Kinda well-orderedness.

Question

Its not the case that $\mathbb{Z}$ is well-ordered (under the usual order). However, observe that

1. Every non-empty subset of $\mathbb{Z}$ that is bounded below has a least element, and
2. Every non-empty subset of $\mathbb{Z}$ that is bounded below inherits a well-ordering.

My question is, is it the case that for all partially ordered sets, the above conditions are equivalent? And if so, does this property have a name?

Answer 1

Let $\langle X,\le\rangle$ be a set with a pre-order. If there are distinct $x,y\in X$ such that $x\le y$ and $y\le x$, then $\{x,y\}$ is bounded below and neither has a least element nor inherits a well-order, so we might as well assume that $\le$ is a partial order. (You also want to restrict yourself to non-empty subsets that are bounded below, as the empty set clearly has no least element.)

It’s immediate that (2) implies (1), and it’s a straightforward (possibly transfinite) induction to show that (1) implies (2).

However, $\langle X,\le\rangle$ can be quite different from $\Bbb Z$. Here’s an example:

Let $X=\Bbb N\times\{0,1\}$, and define a strict partial order $\prec$ on $X$ as follows: $\langle m,i\rangle\prec\langle n,k\rangle$ iff $m<n$ and $k=0$. Let $A\subseteq X$ be non-empty. If $A\subseteq\Bbb N\times\{0\}$, then $A$ has a least element and inherits a well-order. Suppose, however, that $A\cap(\Bbb N\times\{1\})\ne\varnothing$. Let $m$ be minimal such that $\langle m,1\rangle\in A$. There is no $\langle n,i\rangle\in X$ such that $\langle n,i\rangle\prec\langle m,1\rangle$, so either $\langle m,1\rangle$ is the least element of $A$, or $A$ is not bounded below. Thus, every (non-empty) subset of $X$ that is bounded below has a least element, and there are no infinite descending chains, but there are lots of sets that aren’t bounded below.

Answer 2

If $P$ is a poset then the conditions

1) every nonempty bounded below subset of $P$ has a least element

2) every nonempty bounded below subset of $P$, with the inherited order structure from $P$, is well ordered

are equivalent. That $2\implies 1$ is clear since if $S\subseteq P$ is nonepmty and bounded below, then it's a well order, thus has itself a least element in the induced order, thus condition 1 holds.

For the direction $1 \implies 2$: Let $S\subseteq P$ be nonempty and bounded below. To show it is well ordered, we need to show every nonempty subset $T\subseteq S$ has a least element. But $T$ is then clearly a bounded below, nonempty subset of $P$, and so, by condition 1, has a least element. Since the order in $S$ is the induced order, that least element is the desired least element showing $S$ is well ordered. (This argument does not require transfinite induction/recursion).

I don't know of a name for this condition.