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

169 Views Asked by At

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 $c \in C$.

Many thanks for your help!

1

There are 1 best solutions below

0
On BEST ANSWER

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 $A$.