Higman's lemma and a manuscript of Erdős and RadoPairwise intersecting sets of fixed sizeUnpublished Higman...



Higman's lemma and a manuscript of Erdős and Rado


Pairwise intersecting sets of fixed sizeUnpublished Higman manuscriptCriteria for ghost-Witt vectors: looking for history and referencesSperner's lemma and Tucker's lemmaBetter bounds for exact-intersection Erdős–Ko–Rado system?Where is the Erdős–Rado theorem stated in Erdős and Rado's Bull AMS paper?Erdős-Ko-Rado with intersections of size at least twoNewman's Lemma or Diamond LemmaA generalization of Erdős-Ko-Rado theoremIs there a two-dimensional Higman's lemma?













6












$begingroup$


Motivated by a problem in factorization theory, I've recently proved the following:




Theorem. If $X$ is a non-empty finite alphabet and $mathcal W$ an infinite subset of the free semigroup, $X^ast$, over $X$, then there exists a sequence $(w_n)_{n ge 1}$ with values in $mathcal W$ such that $w_n$ is a proper subword of $w_{n+1}$ for every $n in mathbf N^+$.




The theorem implies at once Higman's lemma. The proof is elementary and self-contained (the most advanced thing one is using, is the pigeonhole principle), but I wouldn't call it trivial: The basic idea is to introduce a non-standard factorization of the elements of $X^ast$ that is well suited to an induction on $|X|$, and then distinguish two cases depending on a certain invariant associated with this factorization.



Unfortunately, it turned out that the result is nothing new and comes down to a special case of Theorems 2.1 and 4.3 of G. Higman's paper Ordering by divisibility in abstract algebras [Proc. Lond. Math. Soc., III. Ser. 2 (1952), 326-336]. In particular, Theorem 2.1 in Higman's paper states that the following conditions are equivalent for a quasi-ordered set $(A, preceq)$:



(a) Every sequence of elements of $A$ has a subsequence that is strictly increasing wrt $preceq$.



(b) If $(a_n)_{n ge 1}$ is a sequence of elements of $A$, there exist $i,j in mathbf N^+$ such that $i < j$ and $a_i preceq a_j$.



For the proof of the equivalence, Higman cites an unpublished manuscript of P. Erdős and R. Rado.




Question. Which manuscript of Erdős and Rado does Higman refer to? Has the manuscript been eventually published? If not, is there a book, article, etc. with the details of a proof of the equivalence between (a) and (b)?




I browsed through the joint papers of Erdős and Rado listed at https://www.renyi.hu/~p_erdos/Erdos.html, but I couldn't find what I'm looking for.










share|cite|improve this question











$endgroup$














  • $begingroup$
    Unless I've misunderstood something, this theorem of Higman is an immediate consequence of Ramsey's theorem (which might have been rediscovered by Erdös and Rado).
    $endgroup$
    – Andreas Blass
    11 hours ago










  • $begingroup$
    @AndreasBlass Do you mean the infinite Ramsey thm (en.wikipedia.org/wiki/Ramsey's_theorem)? If so, could you explain in some detail how to use it to prove the equivalence between (a) and (b), if this is what you mean by "this thm of Higman"?
    $endgroup$
    – Salvo Tringali
    10 hours ago










  • $begingroup$
    Yes, I mean the infinite Ramsey theorem. I've put the proof into an answer, since it's too long for a comment.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    It follows immediately from Dilworth theorem: every infinite poset contains either an infinite chain or an infinite antichain. This is an easy (high school level) statement.
    $endgroup$
    – Mark Sapir
    4 hours ago












  • $begingroup$
    @MarkSapir Could you please explain how it follows from Dilworth's theorem? Dilworth's theorem (the one I know of, I'm sure he had many others) says that, if there is a finite bound $k$ on the sizes of antichains in a poset $P$, then $P$ is the union of $k$ chains. If the antichains are all finite but of unbounded finite size, then the number of antichains needed to cover $P$ can be an arbitrarily large infinite cardinal
    $endgroup$
    – bof
    2 hours ago
















6












$begingroup$


Motivated by a problem in factorization theory, I've recently proved the following:




Theorem. If $X$ is a non-empty finite alphabet and $mathcal W$ an infinite subset of the free semigroup, $X^ast$, over $X$, then there exists a sequence $(w_n)_{n ge 1}$ with values in $mathcal W$ such that $w_n$ is a proper subword of $w_{n+1}$ for every $n in mathbf N^+$.




The theorem implies at once Higman's lemma. The proof is elementary and self-contained (the most advanced thing one is using, is the pigeonhole principle), but I wouldn't call it trivial: The basic idea is to introduce a non-standard factorization of the elements of $X^ast$ that is well suited to an induction on $|X|$, and then distinguish two cases depending on a certain invariant associated with this factorization.



Unfortunately, it turned out that the result is nothing new and comes down to a special case of Theorems 2.1 and 4.3 of G. Higman's paper Ordering by divisibility in abstract algebras [Proc. Lond. Math. Soc., III. Ser. 2 (1952), 326-336]. In particular, Theorem 2.1 in Higman's paper states that the following conditions are equivalent for a quasi-ordered set $(A, preceq)$:



(a) Every sequence of elements of $A$ has a subsequence that is strictly increasing wrt $preceq$.



(b) If $(a_n)_{n ge 1}$ is a sequence of elements of $A$, there exist $i,j in mathbf N^+$ such that $i < j$ and $a_i preceq a_j$.



For the proof of the equivalence, Higman cites an unpublished manuscript of P. Erdős and R. Rado.




