# Is it possible to prove this more general variant of Zorn's lemma?

I would like to ask if it is possible to prove this more general variant of Zorn's lemma? If possible, please don't post your proof. I will try to prove it by myself first.

1. Zorn's Lemma
Suppose that $(A,\leq)$ is a partially ordered set in which each chain has an upper bound. Then $A$ has a maximal element.
1. My variant
Suppose that $(A,\leq)$ is a partially ordered set in which each chain has an upper bound, and that $C$ is a chain in $A$. Then there exists a maximal $m$ of $A$ such that $c \leq m$ for all $c \in C$.

• Yes. This is possible, but how do you know to trust me? I haven't given you the proof of my claim. Either try and prove this, or ask for a proof. There's little point in asking if this can be proved or not... – Asaf Karagila Mar 13 '18 at 7:26
• @AsafKaragila, I trust you without reservation! I found this as an exercise in my textbook, but I'm not sure if this exercise is correct or not. I try to prove and post my proof to receive your comment :v – Akira Mar 13 '18 at 7:32
• Don't trust me without reservation, and you can trust me without reservation when I tell you that. :) – Asaf Karagila Mar 13 '18 at 7:33
• And in that case if you can't prove it, then you've done a very good job of not lying to yourself. Mathematics is a lot about what you might call a wasted effort, but these efforts are not wasted, they help you build confidence in your own abilities, as well as intuition and other tools of facing difficulties. – Asaf Karagila Mar 13 '18 at 7:35
• Even simpler: Take some upper bound $u$ of the chain $c$ and verify that the assumptions of Zorn's lemma apply to the subset of elements larger or equal to $u$. Show that a maximal element of this subset will do the trick. – Michael Greinecker Mar 13 '18 at 9:19

On the basis of @drhab and @Michael comments, I have formalized my proofs. Please have a check if these two proofs are fine or not. Thank you so much!

1. @drhab approach

Let $T=\{D \subseteq A \mid C \subseteq D \wedge D \text{ is a chain}\}$. Then $(T,\subseteq)$ is a partially ordered set.

Assume $X$ is a chain in $T$. Let $Y=\cup_{x \in X}x$. Then $Y \in T$ and Y is an upper bound of $X$. As a result, $(T,\subseteq)$ satisfies condition of Zorn’s lemma. Hence $T$ has a maximal element $\bar{C}$.

Let $c$ be an upper bound of $\bar{C}$. Then $c$ is also an upper bound of $C$.

We now prove that $c$ is a maximal element of $A$. Assume $c < c'$. Then $\bar{C} \cup \{c’\} \in T$ and $\bar{C} \subsetneq \bar{C} \cup \{c’\}$. This contradicts the fact that $\bar{C}$ is a maximal element of $T$. As a result, $c \nless c'$ for all $c' \in A$.

To sum up, $c$ is an upper bound of $C$ and a maximal element of $A$.

1. @Michael approach

Let $u$ be an upper bound of $C$ and $T=\{x \in A \mid u \leq x\}$.

Assume $X$ is a chain in $T$. Then $X$ has an upper bound in $T$. As a result, $(T,\leqslant)$ satisfies condition of Zorn’s lemma. Hence $T$ has a maximal element $c$. $c \in T \implies c$ is an upper bound of $C$.

We now prove that $c$ is a maximal element of $A$. Assume $c < c'$. This contradicts the fact that $c$ is a maximal element of $T$. As a result, $c \nless c'$ for all $c' \in A$.

To sum up, $c$ is an upper bound of $C$ and a maximal element of $A$.