# Zorn's Lemma: Need help finding errors in proof

• I

## Main Question or Discussion Point

Proposition(Zorn's Lemma): Let $X\neq\emptyset$ be of partial order with the property that $\forall Y\subseteq X$ such that $Y$ is of total-order then $Y$ has an upperbound, then $X$ contains a maximal element.

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 $\tau <x$.

Related Set Theory, Logic, Probability, Statistics News on Phys.org
fresh_42
Mentor
Before someone will fight their way through your proof, will say I'm not sure yet, whether I'm the one, since it's far too hot here and this increases my chances to be wrong, please tell us your assumption. Zorn's Lemma cannot be proven. It only has various equivalent formulations (four others in my book), so either equivalence is proven or a deduction from one of the other formulations. Which one?

mathwonk
Homework Helper
first of all., in line 6, the letter X1 should be I1. Then in the definition of T(y), as the set of all elements of X which are comparable to y, there is no reason for T(y) to be itself totally ordered. I.e. just because every pair of elements in T(y) are comparable to y, there is no reason for them to be comparable to each other. I stopped there.

Last edited:
I assumed Axiom of Choice. In particular, when choosing a maximal element of each chain $\Bbb{T}(y)$ constructed in the proof.

there is no reason for T(y) to be itself totally ordered.
thank you! I missed that. Is there anyway I could edit this post?

fresh_42
Mentor
thank you! I missed that. Is there anyway I could edit this post?
No, but you can "reply" it, remove the quotation marks and post a new version. To edit it afterwards isn't polite. It would make @mathwonk's answer look stupid, which is why there is a timer on the edit button.

No, but you can "reply" it, remove the quotation marks and post a new version. To edit it afterwards isn't polite.
Noted. This is what I like about the community here. Most people here considers that human beings have feelings too. lol