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!
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.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$