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I am interested in proving the titular claim:

$R$ well-founded relation and $\forall y$, $\{x:xRy\}$ is finite implies $\forall y$, $\{x:xR^t y\}$ is finite (where $R^t$ is the transitive closure)

After drawing a couple of pictures/graphs it reminded me of Konig's Lemma: "every tree that contains infinitely many vertices, each having finite degree, has at least one infinite simple path."

Assuming I have no knowledge of graphs and trees, is there a proof of the claim using the definition of well-founded?

This previous question I asked is related/useful: Transitive Closure of a Well-Founded Relation is Well-Founded (without Axiom of Choice)

Community
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Alberto Takase
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2 Answers2

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Assume that for some $y$ the set $\left\{ x\mid xR^{t}y\right\} $ is not finite and set $a_{0}:=y$.

We have $R^{t}=R\cup (R^{t}\circ R)$ and denoting $\hat{y}:=\left\{ x\mid xRy\right\} $we can write $\left\{ x\mid xR^{t}y\right\} =\hat{y}\cup\bigcup_{z\in\hat{y}}\left\{ x\mid xR^{t}z\right\} $.

Set $\hat{y}$ is finite so $\left\{ x\mid xR^{t}z\right\} $ must be infinite for at least one $z\in\hat{y}$.

Pick out such $z$ and name it $a_{1}$.

This can be repeated and a set $A:=\left\{ a_{n}\mid n=0,1,\dots\right\} $ is constructed with $a_{n+1}Ra_{n}$ for $n=0,1,\dots$.

Then $A$ has no $R$-minimal element which contradicts that $R$ is well-founded.

drhab
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The proof is straightforward using well-founded induction. So, suppose $y \in A$ (the domain of $R$), and assume the inductive hypothesis that whenever $z \mathrel{R} y$, then $\{ x \mid x \mathrel{R^t} z \}$ is finite. Then whenever $x \mathrel{R^t} y$, then either $x \mathrel{R} y$ or there exists $z$ such that $x \mathrel{R^t} z$ and $z \mathrel{R} y$. Therefore, $$\{ x \mid x \mathrel{R^t} y \} = \{ x \mid x \mathrel{R} y \} \cup \bigcup_{z \mid z \mathrel{R} y} \{ x \mid x \mathrel{R^t} z \}.$$ In this expression, the first set is finite by assumption, and the second term is a finite union of finite sets by the inductive hypothesis. It follows that $\{ x \mid x \mathrel{R^t} y \}$ is also finite, completing what we needed to show to apply well-founded induction.

Daniel Schepler
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