Question. Which manuscript of Erdős and Rado does Higman refer to? Has the manuscript been eventually published? If not, is there a book, article, etc. with the details of a proof of the equivalence between (a) and (b)?




I browsed through the joint papers of Erdős and Rado listed at https://www.renyi.hu/~p_erdos/Erdos.html, but I couldn't find what I'm looking for.










share|cite|improve this question











$endgroup$














  • $begingroup$
    Unless I've misunderstood something, this theorem of Higman is an immediate consequence of Ramsey's theorem (which might have been rediscovered by Erdös and Rado).
    $endgroup$
    – Andreas Blass
    11 hours ago










  • $begingroup$
    @AndreasBlass Do you mean the infinite Ramsey thm (en.wikipedia.org/wiki/Ramsey's_theorem)? If so, could you explain in some detail how to use it to prove the equivalence between (a) and (b), if this is what you mean by "this thm of Higman"?
    $endgroup$
    – Salvo Tringali
    10 hours ago










  • $begingroup$
    Yes, I mean the infinite Ramsey theorem. I've put the proof into an answer, since it's too long for a comment.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    It follows immediately from Dilworth theorem: every infinite poset contains either an infinite chain or an infinite antichain. This is an easy (high school level) statement.
    $endgroup$
    – Mark Sapir
    4 hours ago












  • $begingroup$
    @MarkSapir Could you please explain how it follows from Dilworth's theorem? Dilworth's theorem (the one I know of, I'm sure he had many others) says that, if there is a finite bound $k$ on the sizes of antichains in a poset $P$, then $P$ is the union of $k$ chains. If the antichains are all finite but of unbounded finite size, then the number of antichains needed to cover $P$ can be an arbitrarily large infinite cardinal
    $endgroup$
    – bof
    2 hours ago














6












6








6





$begingroup$


Motivated by a problem in factorization theory, I've recently proved the following:




Theorem. If $X$ is a non-empty finite alphabet and $mathcal W$ an infinite subset of the free semigroup, $X^ast$, over $X$, then there exists a sequence $(w_n)_{n ge 1}$ with values in $mathcal W$ such that $w_n$ is a proper subword of $w_{n+1}$ for every $n in mathbf N^+$.




The theorem implies at once Higman's lemma. The proof is elementary and self-contained (the most advanced thing one is using, is the pigeonhole principle), but I wouldn't call it trivial: The basic idea is to introduce a non-standard factorization of the elements of $X^ast$ that is well suited to an induction on $|X|$, and then distinguish two cases depending on a certain invariant associated with this factorization.



Unfortunately, it turned out that the result is nothing new and comes down to a special case of Theorems 2.1 and 4.3 of G. Higman's paper Ordering by divisibility in abstract algebras [Proc. Lond. Math. Soc., III. Ser. 2 (1952), 326-336]. In particular, Theorem 2.1 in Higman's paper states that the following conditions are equivalent for a quasi-ordered set $(A, preceq)$:



(a) Every sequence of elements of $A$ has a subsequence that is strictly increasing wrt $preceq$.



(b) If $(a_n)_{n ge 1}$ is a sequence of elements of $A$, there exist $i,j in mathbf N^+$ such that $i < j$ and $a_i preceq a_j$.



For the proof of the equivalence, Higman cites an unpublished manuscript of P. Erdős and R. Rado.




Question. Which manuscript of Erdős and Rado does Higman refer to? Has the manuscript been eventually published? If not, is there a book, article, etc. with the details of a proof of the equivalence between (a) and (b)?




I browsed through the joint papers of Erdős and Rado listed at https://www.renyi.hu/~p_erdos/Erdos.html, but I couldn't find what I'm looking for.










share|cite|improve this question











$endgroup$




Motivated by a problem in factorization theory, I've recently proved the following:




Theorem. If $X$ is a non-empty finite alphabet and $mathcal W$ an infinite subset of the free semigroup, $X^ast$, over $X$, then there exists a sequence $(w_n)_{n ge 1}$ with values in $mathcal W$ such that $w_n$ is a proper subword of $w_{n+1}$ for every $n in mathbf N^+$.




The theorem implies at once Higman's lemma. The proof is elementary and self-contained (the most advanced thing one is using, is the pigeonhole principle), but I wouldn't call it trivial: The basic idea is to introduce a non-standard factorization of the elements of $X^ast$ that is well suited to an induction on $|X|$, and then distinguish two cases depending on a certain invariant associated with this factorization.



Unfortunately, it turned out that the result is nothing new and comes down to a special case of Theorems 2.1 and 4.3 of G. Higman's paper Ordering by divisibility in abstract algebras [Proc. Lond. Math. Soc., III. Ser. 2 (1952), 326-336]. In particular, Theorem 2.1 in Higman's paper states that the following conditions are equivalent for a quasi-ordered set $(A, preceq)$:



(a) Every sequence of elements of $A$ has a subsequence that is strictly increasing wrt $preceq$.



(b) If $(a_n)_{n ge 1}$ is a sequence of elements of $A$, there exist $i,j in mathbf N^+$ such that $i < j$ and $a_i preceq a_j$.



For the proof of the equivalence, Higman cites an unpublished manuscript of P. Erdős and R. Rado.




Question. Which manuscript of Erdős and Rado does Higman refer to? Has the manuscript been eventually published? If not, is there a book, article, etc. with the details of a proof of the equivalence between (a) and (b)?




