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Computability and Logic

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324 RAMSEY’S THEOREM<br />

as follows. For any size-r subset x of ω, let p be its largest element. Then for any k<br />

large enough that p < n + k, x will be in the domain of f k , <strong>and</strong> we have<br />

f k (x) = f k+1 (x) = f k+2 (x) =···.<br />

Let F(x) be this common value.<br />

By the infinitary version of Ramsey’s theorem, F has an infinite homogeneous set.<br />

That is, there is an infinite Y <strong>and</strong> a j with 1 ≤ j ≤ s such that F:[Y ] r →{j}. Let Z<br />

be the set of the first n elements of Y , <strong>and</strong> take k large enough that the largest element<br />

of Z is less than n + k. Then Z will be a size-n subset of {0,...,n + k − 1}, with<br />

f k (x) = F(x) = j for all size-r subsets x of Z. In other words, Z will be a size-n<br />

homogeneous set for f k , which is impossible, since f k is in T . This contradiction<br />

completes the proof.<br />

Proof of Lemma 26.3: To prove König’s lemma we are going to use the compactness<br />

theorem. Let L T be the language with one one-place predicate B <strong>and</strong> with one<br />

constant t for each node t in the tree T . Let Ɣ consist of the following quantifier-free<br />

sentences:<br />

(1)<br />

Bs 1 ∨···∨Bs k<br />

where s 1 , ..., s k are all the nodes in T 0 ;<br />

(2)<br />

∼(Bs & Bt)<br />

for all pairs of nodes s, t belonging to the same level; <strong>and</strong><br />

(3)<br />

∼Bs ∨ Bu 1 ∨···∨Bu m<br />

for every node s, where u 1 , ..., u m are all the nodes immediately above s. [If there<br />

are no nodes above s, the sentence (3) is just ∼Bs.]<br />

We first show that if Ɣ has a model M, then T has an infinite branch. By (1) there<br />

will be at least one node r in T 0 such that Br is true in M. By (2) there will in fact<br />

be exactly one such node, call it r 0 . By (3) applied with r 0 as s, there will be at least<br />

one node r immediately above r 0 such that Br is true in M. By (2) there will in fact be<br />

exactly one such node, call it r 1 . Repeating the process, we obtain r 0 , r 1 , r 2 , ..., each<br />

immediately above the one before, which is to say that we obtain an infinite branch.<br />

We next show that Ɣ does have a model. By the compactness theorem, it is enough<br />

to show that any finite subset of Ɣ has a model. In fact, we can show that for any<br />

k, the set Ɣ k containing the sentence (1), all the sentences (2), <strong>and</strong> all the sentences<br />

(3) for s of level less than k has a model. We can then, given a finite , apply this fact<br />

to the least k such that all s occurring in are of level

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