Let $(X, \leq)$ be an ordered set. I would like to know if the following two conditions are equivalent:
- Every totally ordered subset of $X$ has an upper bound.
- Every increasing sequence of $X$ has an upper bound.
$1. \implies 2.$ is clear, because any increasing sequence is totally ordered. My question is thus whether $2. \implies 1.$ is true.
I tried to prove this by contradiction, suppose we have a subset $A$ of $X$ that doesn't have upper bound, then we can build an strictly increasing sequence $(a_{n})_{n \in \mathbb{N}}$ of elements in $A$. By 2., we have an element $x$ which is an upper bound of $(a_{n})_{n \in \mathbb{N}}$. Then there exists an element $a \in A$, strictly greater than $x$. But, since we are not sure if there exists $n \in \mathbb{N}$ such that $a \leq a_{n}$, so we can't conclude.
This question arises because I saw two references use different definitions of inductive set in Zorn's lemma, one version adopts the condition 1., another adopts the condition 2. So I try to establish the equivalence between them.
If you are thinking that increasing sequences are indexed by $\mathbb{N}$, and so are countable, then (2) does not imply (1). Ordinals are totally ordered. But any ordinal that has uncountable cofinality (such as $\omega_1$) provides a counterexample, as there is no countable sequence leading up to it.