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Topic: Thin sets for integer functions (Read 1125 times) |
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Pietro K.C.
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Thin sets for integer functions
« on: Oct 16th, 2007, 11:16pm » |
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Let N denote the natural numbers, and let
f : NČ ---> N
be any function. Prove that there is an infinite set A, contained in N, such that f(AČ) is not all of N.
Can you generalize to f : N^k ---> N? (For k=1 it's trivial.)
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Eigenray
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Re: Thin sets for integer functions
« Reply #1 on: Jan 8th, 2008, 6:51am » |
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Well now this was tricky. I started thinking about whether it was true with only finitely many colors. For example, if we color N2 using 3 colors, for {x<y}, {x=y}, and {x>y}, then clearly any infinite set A2 must use all three colors. But what about 4? Will a finite number of colors suffice? Put like this, it seemed a bit like Ramsey theory:
Infinite Ramsey theorem (IRT): Let X be some countably infinite set and colour the elements of X(n) (the subsets of X of size n) in c [finite!] different colours. Then there exists some infinite subset M of X such that the size n subsets of M all have the same colour.
Now we can use this to prove the following: if f:N2  {1,2,3,4}, then there is an infinite subset A  N such that f(A2) misses a value (and as we saw above, this is optimal).
For a<b, define fi : N(2) {1,2,3,4} by f1({a,b})=f(a,b), f2({a,b})=f(b,a), and f3({a,b})=f(a,a).
By IRT, we can pick an infinite set M1  N such that f1(M1(2)) = {x} is a single value. Now, restricting f2 to M1(2), we again use IRT to pick an infinite subset M2 M1 such that f2(M2(2)) = {y} for some y, and then finally A M2 with f3(A(2)) = {z}.
Now if (a,b)  A2, either a < b, a > b, or a = b. In the first case, f(a,b) = f1({a,b}) = x; in the second, f(a,b) = f2({a,b}) = y. And in the third case we can pick b' A with b' > a, so that f(a,a) = f3({a,b'}) = z. So f(A2) = {x,y,z}.
For arbitrary k, we can show that if f : Nk {1,2,...,R+1}, then we can pick A such that f(Ak) misses a value, where R is the number of ways to rank k things (allowing ties). That is, we define R functions hi : N(k) Nk, one using each order type, so that for any element (x1,...,xk) Nk, we have that for some i, hi({x1,...,xk} = (x1,...,xk), roughly speaking. The complication is that if {x1,...,xk} contains fewer than k elements, we can pad the set with sufficiently large numbers. Then take fi = f hi and argue as before.
For example, with k=3, and x<y<z, we might have
h1({x,y,z}) = (x,y,z),
h2({x,y,z}) = (x,z,y),
...
h6({x,y,z}) = (z,y,x),
easy enough. Then we have to consider rankings where two are the same, say a<(b=c). This leads to
h7({x,y,z}) = (x,y,y),
so that (a,b,c) = h7({a,b,large}). Or if b < (a=c), we do
h8({x,y,z}) = (y,x,y),
so that (a,b,c) = h8({b,a,large}).
Finally
h13({x,y,z}) = (x,x,x).
In general: if T is the order type where
a1,...,ak take on r distinct values, say v1 < ... < vr, we would set the corresponding
hT({x1<x2<...<xk}) = (xj1,...,xjk),
where ji {1,...,r} is defined by ai = vji. Clearly the definition of hT only depends on the order type T.
Now by construction, if the k-tuple (a1,...,ak) has order type T, then for L1,...,Lk-r large enough (bigger than vr), if we sort the set
S={a1,...,ak,L1,...,Lk-r} = {v1 < v2 < ... < vr < L1 < ... < Lk-r},
we get by definition
hT(S) = (vj1,...,vjk) = (a1,...,ak).
[I don't know if that actually makes it any clearer, but I needed to write it out to convince myself!]
Now of course if f : Nk U, where U is any set with more than R elements, then we can pick a surjection s : U {1,...,R+1}, and find that sf misses a value, so f did as well.
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| « Last Edit: Jan 8th, 2008, 7:08am by Eigenray » |
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Grimbal
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Re: Thin sets for integer functions
« Reply #2 on: Jan 8th, 2008, 7:03am » |
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on Oct 16th, 2007, 11:16pm, Pietro K.C. wrote:| Prove that there is an infinite set A, contained in N, such that f(AČ) is not all of N. |
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Is f(AČ) = { f(a,b) | a, b in A } ?
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Eigenray
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Re: Thin sets for integer functions
« Reply #3 on: Jan 8th, 2008, 7:13am » |
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on Jan 8th, 2008, 7:03am, Grimbal wrote:
Is f(AČ) = { f(a,b) | a, b in A } ? |
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Yes, as a corollary of A2 = {(a,b) | a,b in A}, and f(S) = { f(x) | x in S }.
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