Transitive Closure of a Well-Founded Relation is Well-Founded (without Axiom of Choice)

Question

I am interested in proving the titular claim:

Transitive Closure of a Well-Founded Relation is Well-Founded (without Axiom of Choice)

My approach:

Let $R$ be a well-founded relation. We construct the transitive closure $R^t$ as the union of the range of $F:\mathbb{N}\to \mathrm{fld}(R)\times\mathrm{fld}(R)$ defined (inductively) by $$ F(0)=R\qquad\text{and}\qquad F(n+1)=F(n)\circ R $$ for every $n\in\mathbb{N}$. That is to say, $R^t\overset{\mathrm{def}}{=}\bigcup\mathrm{ran}(F)=\bigcup_{i=0}^\infty F(i)$...

My thought process:

I can show that for each natural number $n$, $\bigcup_{i=0}^nF(i)$ is well-founded given that $R$ is well-founded (by induction on $n$), but that doesn't imply that $R^t$ is well-founded. It seems I need some sort of "transfinite induction" argument where if all the "finite" predecessors $\bigcup_{i=0}^nF(i)$ are well-founded, then $\bigcup_{i=0}^\infty F(i)$ is well-founded (but I don't think we necessarily have a well-ordered set to make such an argument).

Any hints/advice is much appreciated!

(The reason I want to work in ZF is because Enderton's Elements of Set Theory pg 244 gives a proof using AC, but remarks as an exercise to give a proof without it)

score 2 · Accepted Answer · answered Jul 03 '15 at 14:46

2

Suppose that $R$ is well-founded, and let $A$ be any nonempty subset of its field. We wish to show that $A$ contains an $R^t$-minimal element.

Let $B$ consist of all elements of $A$, plus every element of every finite $R$-sequence that connects two elements of $A$.

Because $R$ is well-founded, $B$ contains an $R$-minimal element $b$. The elements that were added to $B$ as intermediate elements of $R$-sequences cannot, by construction, be $R$-minimal, so we must have $b\in A$.

I claim that $b$ is $R^t$-minimal in $A$ -- because it it were not, it would sit at the top of an $R$-sequence that witnesses $\langle a,b\rangle\in R^t$, and the penultimate element of that sequence would be a predecessor of $b$ in $B$, contradicting $b$ being $R$-minimal in $B$.

Thus, $R^t$ is well founded.

answered Jul 03 '15 at 14:46

hmakholm left over Monica

286,031

Third time's the charm? – hmakholm left over Monica Jul 03 '15 at 14:46
instead of constructing $B$, would your argument hold if we simply considered the field of $R$ (assuming it's nonempty) – Alberto Takase Jul 04 '15 at 23:41
@AlbertoTakase: No, because then we would have no argument that the $R$-minimal element is in fact in $A$. – hmakholm left over Monica Jul 04 '15 at 23:48
Really great argument! I'll see how to go about constructing B from A via induction or transfinite recursion – Alberto Takase Jul 04 '15 at 23:56
$B\overset{\mathrm{def}}{=}A\cup{z\in\mathrm{fld}(R):\exists a_0,a_1\in A,\exists n,m\in\mathbb{N}[a_0F(n)zF(m)a_1]}$ – Alberto Takase Jul 05 '15 at 00:51
1

I know it has been awhile since I asked this question. I wanted to mention that the set $B$ is the "convex-hull" of $A$. This was a recent connection I made, and I thought it was neat enough to share it. – Alberto Takase Jan 21 '19 at 21:58

score 1 · Answer 2 · answered Jul 02 '15 at 19:13

1

The main use of choice in this proof would be to show there is no decreasing chains. But in the absence of choice we need to return to the minimal elements definitions. For simplicity, denote by $S$ the field of $R$ and $R^t$.

So suppose that $A\subseteq S$ is a non-empty set. It has some $R$-minimal element $a$, now show that from this assumption that $a$ is $R$-minimal, it is impossible that there is any $b\in A$ for which $\langle b,a\rangle\in R^t$. To do so, recall that if $\langle x,y\rangle\in R^t$, then there is a finite chain of elements such that each "link" in the chain belongs to $R$.

answered Jul 02 '15 at 19:13

Asaf Karagila

393,674

Is "$R$-minimal" well-behaved enough for this argument to work if $R$ isn't known to be transitive? For example, over $\mathbb Z$, let $aRb$ iff $a+1=b$ and let $A$ be the set of even numbers. Then every element of $A$ is $R$-minimal, but none of them are $R^t$-minimal. – hmakholm left over Monica Jul 02 '15 at 19:24
To quote a barefooted man, "Ah, bugger." :-) Although $R$ in your comment is not well-founded, so maybe this approach can still be salvaged. – Asaf Karagila Jul 02 '15 at 19:29

hmakholm left over Monica · Answer 3 · 2015-07-02T19:52:58.190

1

I don't think it is necessarily true in the absence of choice.

Assume that countable choice fails -- i.e. there is a family of nonempty sets $(A_n)_{n\in\omega}$ which has no choice function. Without loss of generality assume the $A_n$s are pairwise disjoint and also disjoint from $\omega$.

Now let $$R=\{\langle a,n\rangle\mid n\in\omega, a\in A_{n+1}\}\cup\{\langle n,a\rangle\mid n\in\omega, a\in A_n\}$$

Then $R$ is well-founded, because an infinitely descending $R$-chain would pick an element from each of cofinitely many of the $A_n$s.

On the other hand $R^t$ is not well-founded, due to the descending chain $$ \cdots 3 \mathrel{R^t} 2 \mathrel{R^t} 1 \mathrel{R^t} 0 $$

edited Jul 02 '15 at 19:52

answered Jul 02 '15 at 19:51

hmakholm left over Monica

286,031

The lack of chains does not imply being well-founded. (Not saying that your example is false, yet.) – Asaf Karagila Jul 02 '15 at 19:52
The example is very good. $\bigcup A_n\cup\omega$ is certainly non-empty and does not have any $R$-minimal elements. – Asaf Karagila Jul 02 '15 at 19:56
@AsafKaragila: But that just means that neither $R$ nor $R^t$ is well-founded, in which case it is not a counterexample by the "non-empty subset must contain minimal element" definition. I'll leave the answer up as an example of needing to be careful about definitions here. – hmakholm left over Monica Jul 02 '15 at 20:13
In that case, I think I'll undelete my answer as well. At least until I come up with a better solution. :-) And maybe this whole thing is a hint, that I should go to bed now. – Asaf Karagila Jul 02 '15 at 20:34

Transitive Closure of a Well-Founded Relation is Well-Founded (without Axiom of Choice)

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