Then $\lim\limits_{t\rightarrow x}f(t)$ means the same thing as $\lim_{(\mathbb{R},\preceq)}f(t)$.
($a$) Prove the limits are unique: if $\lim_{\Omega}f=L_{1}$ and $\lim_{\Omega}f=L_{2}$ then $L_{1}=L_{2}$.
Preliminary Work/Remarks: I only need help with part (a) (for the rest of the problem, see the text), as once I get around the part I'm stuck on, I feel confident I can finish the entire problem - this is thus the reason why I'm asking.  As far as the way my proof for (a) goes, see below:
//Proof: We suppose that $\Omega$ is a set that is equipped with some, defined, transitive relation, $\lesssim$. In this sense, the relation $\lesssim$ is called a pre-order (or sometimes called a quasi-order), and the set $\Omega$ is called a pre-ordered set, or a proset. At this point, we suppose that the $\lim_{\Omega}f(\omega)$ exists as well as $\lim_{\Omega}f(\omega)=L_{1}$. This implies that for $\varepsilon>0$ there exists an $\omega_{0}\in\Omega$ such that for all $\omega\in\Omega$ with $\omega_{0}\lesssim\omega$ implies that $\big|f(\omega)-L_{1}\big|<\frac{\varepsilon}{2}$. Furthermore, we also assume that $\lim_{\Omega}f(\omega)=L_{2}$. This means that there exists an $\omega_{1}\in\Omega$ such that for all $\omega\in\Omega$ with $\omega_{1}\lesssim\omega$ implies that $\big|f(\omega)-L_{2}\big|<\frac{\varepsilon}{2}$. We can suppose without loss of generality that the pre-order $\lesssim$ is a weak order or a total pre-order (by ???), so that we can compare the elements in $\Omega$ - i.e., for all $\overline{\omega},\widetilde{\omega}\in\Omega$ we have that either $\overline{\omega}\lesssim\widetilde{\omega}$ or $\widetilde{\omega}\lesssim\overline{\omega}$. Then, in this case, we will either have that $\omega_{0}\lesssim\omega_{1}$ or $\omega_{1}\lesssim\omega_{0}$ implying that either $\omega_{0}\lesssim\omega_{1}\lesssim\omega$ or $\omega_{1}\lesssim\omega_{0}\lesssim\omega$. Hence, for all $\omega\in\Omega$ with either $\omega_{0}\lesssim\omega_{1}\lesssim\omega$ or $\omega_{1}\lesssim\omega_{0}\lesssim\omega$ implies that:
$|L_{1}-L_{2}|=\big|L_{1}-f(\omega)+f(\omega)-L_{2}\big|\leq\big|L_{1}-f(\omega)\big|+\big|f(\omega)-L_{2}\big|<\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon$.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$
I'm pretty sure assuming that the pre-order is a total pre-order is not permitted, but this is my problem. I'm not sure if the problem-statement is missing something (I know I did find a webpage here that mentions that the problem-statement may require the pre-order relation additionally be either antisymmetric, total, or both), but I'm sure there must be a way to solve this. My problem is after we find $\omega_{0},\omega_{1}\in\Omega$, as I've done above, I need a way to compare these elements, so that, between $\omega_{0}$ and $\omega_{1}$, one of them is maximal, or that a maximal element can be found so that both $\omega_{0},\omega_{1}\lesssim\omega_{M}$ where $\omega_{M}$ is a maximal element in $\Omega$. Then I figured to use the transitive property of the relation in order to develop that $\omega_{0}\lesssim\omega_{1}\lesssim\omega$ in $\Omega$. Then we can use the rest of my proof to finish. However, comparability and the existence of a maximal element may not exist under the assumption that $\lesssim$ is a pre-order relation, unless we bring the properties of a partial order to the scenario (or am I incorrect [?]). My apologies if any of my definitions are incorrect above (or anything for that matter). Essentially I need help with this, and any comments, suggestions, answers, recommendations, etc. are GREATLY APPRECIATED!
 A: I posted the proof with overlap on my preliminary work above - sorry about that (I didn't want to stray too much away from my preliminary work); let me know what you guys think...I also included a sub-claim and a proof that any finite and nonempty set equipped with a partial order will always have a minimum/maximum element.
