In some recent work, I need a strengthening of King's lemma to "trees" of arbitrary orderly height. Trees in this context are actually only justified, partially ordered quantities. See, for example, page 114 in Jech's *Set theory.* One has to be careful when modifying the hypothesis of the "finite branching" of King's lemma in this situation due to the existence of arsonzajn trees. The generalization is as follows:

**sentence**: If $ (S, <) $ is a well-founded suborder, so for each atomic number $ beta $ the amount of height points $ beta $designated $ S ( beta), $ Finally, there is a branch in $ S $ with the same height as $ P. $

Here is my short proof of this result.

*proof*: To let $ alpha $ be the height of $ P. $ We recursively define the desired branch as follows. If $ {s_ {i} } _ {i < delta} $ is defined at height $ Delta, $ then we choose $ s _ { delta} $ to be one of the finitely many elements $ t in S ( delta) $ this majorizes these previous points, and that has the added property of having every atomic number $ j $.

$$ ( ast) , , text {if} delta <j < alpha, text {then there are some} s in S (j) text {such that} t <s, $$

if such an item $ t $ exist. Otherwise we finish the recursion.

To let $ beta $ Be the height of the branch we just defined. Suppose that contradicts $ 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) $ satisfied $ ( ast) $ With $ Delta = Beta. $ To let $ t_0 $ be one of them. That's when our recursion ended $ beta $, We know that $ t_0 $ does not make our industry big. To let $ beta_0 $ be the smallest index so that $ s _ { beta_0} not <t_0. $

Among the finitely many elements $ t in S ( beta) $ they make the majority $ s _ { beta_0}, $ there is at least one who satisfies $ ( ast), $ since $ s _ { beta_0} $ self-satisfied $ ( ast). $ To let $ t_1 $ be any such element. To let $ beta_1 $ be the smallest index, so that $ t_1 $ Does not vote $ s _ { beta_1}. $ Clear $ beta_1> beta_0. $

If we repeat this process, we get an infinite list of elements $ t_0, t_1, ldots in S ( beta), $ they are different since the corresponding one $ beta_0, beta_1, ldots $ are different. This contradicts the finiteness of $ S ( beta). $$ boxed {} $

My main question is whether there is a good reference in the literature for this result or not. When searching the internet, I found another proof in this blog. The idea is to move to a slightly simpler structure.

In that direction, when we replace $ S $ with the set of points $ S ^ { ast} $ which satisfy $ ( ast) $, then $ (S ^ { ast}, <) $ is a wpo (ie a well-founded, partially ordered set without infinite antichains) that has the same height as $ S $, So it is enough to prove this sentence for wpo.