Theorem 1.2 Ramsey's Theorem ([7]) for any m,k < ω and f : [ω]k → m there exists
...... Apply the minimal cover assumption to all As, to get Bs ⊇ As with Bs ∈ B.
Infinite Ramsey Theory Math 873 Fall 1996 A.Miller
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Ramsey’s Theorem
Let ω = {0, 1, 2, . . .} and for any set A and n ≤ ω let [A]n = {B ⊆ A : |B| = n} where |B| is the cardinality of the set B. So, for example, [ω]ω , is the set of all infinite subsets of ω. Theorem 1.1 (Pidgeon Hole Principal1 ) Suppose f : ω → k. Then there exists H ∈ [ω]ω such that f H is constant. Theorem 1.2 Ramsey’s Theorem ([7]) for any m, k < ω and f : [ω]k → m there exists H ∈ [ω]ω such that f [H]k is constant. proof: The set H is said to be homogeneous for the function f . We begin with the standard proof for k = 2. Construct a0 < a1 < ... < an−1 and Xn ∈ [ω]ω as follows: Let an = min{Xn−1 } and find Xn ∈ [Xn−1 \ {an }]ω so that for every a ≤ an and u, v ∈ Xn f (a, u) = f (a, v). We can construct such a set by iteratively applying the pidgeon hole principal as follows. Note given any a ∈ ω and Y ∈ [ω]ω there is a Z ∈ [Y ]ω and i < m such that for every z ∈ Z we have f ({a, z}) = i. Now we just iterate this Xn−1 = Z0 ⊇ Z1 ⊇ Z2 . . . ⊇ Zm = Xn taking care of all a ≤ an (so m = an + 1). Finally consider the set K = {an : n ∈ ω}. It is “tail homogeneous”, i.e., for any u, v, w distinct elements of K if u < v and u < w, then f (u, v) = 1
One of my colleagues told me about the pidgeon head principal. If you get into an elevator and there are more buttons pressed than there are people in the elevator, then there must be a pidgeon head.
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f (u, w). Thus we can define g : K → m by g(u) = f (u, v) for any v > u in K. By the pidgeon hole principle there exists H ∈ [K]ω such that g H is constant and therefore f [H]2 is constant. Instead of giving the standard proof for k > 2 for novelty we give a proof2 using model theory. Let f : [ω]k+1 → m be any function. Consider the model A = (ω, an
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f (u1 , . . . , uk , x) = iu1 ,...,uk
u1