I browsed through the joint papers of Erdős and Rado listed at https://www.renyi.hu/~p_erdos/Erdos.html, but I couldn't find what I'm looking for.







reference-request co.combinatorics lattices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago







Salvo Tringali

















asked 14 hours ago









Salvo TringaliSalvo Tringali

4,2491 gold badge16 silver badges36 bronze badges




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  • $begingroup$
    Unless I've misunderstood something, this theorem of Higman is an immediate consequence of Ramsey's theorem (which might have been rediscovered by Erdös and Rado).
    $endgroup$
    – Andreas Blass
    11 hours ago










  • $begingroup$
    @AndreasBlass Do you mean the infinite Ramsey thm (en.wikipedia.org/wiki/Ramsey's_theorem)? If so, could you explain in some detail how to use it to prove the equivalence between (a) and (b), if this is what you mean by "this thm of Higman"?
    $endgroup$
    – Salvo Tringali
    10 hours ago










  • $begingroup$
    Yes, I mean the infinite Ramsey theorem. I've put the proof into an answer, since it's too long for a comment.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    It follows immediately from Dilworth theorem: every infinite poset contains either an infinite chain or an infinite antichain. This is an easy (high school level) statement.
    $endgroup$
    – Mark Sapir
    4 hours ago












  • $begingroup$
    @MarkSapir Could you please explain how it follows from Dilworth's theorem? Dilworth's theorem (the one I know of, I'm sure he had many others) says that, if there is a finite bound $k$ on the sizes of antichains in a poset $P$, then $P$ is the union of $k$ chains. If the antichains are all finite but of unbounded finite size, then the number of antichains needed to cover $P$ can be an arbitrarily large infinite cardinal
    $endgroup$
    – bof
    2 hours ago


















  • $begingroup$
    Unless I've misunderstood something, this theorem of Higman is an immediate consequence of Ramsey's theorem (which might have been rediscovered by Erdös and Rado).
    $endgroup$
    – Andreas Blass
    11 hours ago










  • $begingroup$
    @AndreasBlass Do you mean the infinite Ramsey thm (en.wikipedia.org/wiki/Ramsey's_theorem)? If so, could you explain in some detail how to use it to prove the equivalence between (a) and (b), if this is what you mean by "this thm of Higman"?
    $endgroup$
    – Salvo Tringali
    10 hours ago










  • $begingroup$
    Yes, I mean the infinite Ramsey theorem. I've put the proof into an answer, since it's too long for a comment.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    It follows immediately from Dilworth theorem: every infinite poset contains either an infinite chain or an infinite antichain. This is an easy (high school level) statement.
    $endgroup$
    – Mark Sapir
    4 hours ago












  • $begingroup$
    @MarkSapir Could you please explain how it follows from Dilworth's theorem? Dilworth's theorem (the one I know of, I'm sure he had many others) says that, if there is a finite bound $k$ on the sizes of antichains in a poset $P$, then $P$ is the union of $k$ chains. If the antichains are all finite but of unbounded finite size, then the number of antichains needed to cover $P$ can be an arbitrarily large infinite cardinal
    $endgroup$
    – bof
    2 hours ago
















$begingroup$
Unless I've misunderstood something, this theorem of Higman is an immediate consequence of Ramsey's theorem (which might have been rediscovered by Erdös and Rado).
$endgroup$
– Andreas Blass
11 hours ago




$begingroup$
Unless I've misunderstood something, this theorem of Higman is an immediate consequence of Ramsey's theorem (which might have been rediscovered by Erdös and Rado).
$endgroup$
– Andreas Blass
11 hours ago












$begingroup$
@AndreasBlass Do you mean the infinite Ramsey thm (en.wikipedia.org/wiki/Ramsey's_theorem)? If so, could you explain in some detail how to use it to prove the equivalence between (a) and (b), if this is what you mean by "this thm of Higman"?
$endgroup$
– Salvo Tringali
10 hours ago




$begingroup$
@AndreasBlass Do you mean the infinite Ramsey thm (en.wikipedia.org/wiki/Ramsey's_theorem)? If so, could you explain in some detail how to use it to prove the equivalence between (a) and (b), if this is what you mean by "this thm of Higman"?
$endgroup$
– Salvo Tringali
10 hours ago












$begingroup$
Yes, I mean the infinite Ramsey theorem. I've put the proof into an answer, since it's too long for a comment.
$endgroup$
– Andreas Blass
8 hours ago




$begingroup$
Yes, I mean the infinite Ramsey theorem. I've put the proof into an answer, since it's too long for a comment.
$endgroup$
– Andreas Blass
8 hours ago












$begingroup$
It follows immediately from Dilworth theorem: every infinite poset contains either an infinite chain or an infinite antichain. This is an easy (high school level) statement.
$endgroup$
– Mark Sapir
4 hours ago






$begingroup$
It follows immediately from Dilworth theorem: every infinite poset contains either an infinite chain or an infinite antichain. This is an easy (high school level) statement.
$endgroup$
– Mark Sapir
4 hours ago














$begingroup$
@MarkSapir Could you please explain how it follows from Dilworth's theorem? Dilworth's theorem (the one I know of, I'm sure he had many others) says that, if there is a finite bound $k$ on the sizes of antichains in a poset $P$, then $P$ is the union of $k$ chains. If the antichains are all finite but of unbounded finite size, then the number of antichains needed to cover $P$ can be an arbitrarily large infinite cardinal
$endgroup$
– bof
2 hours ago




$begingroup$
@MarkSapir Could you please explain how it follows from Dilworth's theorem? Dilworth's theorem (the one I know of, I'm sure he had many others) says that, if there is a finite bound $k$ on the sizes of antichains in a poset $P$, then $P$ is the union of $k$ chains. If the antichains are all finite but of unbounded finite size, then the number of antichains needed to cover $P$ can be an arbitrarily large infinite cardinal
$endgroup$
– bof
2 hours ago










2 Answers
2






active

oldest

votes


















3













$begingroup$

I too have come upon that statement in Higman's paper and wondered which manuscript he is referring to. There are two "papers" of Erdős and Rado that I think it might be.



