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I am currently dealing with a set theory problem where I am told to show that any well-founded partial ordering is also a well-ordering (in those words exactly, no other givens). I know I have the ability to solve this problem - the issue is, I am unsure exactly what a 'well-founded partial ordering' is. You see, by definition (at least, as I have learned them):


If $R \subseteq A \times A$ is a partial ordering, it is reflexive, transitive, and antisymmetric.

If $R \subseteq A \times A$ is well-founded, then $$\forall B \subseteq A(B \neq \emptyset \Rightarrow \exists x \in B \forall y \in B (\lnot yRx)).$$

(In other words, every nonempty subset of $A$ has a minimal element.)


The problem? For each $B$ and each minimal element $x \in B$, wouldn't this imply that $\lnot xRx$ for each of these minimal elements spanning across the various nonempty subsets of $A$? However, we stated that $R$ is a partial order, and hence reflexive (ergo $xRx$ for all $x \in A$). I see a contradiction here...

So is this entity known as a 'well-founded partial ordering' necessarily strict (read: irreflexive)? If so, why wouldn't the author have just stated that instead of calling it a 'well-founded partial ordering'?

Anyways, any clarification would be helpful from a set theory minded person. Also, not trying to be sarcastic or anything, I just don't see what good ambiguity brings to learning undergrad mathematics.

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    $\begingroup$ According to wikipedia, "In order theory, a partial order is called well-founded if the corresponding strict order is a well-founded relation." So given a partial order $R$ there is a strict order $S$ which is defined as $xSy$ iff $x\neq y$ and $xRy.$ This essentially means that your expression should end $\forall y\in B(y=x\lor \lnot yRx).$ en.wikipedia.org/wiki/Well-founded_relation $\endgroup$ – Thomas Andrews Oct 1 '19 at 17:24
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    $\begingroup$ On the other hand, it is not true that any well-founded partial ordering is a well-ordering, unless you have an unusual definition of well-ordering. $\endgroup$ – Thomas Andrews Oct 1 '19 at 17:28
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There are two equivalent ways of defining a partial order on a set $A$:

(1) a partial order is a subset $R\subset A\times A$ such that $R$ is reflexive, anti-symmetric (for every $x,y \in A$ if $xRy$ and $yRx$ then $x=y$) and transitive.

(2) a partial order is a subset $R'\subset A\times A$ such that $R'$ is anti-reflexive (for every $x \in A$ it is the case that $x\not R'x$) and transitive.

Indeed, if you have an $R'$, the corresponding $R$ is $R' \cup \{(x,x):x\in A \}$ and if you have an $R$ you just remove the couples of the form $(x,x)$ in order to get the corresponding $R'$.

I guess that your definition of well-founded partial order is based on the definition (2).

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  • $\begingroup$ Thank you. I whipped out Enderton's set theory book and it had 10x the clarification the class notes I was working out of had. In short, if R is a relation then R' is a transitive relation and R is a subset of R'. He also states R is well-founded implies R' is well-founded and a partial ordering on A. Lastly, he also states there are two equivalent ways of defining a partial order on a set A. Thanks again! $\endgroup$ – greycatbird Oct 1 '19 at 17:37

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