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OK so here's a precise question:

Is it true that for every integer $k\geq1$ and every $f:\mathbb{Z}^k\to\mathbb{Z}$, there is some infinite subset $A\subseteq\mathbb{Z}$ such that $f(A^k)$ is not all of $\mathbb{Z}$? Here $A^k$ is the obvious subset of $\mathbb{Z}^k$ consisting of elements all of whose coordinates are in $A$.

This is some delicate question in infinite Ramsey theory. It seems to me that Harvey Friedman knows a lot about this question, and knows various fragments of Peano Arithmetic where this is provable, or not provable, and so on. My understanding is that Friedman calls this "the thin set theorem". But my reading of various bits of Friedman's writing on this subject still leaves me confused about whether this result is provable in what I would call "mathematics" and what he would probably call "ZFC". Can anyone clarify the status of this question in ZFC for me?

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Friedman himself provides a proof here: – user83827 Feb 10 '12 at 21:18
Thanks. Do you understand what is written there? I know of a theorem that I thought was called "Ramsey's theorem" -- for all positive integers $a$ and $b$ there's some large $N$ such that if you colour the edges of the complete graph with $N$ vertices with two colours red and blue, there's either a complete red subgraph with $a$ vertices or a complete blue subgraph with $b$ vertices. This gets me to line 2 of the proof. I am now stuck on "let $p$ be the number of order types of $k$-tuples". Can you give me a hint? – Kevin Buzzard Feb 10 '12 at 21:38
He's using the infinite Ramsey theorem: if you color k-sized subsets of the naturals with finitely many colors, there's an infinite set all of whose k-subsets get the same color. The problem here is that you're coloring (ordered) tuples instead of (unordered) sets, so you can't get away with a single color but by iterating the Ramsey theorem once for each order type of the tuple you can get a set where the color of the tuple depends only on the order type. Here order type means recording where in the tuple the smallest element is, where the next smallest is, and so on. – user83827 Feb 10 '12 at 21:53
So "order type" is just basically an element of the symmetric group of size $k$? Thanks -- you've given me a good start for reading this rather terse proof. – Kevin Buzzard Feb 10 '12 at 22:01
Basically, although there's the added wrinkle of potentially duplicating an element. For example, in 2-tuples there are three order types: smaller first, bigger first, or both the same number. When you allow duplications you are essentially passing to more degenerate form of Ramsey's theorem (coloring smaller subsets). For example, in this case, coloring pairs of the form $(z,z)$ is really just like coloring integers themselves, so you are using the pigeonhole principle rather than the true Ramsey theorem. – user83827 Feb 10 '12 at 22:09
up vote 5 down vote accepted

I'll follow @ccc's advice and answer my own question by explaining Friedman's argument but giving some more details. I've made it community wiki (for the usual reason: I then won't be seen to gain from answering my own question).

Notation: if $X$ is a set and $n\in\mathbb{Z}_{\geq1}$ then $X^{(n)}$ denotes the set of subsets of $X$ of size $n$.

We start with

The infinite Ramsey Theorem: If $X$ is countably infinite and we colour $X^{(n)}$ with $k\in\mathbb{Z}_{\geq1}$ colours (that is, we give a map $X^{(n)}\to\{1,2,\ldots,k\}$) then there's an infinite monochramatic subset of $X$ (that is, there's some infinite $Y\subseteq X$ such that the induced map $Y^{(n)}\to\{1,2,\ldots,k\}$ is constant).

The proof is short but clever, and is at wikipedia for example. The proof does the case $k=2$ by a clever induction on $n$ and then does the general case by induction on $k$. I'll omit the details.

Clearly this is a result with the same flavour as the thin set theorem, but as ccc points out, this isn't quite what we want because in the thin set theorem we have ordered $k$-tuples, not subsets of size $n$. So we need a minor modification to deal with this. Here's how it goes. Define an equivalence relation on $\mathbf{Z}^k$, called "being of the same order type", thus: say $a=(a_1,a_2,\ldots,a_k)$ and $b=(b_1,b_2,\ldots,b_k)$ are equivalent if for all $1\leq i,j\leq k$ we have $a_i<a_j$ iff $b_i<b_j$ (note that this implies $a_i=a_j$ iff $b_i=b_j$). Clearly there are only finitely many equivalence classes. Say there $p=p(k)$ equivalence classes. We call the equivalence classes "order types".

Now here's Friedman's proof of the thin set theorem. Given $f$ as in the theorem, define $H:\mathbf{Z}^k\to\{1,2,\ldots,p+1\}$ by $H(x)=f(x)$ if $1\leq f(x)\leq p$, and $H(x)=p+1$ otherwise. Now we apply the infinite Ramsey Theorem $p$ times! We first apply it to the colouring of $\mathbf{Z}^{(k)}$ given thus: given a subset of $\mathbf{Z}$ of size $k$, order the elements as $x_1<x_2<\ldots<x_k$. Now colour this set with colour $H((x_1,x_2,\ldots,x_k))$. Applying Ramsey we get an infinite subset $A_1$ of $\mathbf{Z}$ which is monochromatic. Now we apply it to the colouring of $(A_1)^{(k)}$ given thus: given a subset of size $k$ write it as $x_2<x_1<x_3<x_4<\ldots<x_k$ and colour the subset with the colour $H((x_1,x_2,\ldots,x_k))$. We get an infinite subset $A_2$ of $A_1$ which is monochromatic. We continue in this way, applying Ramsey to each order type. For example we might next do the order type $x_1=x_2<x_3<x_4<\ldots$ and we deal with this by colouring $k-1$-tuples of integers by ordering them and repeating the first element and then applying $H$.

The upshot is an infinite subset $A_p$ of the integers with the property that $H(a_1,a_2,\ldots,a_k)$ depends only on the order type of $(a_1,a_2,\ldots,a_k)$. But there are only $p$ order types, and $H$ takes $p+1$ values, so $H$ on $A^k$ is not surjective and hence $f$ isn't either.

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Dear Kevin, In your second para. of the proof of the infinite Ramsey Theorem, you mention a parameter $k$ which doesn't seem to have appeared before. [Added: I just looked at wikipedia, and see that $k$ should be $c$, as I suspected.] Cheers, – Matt E Apr 4 '12 at 1:12
Thanks Matt. I fixed it. You could have edited it yourself, right? (rather than wait 3 months for me to log into a stackexchange site :-) ) – Kevin Buzzard Jul 10 '12 at 17:42
Dear Kevin, Sure, but I prefer just to leave a note than to edit someones else's work. Cheers, – Matt E Jul 10 '12 at 17:45

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