//Proof (a): We suppose that the $\lim_{\Omega}f(\omega)$ exists as well as $\lim_{\Omega}f(\omega)=L_{1}$. This implies that for $\varepsilon>0$ there exists an $\omega_{0}\in\Omega$ such that for all $\omega\in\Omega$ with $\omega_{0}\lesssim\omega$ implies that $\big|f(\omega)-L_{1}\big|<\frac{\varepsilon}{2}$. Furthermore, we also assume that $\lim_{\Omega}f(\omega)=L_{2}$. This means that there exists an $\omega_{1}\in\Omega$ such that for all $\omega\in\Omega$ with $\omega_{1}\lesssim\omega$ implies that $\big|f(\omega)-L_{2}\big|<\frac{\varepsilon}{2}$. Define the finite set $\Phi=\big\{\omega_{0},\omega_{1}\big\}\subset\Omega$. Now, whether we choose to assume the Axiom of Choice, or not, $\Phi\subset\Omega$ will have a minimal/maximal element whenever we partially order the elements in $\Phi$. To see why this is true, let $S\neq\varnothing$ be a finite set coupled with a partial ordering $\leq$. We now claim $S$ has a maximal (or minimal) element. We can suppose without loss of generality that $I=\big\{1,2,...,n\big\}\subset\mathbb{N}$ so that $S$ is of size $n$ since $S$ is non-empty and finite. Proceeding by induction on $n\geq 1$, the base case is vacuously true since $n=1$ gives $S=\big\{s_{1}\big\}$, and then $s_{1}\leq s_{1}$ by reflexivity. For the inductive hypothesis, suppose the statement holds for $n>1$ so that $S=\big\{s_{i}:i\in I=\{1,2,...,n\}\big\}$ has a maximal element. To consider the next inductive step, we let $S'=S\cup\big\{s_{n+1}\big\}=\big\{s_{i}:i\in I=\{1,2,...,n+1\}\big\}$. For any $s\in S'$, if we restrict the partial order to $S'\big\backslash\big\{s\big\}$, we have by the inductive hypothesis that $S'\big\backslash\big\{s\big\}$ has a maximal element, say it is $s_{0}\in S'\big\backslash\big\{s\big\}$. Then, we have for any $s'\in S'$ that either $s_{0}<s'$ or $s'\leq s_{0}$ in $S'$. If $s_{0}<s'$, then $s<s'$ for $s\in S$, hence $s'$ is a maximal element of $S'$. Otherwise, $s'\leq s_{0}$ and $s_{0}$ is a maximal element of $S'$. Thus $S'$ has a maximal element from either case, and therefore, by induction, we have that $S=\big\{s_{i}:i\in I=\{1,2,...,n\}\big\}$ has a maximal element for all $n\in\mathbb{N}$ proving the claim - to prove the minimum, we simply use a symmetric argument in the inductive step. Then, in this case, we will either have that $\omega_{0}\lesssim\omega_{1}$ or $\omega_{1}\lesssim\omega_{0}$, as one of the two must be a maximal element, for example, when comparing the two (by extending the pre-order on $\Omega$ to the partial order on $\Phi$). This implies that either $\omega_{0}\lesssim\omega_{1}\lesssim\omega$ or $\omega_{1}\lesssim\omega_{0}\lesssim\omega$. Hence, for all $\omega\in\Omega$ with either $\omega_{0}\lesssim\omega_{1}\lesssim\omega$ or $\omega_{1}\lesssim\omega_{0}\lesssim\omega$ implies that:
$|L_{1}-L_{2}|=\big|L_{1}-f(\omega)+f(\omega)-L_{2}\big|\leq\big|L_{1}-f(\omega)\big|+\big|f(\omega)-L_{2}\big|<\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon$.
Thus, we have that $|L_{1}-L_{2}|<\varepsilon$, so, letting $\varepsilon\rightarrow 0$ since $\varepsilon>0$ is arbitrary implies that $0\leq|L_{1}-L_{2}|\leq 0$ so $L_{1}-L_{2}=0$ and therefore we have that $L_{1}=L_{2}$. We conclude that $\lim_{\Omega}f(\omega)$ is unique, whenever it exists.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$
