- #1

- 317

- 26

Proof:

Case 1: ##B\neq\emptyset## such that ##B##=##\{####b\in X##: ##b## has an undefined order relation with all elements in ##X####\}##. Thus, ##\forall b\in B##, ##b## is maximal since there is no ##x\in X## where ##b<x##.

Case 2: ##B=\emptyset##; Let ##I_1## be some index set and let ##Y_\alpha,\forall\alpha\in X_1## denote all the totally-ordered sets in ##X##. Consider that ##\cup_{\alpha\in I_1}Y_\alpha## is either a poset or a totally-ordered set in ##X##. If ##\cup_{\alpha\in I_1}Y_\alpha## is totally ordered, then, by the premise of the lemma, ##\cup_{\alpha\in I_1}Y_\alpha## has an upperbound ##\hat{y}##. Since ##\hat{y}## would then be an upperbound of all totally-ordered subsets in ##X##, then no ##x\in X## satisfies ##\hat{y}<x##. Hence, ##\hat{y}## is a maximal element in ##X##. If ##\cup_{\alpha\in I_1}Y_\alpha## is a poset, then ##\exists y_1,y_2\in\cup_{\alpha\in I_1}Y_\alpha## such that the order relation ##y_1## and ##y_2## is undefined. In order to resolve this issue(of being unable to "compare"), we construct a set ##\mathscr{T}## of totally-ordered sets ##\Bbb{T}(y)## where ##\forall\overline{y}\in\cup_{\alpha\in I_1}Y_\alpha##, we let \begin{align}\Bbb{T}(\overline{y})=\{y\in\cup_{\alpha\in I_1}Y_\alpha:y\leq_X\overline{y}\quad \lor\quad \overline{y}\leq_X y\}\end{align}

Note that the correspondence of each ##\overline{y}\in\cup_{\alpha\in I_1}Y_\alpha## to a set ##\Bbb{T}(\overline{y})## is not one-to-one. Nevertheless, we have a collection ##\mathscr{T}=\{\Bbb{T}(y):y\in\cup_{\alpha\in I_1}Y_\alpha\}=\{\Bbb{T}(y):y\in X\}##. Each ##\Bbb{T}(y)## is constructed in such a way so that ##\forall\alpha\in I_1, \exists\Bbb{T}(y)\in\mathscr{T}## such that ##Y_\alpha\subseteq\Bbb{T}(y)##. Moreover, since ##B=\emptyset## and ##\forall y\in\cup_{\alpha\in I_1}Y_\alpha##, ##\Bbb{T}(y)## contains the maximum number of elements from ##X## while remaining totally-ordered, then ##\forall x_i\in X## where ##x_i\notin\Bbb{T}(y)##, it must be that ##x_i\in\Bbb{T}(y')## where ##y## and ##y'## have an undefined order-relation. Lastly, since we know that ##\forall y\in X, \Bbb{T}(y)\subseteq X## and is totally-ordered, then, by our premise, ##\Bbb{T}(y)## has an upperbound ##\tau## which I think is some maximal of ##X## because all upperbounds of each distinct ##\Bbb{T}(y^*)\in\mathscr{T}## are pair-wise undefined order relation(or incomparable). So for all upperbound ##\tau## in some ##\Bbb{T}(y)## there does not exist ##x\in X## such that ##\tau <x##.