First, it could be a reference to what became their solution to problem 4358 in the Monthly in 1952.



Second, it could instead be their 1959 paper "A theorem on partial well-ordering of sets of vectors" (J. London Math. Soc. (1959), 222–224).






share|cite|improve this answer









$endgroup$















  • $begingroup$
    Thanks. Erdos and Rado's 1959 paper is freely available at renyi.hu/~p_erdos/1959-02.pdf, but I'm having some difficulty in understanding its relevance to the OP: In the paper, 𝑊𝑆(<𝜔) is the free monoid over 𝑆 (is it?), and assuming that (𝑆,≤) is a partial order, Erdős and Rado refer back to Higman's 1952 paper for a proof that, if 𝑆 is partially well-ordered wrt ≤, then so is 𝑊𝑆(<𝜔) wrt the "subword order" induced from ≤ (as defined in the 2nd paragraph of p. 222). But then Erdős and Rado move on to (settle a conjecture of Rado and) prove a generalization of [...]
    $endgroup$
    – Salvo Tringali
    13 hours ago












  • $begingroup$
    [...] of Higman's result, where finite words are replaced with words of "length" smaller than $omega^omega$ which have only a finite number of distinct letters (Erdős and Rado use the term "vector" for "word" and "component" for "letter"); incidentally, they mention that the same result was also obtained by J. Kruskal in an unpublished manuscript. Am I missing anything?
    $endgroup$
    – Salvo Tringali
    12 hours ago












  • $begingroup$
    As for Erdős and Rado's solution of Problem 4358 in the AMM, there is a note on partially-well-ordered posets on pp. 256-257: In particular, the note credits Higman and B.H. Neumann for having found, independently of each other, a proof that all words over a partially-well-ordered alphabet $S$, is partially well ordered wrt the "subword order" induced from $S$ (this is Thm 4.3 in Higman's paper). But, again, there is no reference to the equivalence of conditions (a) and (b) from the OP.
    $endgroup$
    – Salvo Tringali
    10 hours ago





















3













$begingroup$

This is an answer not for the original question but for how to get the equivalence of (a) and (b) from the infinite Ramsey theorem (as requested in a comment). The implication from (a) to (b) is trivial. For the converse, I'm going to assume that "sequence" means one-to-one sequence (i.e., no repetitions), because otherwise a one-element set $A$ is a counterexample (because of the "strictly" in (a) and the non-strict $preceq$ in (b)).



So assume (b) and let $(a_i)_{iinmathbb N}$ be a sequence of distinct elements of $A$. Partition the set $[mathbb N]^2$ of $2$-element subsets ${i<j}$ of $mathbb N$ into two parts by putting ${i<j}$ into the first part if $a_iprec a_j$ and into the second part otherwise. By Ramsey's theorem, there is an infinite subset $H$ of $mathbb N$ all of whose $2$-element subsets are in the same part.



If they're all in the first part, that means that, for all $i<j$ in $H$, we have $a_iprec a_j$. In other words, the subsequence $(a_i)_{iin H}$ of our original sequence is increasing with respect to $prec$, as required for (a).



If they're all in the second part, that means that, for all $i<j$ in $H$, we have $a_inotprec a_j$. But then the sequence $(a_i)_{iin H}$ violates our assumption (b).



(The last step is where I need that the sequence is one-to-one, because I need $a_inotpreceq a_j$ in order to violate (b). If one allows sequences that are not one-to-one, then the best I can do is to partition $[mathbb N]^2$ into three parts, according to whether $a_iprec a_j$, $a_i=a_j$ or $a_inotpreceq a_j$. If $H$ is an infinite homogeneous set as given by Ramsey's theorem, then the result is that $(a_i)_{iin H}$ either is increasing as required for (a), or is constant, or violates (b). So what can go wrong in the not one-to-one case is essentially just the possibility of a constant sequence, as I indicated above in my remark about a one-element set $A$.)






share|cite|improve this answer









$endgroup$















  • $begingroup$
    @SalvoTringali OK. Now that "strictly" is gone, my argument works even when sequences are allowed to have repetitions. But I think I'll leave my answer as it is, since the last paragraph (in parentheses) might be useful for some readers.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    I see, many thanks for the details. The "strictly" in (a) is a mistake from my side, I'll edit the OP to fix it: Higman in his paper writes of "an infinite ascending subsequence." Edit. You posted your comment while I was editing mine (and I had to delete mine, since the edit came too late).
    $endgroup$
    – Salvo Tringali
    8 hours ago
















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2 Answers
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active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3













$begingroup$

I too have come upon that statement in Higman's paper and wondered which manuscript he is referring to. There are two "papers" of Erdős and Rado that I think it might be.



First, it could be a reference to what became their solution to problem 4358 in the Monthly in 1952.



