why can't a set with partial ordering and an upper bound have 2 infinitely long chains?

Question

If a set has an ordering and an upper bound we could be in a situation with a total ordering so all elements are in one chain and the chain approaches the upper bound forever so no maximal element. But if the ordering is partial there will be chains that are independent of each other where the order in one chain does not tell us the order of all of those elements in another chain. But by Zorn's Lemma we know there are not two infinitely ascending chains in a set with a partial ordering. Why can't that happen?

Answer 1

The claim

by Zorn's Lemma we know there are not two infinitely ascending chains in a set with a partial ordering

is false (and I think stems from a misreading of Zorn's lemma). For example, consider the partial order consisting of all natural numbers (including $0$) ordered by divisibility: $$a\trianglelefteq b\quad\iff\quad\exists n\in\mathbb{N}\mbox{ such that } an=b.$$ This is a partial order with a maximal element (namely, $0$), and has lots of infinite chains; for example, the set of powers of $2$ and the set of powers of $3$ are disjoint infinite chains, each of which has no upper bound in the partial ordering other than $0$ itself.