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For a set $A \subseteq \omega$, let $[A]^n$ denote the set of subsets of $A$ of size $n$. I am trying to prove Ramsey's Theorem, and it seems like the following fact is used in the proof I am reading (Jech 9.1):

Suppose for every function $F:[\omega]^n \to \{ 1, \ldots, k \}$, there exists an infinite set $H \subseteq \omega$ such that F is constant on $[H]^n$. Then for any infinite subset $A\subseteq \omega$, and any function $F_A: [A]^n \to \{1, \ldots, k\}$, there exists an infinite subset $H_A \subseteq A$ such that $F_A$ is constant on $H_A$.

Is this true, and how do I prove it? I can't work out how to define $H_A$. If I extend $F_A$ to all of $[\omega]^n$ somehow, I know I can deduce there is an infinite set $H$ such that the extension is constant on $[H]^n$, but I don't know how to reduce this to an infinite set $H_A \subseteq A$, or how to extend the function. If I just define $H_A := A \cap H$, then I don't see why this is necessarily infinite.

Any help would be appreciated.

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If $A \subseteq \omega$ is infinite, choose a bijection $A \cong \omega$ and apply your assumption. HTH, AB –  martini Mar 13 '12 at 21:34
    
@martini: HTH,AB? –  Asaf Karagila Mar 13 '12 at 21:39
    
Brilliant, thanks. Asaf HTH = hope this helps! –  Paul Slevin Mar 13 '12 at 21:41
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@Paul: If you solved this, you should post an answer. –  Asaf Karagila Mar 14 '12 at 0:29
    
done so, thanks –  Paul Slevin Mar 14 '12 at 1:18

1 Answer 1

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Since $A$ is infinite, it has cardinality $\omega$, so there exists a bijection $f:\omega \to A$. Given a function $F: [A]^n \to \{1, \ldots, k \}$, define a function $\tilde{F}:[\omega]^n \to \{ 1, \ldots, k\}$ by $\tilde{F}(X) = F( f^{-1} (X))$. By hypothesis there exists an infinite set $H \subseteq \omega$ such that $\tilde{F}$ is constant on $H$. It is clear that $F$ is now constant on $H_A := f(H)$.

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