Second, it could instead be their 1959 paper "A theorem on partial well-ordering of sets of vectors" (J. London Math. Soc. (1959), 222–224).






share|cite|improve this answer









$endgroup$















  • $begingroup$
    Thanks. Erdos and Rado's 1959 paper is freely available at renyi.hu/~p_erdos/1959-02.pdf, but I'm having some difficulty in understanding its relevance to the OP: In the paper, 𝑊𝑆(<𝜔) is the free monoid over 𝑆 (is it?), and assuming that (𝑆,≤) is a partial order, Erdős and Rado refer back to Higman's 1952 paper for a proof that, if 𝑆 is partially well-ordered wrt ≤, then so is 𝑊𝑆(<𝜔) wrt the "subword order" induced from ≤ (as defined in the 2nd paragraph of p. 222). But then Erdős and Rado move on to (settle a conjecture of Rado and) prove a generalization of [...]
    $endgroup$
    – Salvo Tringali
    13 hours ago












  • $begingroup$
    [...] of Higman's result, where finite words are replaced with words of "length" smaller than $omega^omega$ which have only a finite number of distinct letters (Erdős and Rado use the term "vector" for "word" and "component" for "letter"); incidentally, they mention that the same result was also obtained by J. Kruskal in an unpublished manuscript. Am I missing anything?
    $endgroup$
    – Salvo Tringali
    12 hours ago












  • $begingroup$
    As for Erdős and Rado's solution of Problem 4358 in the AMM, there is a note on partially-well-ordered posets on pp. 256-257: In particular, the note credits Higman and B.H. Neumann for having found, independently of each other, a proof that all words over a partially-well-ordered alphabet $S$, is partially well ordered wrt the "subword order" induced from $S$ (this is Thm 4.3 in Higman's paper). But, again, there is no reference to the equivalence of conditions (a) and (b) from the OP.
    $endgroup$
    – Salvo Tringali
    10 hours ago


















3













$begingroup$

I too have come upon that statement in Higman's paper and wondered which manuscript he is referring to. There are two "papers" of Erdős and Rado that I think it might be.



First, it could be a reference to what became their solution to problem 4358 in the Monthly in 1952.



Second, it could instead be their 1959 paper "A theorem on partial well-ordering of sets of vectors" (J. London Math. Soc. (1959), 222–224).






share|cite|improve this answer









$endgroup$















  • $begingroup$
    Thanks. Erdos and Rado's 1959 paper is freely available at renyi.hu/~p_erdos/1959-02.pdf, but I'm having some difficulty in understanding its relevance to the OP: In the paper, 𝑊𝑆(<𝜔) is the free monoid over 𝑆 (is it?), and assuming that (𝑆,≤) is a partial order, Erdős and Rado refer back to Higman's 1952 paper for a proof that, if 𝑆 is partially well-ordered wrt ≤, then so is 𝑊𝑆(<𝜔) wrt the "subword order" induced from ≤ (as defined in the 2nd paragraph of p. 222). But then Erdős and Rado move on to (settle a conjecture of Rado and) prove a generalization of [...]
    $endgroup$
    – Salvo Tringali
    13 hours ago












  • $begingroup$
    [...] of Higman's result, where finite words are replaced with words of "length" smaller than $omega^omega$ which have only a finite number of distinct letters (Erdős and Rado use the term "vector" for "word" and "component" for "letter"); incidentally, they mention that the same result was also obtained by J. Kruskal in an unpublished manuscript. Am I missing anything?
    $endgroup$
    – Salvo Tringali
    12 hours ago












  • $begingroup$
    As for Erdős and Rado's solution of Problem 4358 in the AMM, there is a note on partially-well-ordered posets on pp. 256-257: In particular, the note credits Higman and B.H. Neumann for having found, independently of each other, a proof that all words over a partially-well-ordered alphabet $S$, is partially well ordered wrt the "subword order" induced from $S$ (this is Thm 4.3 in Higman's paper). But, again, there is no reference to the equivalence of conditions (a) and (b) from the OP.
    $endgroup$
    – Salvo Tringali
    10 hours ago
















3














3










3







$begingroup$

I too have come upon that statement in Higman's paper and wondered which manuscript he is referring to. There are two "papers" of Erdős and Rado that I think it might be.



First, it could be a reference to what became their solution to problem 4358 in the Monthly in 1952.



Second, it could instead be their 1959 paper "A theorem on partial well-ordering of sets of vectors" (J. London Math. Soc. (1959), 222–224).






share|cite|improve this answer









$endgroup$



I too have come upon that statement in Higman's paper and wondered which manuscript he is referring to. There are two "papers" of Erdős and Rado that I think it might be.



First, it could be a reference to what became their solution to problem 4358 in the Monthly in 1952.



Second, it could instead be their 1959 paper "A theorem on partial well-ordering of sets of vectors" (J. London Math. Soc. (1959), 222–224).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 13 hours ago









Vince VatterVince Vatter

9086 silver badges24 bronze badges




9086 silver badges24 bronze badges















  • $begingroup$
    Thanks. Erdos and Rado's 1959 paper is freely available at renyi.hu/~p_erdos/1959-02.pdf, but I'm having some difficulty in understanding its relevance to the OP: In the paper, 𝑊𝑆(<𝜔) is the free monoid over 𝑆 (is it?), and assuming that (𝑆,≤) is a partial order, Erdős and Rado refer back to Higman's 1952 paper for a proof that, if 𝑆 is partially well-ordered wrt ≤, then so is 𝑊𝑆(<𝜔) wrt the "subword order" induced from ≤ (as defined in the 2nd paragraph of p. 222). But then Erdős and Rado move on to (settle a conjecture of Rado and) prove a generalization of [...]
    $endgroup$
    – Salvo Tringali
    13 hours ago












  • $begingroup$
    [...] of Higman's result, where finite words are replaced with words of "length" smaller than $omega^omega$ which have only a finite number of distinct letters (Erdős and Rado use the term "vector" for "word" and "component" for "letter"); incidentally, they mention that the same result was also obtained by J. Kruskal in an unpublished manuscript. Am I missing anything?
    $endgroup$
    – Salvo Tringali
    12 hours ago












