Well-founded ordering on natural numbers

Let us define the following ordering on $\mathbb{N} = \{1,2,3,\ldots\}$.

$\forall x,y \in \mathbb{N}, \ x \succ y \Leftrightarrow \exists k \in \mathbb{N}-\{1\}\text{ s.t. } x = ky$

Is this ordering well-founded?

Our definition of a well-founded ordering, is that it is a strict partial order, and there are no infinite descending chains.

I have already shown that this defines a strict partial order. I have also read that the second condition is equivalent to: "For every subset $S \subseteq \mathbb{N}$, there exists a minimal element."

For this relation, for every $S$, there exists an element $m$ such that there does not exist an $x \in S$ such that $m \succ x$.

It is also the case that, there does not exist an $m \in S$ such that $x \succ m$ for every $x \in S$ ($m \neq x$).

Obviously, I am a bit unclear on the precise definition of well-founded ordering! Which argument is correct?

• The minimal element need not be distinct {2,5, 8,30, 15} has two minimal elements. 2 and 5. we don't need to froce one to be less than the other; just that the are unrelatable. (thi is possible as the order is partial-- not strict.). Feb 11, 2017 at 3:31
• @fleablood Do you mean the order is partial -- not total? Can't an order be strict AND partial? Feb 11, 2017 at 3:45
• I did mean total. I can't conceive why nor do I remember that I did write "strict". Yet obviously I did type "strict" so it must have been a moment a typing aphasia. I certainly meant to type "total" and I have distinct memory of doing so. ... but apparently I didn't. Strange. Really strange. Feb 11, 2017 at 4:44

"For this relation, for every S, there exists an element m such that there does not exist an x∈S such that m≻x. "

Yes, and there may be more than 1. They are all minimal. minimal means there is no $y < m$. It does not mean that $m < w$ for all $w$. As $>$ is a partial, not total, this is perfectly acceptable.

Let $S$ be all multiples of $5$ and $7$. Then there is no $x < 5$ and there is no $x < 7$. They are both minimal. $5$ and $7$ can not be related to each other.

"It is also the case that, there does not exist an m∈S such that x≻m for every x∈S (m≠x)."

Um, it depends on $S$. If $S$ is all multiples of $3$ then $3 < x$ for every $s \in S$. But in general, no there need not be any such element.

Given your analysis so far, you should be able to see that the property equivalent to lack of infinite descending chains is the one that says that for every nonempty subset there exists a minimal element.

That is, an element $m$ such that no other element in the subset precedes it in the strict partial order.

Maybe the quickest way to answer the well-foundedness question in this case begins by noting that if $x\succ y$ then $x>y.$ (Exercise: Figure out what that is true.) For any $\varnothing\ne S\subseteq\mathbb N,$ find a member of $S$ that is smallest with respect to the familiar order. Show that it is minimal, i.e. no other element is smaller than it in this proposed order, although some may be incomparable with it.