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In some recent work, I need a strengthening of König's Lemma to "trees" of arbitrary ordinal heights. Trees, in this context, are really just well-founded partially ordered sets. See, for instance, page 114 in Jech's Set Theory. One must be careful when modifying the "finite branching" hypothesis of König's Lemma in this situation, due to the existence of Aronszajn trees. The generalization is as follows:

Theorem: If $(S,<)$ is a well-founded partial order, such that for each ordinal $\beta$ the set of points of height $\beta$, denoted $S(\beta),$ is finite, then there is a branch in $S$ with the same height as $S.$

Here is my short proof of this result.

Proof: Let $\alpha$ be the height of $S.$ We recursively define the branch we want as follows. If $\{s_{i}\}_{i<\delta}$ is defined to height $\delta,$ then we choose $s_{\delta}$ to be any one of the finitely many elements $t\in S(\delta)$ that majorizes these previous points, and which has the additional property that for any ordinal $j$, $$(\ast)\,\, \text{ if }\delta<j<\alpha,\text{ then there exists some } s\in S(j) \text{ such that } t<s,$$ if such an element $t$ exists. Otherwise we end the recursion.

Let $\beta$ be the height of the branch we just defined. Suppose, by way of contradiction, that $\beta<\alpha.$ Then, from the fact that the height of $S$ is $\alpha,$ at least one of the finitely many elements $t\in S(\beta)$ satisfies $(\ast)$ with $\delta=\beta.$ Let $t_0$ be any one of them. Since our recursion ended at $\beta$, we know that $t_0$ does not majorize our branch. Let $\beta_0$ be the smallest index such that $s_{\beta_0}\not< t_0.$

Among the finitely many elements $t\in S(\beta)$ which do majorize $s_{\beta_0},$ there is at least one which satisfies $(\ast),$ since $s_{\beta_0}$ itself satisfies $(\ast).$ Let $t_1$ be any such element. Let $\beta_1$ be the smallest index such that $t_1$ doesn't majorize $s_{\beta_1}.$ Clearly $\beta_1>\beta_0.$

Repeating this process, we get an infinite list of elements $t_0,t_1,\ldots\in S(\beta),$ which are distinct since the corresponding $\beta_0,\beta_1,\ldots$ are distinct. This contradicts the finiteness of $S(\beta).$$\boxed{}$

My main question is whether or not there is a good reference for this result in the literature. While searching the internet, I found a different proof for it at this blog. The idea there is to pass to a slightly simpler structure.

Along those lines, if we replace $S$ with the set of points $S^{\ast}$ which satisfy $(\ast)$, then $(S^{\ast},<)$ is a wpo (that is, a well-founded partially ordered set with no infinite antichains) that has the same height as $S$. So it also suffices to prove this theorem for wpo's.

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  • $\begingroup$ You probably want to take a look at this: blog.math.toronto.edu/GraduateBlog/files/2014/03/… and the references in there. $\endgroup$ Commented Aug 20, 2019 at 17:29
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    $\begingroup$ Another argument: 1. If there is a highest level $S(\phi)$, then for each $x\in S(\phi)$ the set $\{y:y\leq x\}$ is a branch intersecting all levels. 2. If there is no highest level, then let $U$ be an ultrafilter such that $\bigcup\{S(\beta):\beta<\alpha\}\notin U$`` for any $\alpha<\phi$, where $\phi$ is least that $S(\phi)=\emptyset$. For each level $S(\beta)$ there is a unique element $x_\beta\in S(\beta)$ such that $U(x_\beta)=\{z:z\geq x_\beta\}\in U$ and one can easily see that $B=\{x_\beta:\beta<\phi\}$ is a branch. $\endgroup$ Commented Aug 21, 2019 at 15:07
  • $\begingroup$ König the father wrote his name with ö; but Kőnig the son wrote his name with ő. $\endgroup$
    – Asaf Karagila
    Commented Aug 29, 2019 at 11:22
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    $\begingroup$ @Monroe: Ancient Sumerian, yes. $\endgroup$
    – Asaf Karagila
    Commented Aug 29, 2019 at 19:51
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    $\begingroup$ The names of Konigs: this is a strange thing. In old Hungary, quite some jews were named Konig, they used double strokes as in the name of Erdos. When they published in German (and foreign language publications were almost exclusively German those days) father and son used umlaut. Denes Konig (the son) was a teacher of Erdos. He commited suicide in October 1944 when a Nazi party (the Arrow Cross Party) took power in Hungary and murdering Jews was just very general.. $\endgroup$ Commented Sep 1, 2019 at 6:21

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I found the paper "Generalizing König's Infinity Lemma" by Robert H. Cowan. Theorem 3 gets this result in a nice, more general form. Cowan's theorem is equivalent, over ZF, to the ultrafilter lemma, tying nicely into Péter Komjáth's comment.

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    $\begingroup$ Since the "weak Kőnig's lemma" is equivalent to the compactness of the Cantor set, and in general the compactness of $2^X$ is equivalent to the compactness theorem of first-order logic which is equivalent to the ultrafilter lemma, it's not very surprising. $\endgroup$
    – Asaf Karagila
    Commented Aug 29, 2019 at 9:54
  • $\begingroup$ Various tree versions have also been used by Stevo and Saharon. $\endgroup$ Commented Aug 29, 2019 at 12:49
  • $\begingroup$ (That is, Stevo Todorcevic and Saharon Shelah.) $\endgroup$ Commented Aug 29, 2019 at 20:54

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