  • $begingroup$
    As for Erdős and Rado's solution of Problem 4358 in the AMM, there is a note on partially-well-ordered posets on pp. 256-257: In particular, the note credits Higman and B.H. Neumann for having found, independently of each other, a proof that all words over a partially-well-ordered alphabet $S$, is partially well ordered wrt the "subword order" induced from $S$ (this is Thm 4.3 in Higman's paper). But, again, there is no reference to the equivalence of conditions (a) and (b) from the OP.
    $endgroup$
    – Salvo Tringali
    10 hours ago




















  • $begingroup$
    Thanks. Erdos and Rado's 1959 paper is freely available at renyi.hu/~p_erdos/1959-02.pdf, but I'm having some difficulty in understanding its relevance to the OP: In the paper, 𝑊𝑆(<𝜔) is the free monoid over 𝑆 (is it?), and assuming that (𝑆,≤) is a partial order, Erdős and Rado refer back to Higman's 1952 paper for a proof that, if 𝑆 is partially well-ordered wrt ≤, then so is 𝑊𝑆(<𝜔) wrt the "subword order" induced from ≤ (as defined in the 2nd paragraph of p. 222). But then Erdős and Rado move on to (settle a conjecture of Rado and) prove a generalization of [...]
    $endgroup$
    – Salvo Tringali
    13 hours ago












  • $begingroup$
    [...] of Higman's result, where finite words are replaced with words of "length" smaller than $omega^omega$ which have only a finite number of distinct letters (Erdős and Rado use the term "vector" for "word" and "component" for "letter"); incidentally, they mention that the same result was also obtained by J. Kruskal in an unpublished manuscript. Am I missing anything?
    $endgroup$
    – Salvo Tringali
    12 hours ago












  • $begingroup$
    As for Erdős and Rado's solution of Problem 4358 in the AMM, there is a note on partially-well-ordered posets on pp. 256-257: In particular, the note credits Higman and B.H. Neumann for having found, independently of each other, a proof that all words over a partially-well-ordered alphabet $S$, is partially well ordered wrt the "subword order" induced from $S$ (this is Thm 4.3 in Higman's paper). But, again, there is no reference to the equivalence of conditions (a) and (b) from the OP.
    $endgroup$
    – Salvo Tringali
    10 hours ago


















$begingroup$
Thanks. Erdos and Rado's 1959 paper is freely available at renyi.hu/~p_erdos/1959-02.pdf, but I'm having some difficulty in understanding its relevance to the OP: In the paper, 𝑊𝑆(<𝜔) is the free monoid over 𝑆 (is it?), and assuming that (𝑆,≤) is a partial order, Erdős and Rado refer back to Higman's 1952 paper for a proof that, if 𝑆 is partially well-ordered wrt ≤, then so is 𝑊𝑆(<𝜔) wrt the "subword order" induced from ≤ (as defined in the 2nd paragraph of p. 222). But then Erdős and Rado move on to (settle a conjecture of Rado and) prove a generalization of [...]
$endgroup$
– Salvo Tringali
13 hours ago






$begingroup$
Thanks. Erdos and Rado's 1959 paper is freely available at renyi.hu/~p_erdos/1959-02.pdf, but I'm having some difficulty in understanding its relevance to the OP: In the paper, 𝑊𝑆(<𝜔) is the free monoid over 𝑆 (is it?), and assuming that (𝑆,≤) is a partial order, Erdős and Rado refer back to Higman's 1952 paper for a proof that, if 𝑆 is partially well-ordered wrt ≤, then so is 𝑊𝑆(<𝜔) wrt the "subword order" induced from ≤ (as defined in the 2nd paragraph of p. 222). But then Erdős and Rado move on to (settle a conjecture of Rado and) prove a generalization of [...]
$endgroup$
– Salvo Tringali
13 hours ago














$begingroup$
[...] of Higman's result, where finite words are replaced with words of "length" smaller than $omega^omega$ which have only a finite number of distinct letters (Erdős and Rado use the term "vector" for "word" and "component" for "letter"); incidentally, they mention that the same result was also obtained by J. Kruskal in an unpublished manuscript. Am I missing anything?
$endgroup$
– Salvo Tringali
12 hours ago






$begingroup$
[...] of Higman's result, where finite words are replaced with words of "length" smaller than $omega^omega$ which have only a finite number of distinct letters (Erdős and Rado use the term "vector" for "word" and "component" for "letter"); incidentally, they mention that the same result was also obtained by J. Kruskal in an unpublished manuscript. Am I missing anything?
$endgroup$
– Salvo Tringali
12 hours ago














$begingroup$
As for Erdős and Rado's solution of Problem 4358 in the AMM, there is a note on partially-well-ordered posets on pp. 256-257: In particular, the note credits Higman and B.H. Neumann for having found, independently of each other, a proof that all words over a partially-well-ordered alphabet $S$, is partially well ordered wrt the "subword order" induced from $S$ (this is Thm 4.3 in Higman's paper). But, again, there is no reference to the equivalence of conditions (a) and (b) from the OP.
$endgroup$
– Salvo Tringali
10 hours ago






$begingroup$
As for Erdős and Rado's solution of Problem 4358 in the AMM, there is a note on partially-well-ordered posets on pp. 256-257: In particular, the note credits Higman and B.H. Neumann for having found, independently of each other, a proof that all words over a partially-well-ordered alphabet $S$, is partially well ordered wrt the "subword order" induced from $S$ (this is Thm 4.3 in Higman's paper). But, again, there is no reference to the equivalence of conditions (a) and (b) from the OP.
$endgroup$
– Salvo Tringali
10 hours ago













3













$begingroup$

This is an answer not for the original question but for how to get the equivalence of (a) and (b) from the infinite Ramsey theorem (as requested in a comment). The implication from (a) to (b) is trivial. For the converse, I'm going to assume that "sequence" means one-to-one sequence (i.e., no repetitions), because otherwise a one-element set $A$ is a counterexample (because of the "strictly" in (a) and the non-strict $preceq$ in (b)).



So assume (b) and let $(a_i)_{iinmathbb N}$ be a sequence of distinct elements of $A$. Partition the set $[mathbb N]^2$ of $2$-element subsets ${i<j}$ of $mathbb N$ into two parts by putting ${i<j}$ into the first part if $a_iprec a_j$ and into the second part otherwise. By Ramsey's theorem, there is an infinite subset $H$ of $mathbb N$ all of whose $2$-element subsets are in the same part.



If they're all in the first part, that means that, for all $i<j$ in $H$, we have $a_iprec a_j$. In other words, the subsequence $(a_i)_{iin H}$ of our original sequence is increasing with respect to $prec$, as required for (a).



If they're all in the second part, that means that, for all $i<j$ in $H$, we have $a_inotprec a_j$. But then the sequence $(a_i)_{iin H}$ violates our assumption (b).



(The last step is where I need that the sequence is one-to-one, because I need $a_inotpreceq a_j$ in order to violate (b). If one allows sequences that are not one-to-one, then the best I can do is to partition $[mathbb N]^2$ into three parts, according to whether $a_iprec a_j$, $a_i=a_j$ or $a_inotpreceq a_j$. If $H$ is an infinite homogeneous set as given by Ramsey's theorem, then the result is that $(a_i)_{iin H}$ either is increasing as required for (a), or is constant, or violates (b). So what can go wrong in the not one-to-one case is essentially just the possibility of a constant sequence, as I indicated above in my remark about a one-element set $A$.)






share|cite|improve this answer









$endgroup$















  • $begingroup$
    @SalvoTringali OK. Now that "strictly" is gone, my argument works even when sequences are allowed to have repetitions. But I think I'll leave my answer as it is, since the last paragraph (in parentheses) might be useful for some readers.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    I see, many thanks for the details. The "strictly" in (a) is a mistake from my side, I'll edit the OP to fix it: Higman in his paper writes of "an infinite ascending subsequence." Edit. You posted your comment while I was editing mine (and I had to delete mine, since the edit came too late).
    $endgroup$
    – Salvo Tringali
    8 hours ago


















3













$begingroup$

This is an answer not for the original question but for how to get the equivalence of (a) and (b) from the infinite Ramsey theorem (as requested in a comment). The implication from (a) to (b) is trivial. For the converse, I'm going to assume that "sequence" means one-to-one sequence (i.e., no repetitions), because otherwise a one-element set $A$ is a counterexample (because of the "strictly" in (a) and the non-strict $preceq$ in (b)).



So assume (b) and let $(a_i)_{iinmathbb N}$ be a sequence of distinct elements of $A$. Partition the set $[mathbb N]^2$ of $2$-element subsets ${i<j}$ of $mathbb N$ into two parts by putting ${i<j}$ into the first part if $a_iprec a_j$ and into the second part otherwise. By Ramsey's theorem, there is an infinite subset $H$ of $mathbb N$ all of whose $2$-element subsets are in the same part.



If they're all in the first part, that means that, for all $i<j$ in $H$, we have $a_iprec a_j$. In other words, the subsequence $(a_i)_{iin H}$ of our original sequence is increasing with respect to $prec$, as required for (a).



If they're all in the second part, that means that, for all $i<j$ in $H$, we have $a_inotprec a_j$. But then the sequence $(a_i)_{iin H}$ violates our assumption (b).



(The last step is where I need that the sequence is one-to-one, because I need $a_inotpreceq a_j$ in order to violate (b). If one allows sequences that are not one-to-one, then the best I can do is to partition $[mathbb N]^2$ into three parts, according to whether $a_iprec a_j$, $a_i=a_j$ or $a_inotpreceq a_j$. If $H$ is an infinite homogeneous set as given by Ramsey's theorem, then the result is that $(a_i)_{iin H}$ either is increasing as required for (a), or is constant, or violates (b). So what can go wrong in the not one-to-one case is essentially just the possibility of a constant sequence, as I indicated above in my remark about a one-element set $A$.)






share|cite|improve this answer









$endgroup$















  • $begingroup$
    @SalvoTringali OK. Now that "strictly" is gone, my argument works even when sequences are allowed to have repetitions. But I think I'll leave my answer as it is, since the last paragraph (in parentheses) might be useful for some readers.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    I see, many thanks for the details. The "strictly" in (a) is a mistake from my side, I'll edit the OP to fix it: Higman in his paper writes of "an infinite ascending subsequence." Edit. You posted your comment while I was editing mine (and I had to delete mine, since the edit came too late).
    $endgroup$
    – Salvo Tringali
    8 hours ago
















3














3










3







$begingroup$

This is an answer not for the original question but for how to get the equivalence of (a) and (b) from the infinite Ramsey theorem (as requested in a comment). The implication from (a) to (b) is trivial. For the converse, I'm going to assume that "sequence" means one-to-one sequence (i.e., no repetitions), because otherwise a one-element set $A$ is a counterexample (because of the "strictly" in (a) and the non-strict $preceq$ in (b)).



So assume (b) and let $(a_i)_{iinmathbb N}$ be a sequence of distinct elements of $A$. Partition the set $[mathbb N]^2$ of $2$-element subsets ${i<j}$ of $mathbb N$ into two parts by putting ${i<j}$ into the first part if $a_iprec a_j$ and into the second part otherwise. By Ramsey's theorem, there is an infinite subset $H$ of $mathbb N$ all of whose $2$-element subsets are in the same part.



If they're all in the first part, that means that, for all $i<j$ in $H$, we have $a_iprec a_j$. In other words, the subsequence $(a_i)_{iin H}$ of our original sequence is increasing with respect to $prec$, as required for (a).



If they're all in the second part, that means that, for all $i<j$ in $H$, we have $a_inotprec a_j$. But then the sequence $(a_i)_{iin H}$ violates our assumption (b).



(The last step is where I need that the sequence is one-to-one, because I need $a_inotpreceq a_j$ in order to violate (b). If one allows sequences that are not one-to-one, then the best I can do is to partition $[mathbb N]^2$ into three parts, according to whether $a_iprec a_j$, $a_i=a_j$ or $a_inotpreceq a_j$. If $H$ is an infinite homogeneous set as given by Ramsey's theorem, then the result is that $(a_i)_{iin H}$ either is increasing as required for (a), or is constant, or violates (b). So what can go wrong in the not one-to-one case is essentially just the possibility of a constant sequence, as I indicated above in my remark about a one-element set $A$.)






share|cite|improve this answer









$endgroup$



This is an answer not for the original question but for how to get the equivalence of (a) and (b) from the infinite Ramsey theorem (as requested in a comment). The implication from (a) to (b) is trivial. For the converse, I'm going to assume that "sequence" means one-to-one sequence (i.e., no repetitions), because otherwise a one-element set $A$ is a counterexample (because of the "strictly" in (a) and the non-strict $preceq$ in (b)).



So assume (b) and let $(a_i)_{iinmathbb N}$ be a sequence of distinct elements of $A$. Partition the set $[mathbb N]^2$ of $2$-element subsets ${i<j}$ of $mathbb N$ into two parts by putting ${i<j}$ into the first part if $a_iprec a_j$ and into the second part otherwise. By Ramsey's theorem, there is an infinite subset $H$ of $mathbb N$ all of whose $2$-element subsets are in the same part.



If they're all in the first part, that means that, for all $i<j$ in $H$, we have $a_iprec a_j$. In other words, the subsequence $(a_i)_{iin H}$ of our original sequence is increasing with respect to $prec$, as required for (a).



If they're all in the second part, that means that, for all $i<j$ in $H$, we have $a_inotprec a_j$. But then the sequence $(a_i)_{iin H}$ violates our assumption (b).



(The last step is where I need that the sequence is one-to-one, because I need $a_inotpreceq a_j$ in order to violate (b). If one allows sequences that are not one-to-one, then the best I can do is to partition $[mathbb N]^2$ into three parts, according to whether $a_iprec a_j$, $a_i=a_j$ or $a_inotpreceq a_j$. If $H$ is an infinite homogeneous set as given by Ramsey's theorem, then the result is that $(a_i)_{iin H}$ either is increasing as required for (a), or is constant, or violates (b). So what can go wrong in the not one-to-one case is essentially just the possibility of a constant sequence, as I indicated above in my remark about a one-element set $A$.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 8 hours ago









Andreas BlassAndreas Blass

60.3k7 gold badges146 silver badges235 bronze badges




60.3k7 gold badges146 silver badges235 bronze badges















  • $begingroup$
    @SalvoTringali OK. Now that "strictly" is gone, my argument works even when sequences are allowed to have repetitions. But I think I'll leave my answer as it is, since the last paragraph (in parentheses) might be useful for some readers.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    I see, many thanks for the details. The "strictly" in (a) is a mistake from my side, I'll edit the OP to fix it: Higman in his paper writes of "an infinite ascending subsequence." Edit. You posted your comment while I was editing mine (and I had to delete mine, since the edit came too late).
    $endgroup$
    – Salvo Tringali
    8 hours ago




















  • $begingroup$
    @SalvoTringali OK. Now that "strictly" is gone, my argument works even when sequences are allowed to have repetitions. But I think I'll leave my answer as it is, since the last paragraph (in parentheses) might be useful for some readers.
    $endgroup$
    – Andreas Blass
    8 hours ago










  • $begingroup$
    I see, many thanks for the details. The "strictly" in (a) is a mistake from my side, I'll edit the OP to fix it: Higman in his paper writes of "an infinite ascending subsequence." Edit. You posted your comment while I was editing mine (and I had to delete mine, since the edit came too late).
    $endgroup$
    – Salvo Tringali
    8 hours ago


















$begingroup$
@SalvoTringali OK. Now that "strictly" is gone, my argument works even when sequences are allowed to have repetitions. But I think I'll leave my answer as it is, since the last paragraph (in parentheses) might be useful for some readers.
$endgroup$
– Andreas Blass
8 hours ago




$begingroup$
@SalvoTringali OK. Now that "strictly" is gone, my argument works even when sequences are allowed to have repetitions. But I think I'll leave my answer as it is, since the last paragraph (in parentheses) might be useful for some readers.
$endgroup$
– Andreas Blass
8 hours ago












$begingroup$
I see, many thanks for the details. The "strictly" in (a) is a mistake from my side, I'll edit the OP to fix it: Higman in his paper writes of "an infinite ascending subsequence." Edit. You posted your comment while I was editing mine (and I had to delete mine, since the edit came too late).
$endgroup$
– Salvo Tringali
8 hours ago






$begingroup$
I see, many thanks for the details. The "strictly" in (a) is a mistake from my side, I'll edit the OP to fix it: Higman in his paper writes of "an infinite ascending subsequence." Edit. You posted your comment while I was editing mine (and I had to delete mine, since the edit came too late).
$endgroup$
– Salvo Tringali
8 hours ago




















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