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First, some notation:

  • Set variables, $X, Y$, range over sets of natural numbers, $\mathbb{N}={1,2,3,..}$.

  • Square brackets represent sets of natural numbers based on a formula. $$[φ(\mathbf{n})]=\{φ(n)|n∈N\}$$ So, $[2n]=\{2,4,6…\}$ and $[\mathbf{n}^2] = {1, 4, 9, ..}$

  • $X_Y$ = the selection of $X$ based on $Y$: $$X_Y = \{X_k|k\in Y\}$$ where $X_k$ is the $k$-th member of $X$ (in numeric order).

    So, $X_{[2n]}$ consists of every second element of X and $X_{[n^2]} = \{X_1, X_4, X_9,…\}$

Now, the question: Is there a non-principal ultrafilter, $F$, over $\mathbb{N}$ such that:

  1. For each $k \in \mathbb{N}$, $[k\mathbf{n}]\in F$.
  2. If $X \in F$, then, for every $k$, $[k\mathbf{n}]_X \in F$.
Asaf Karagila
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fmkatz
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  • How did you come up with the question? – tomasz Nov 03 '12 at 18:15
  • I am trying to prove the existence of sizings of sets of natural numbers which have satisfy some similarity conditions - for example - that $$\sigma(X) = \sigma(Y) \iff \sigma([kn]X) =\sigma([kn]_Y)$$. By _sizings, I mean the kinds of orderings developed in the citations on my answer to http://math.stackexchange.com/questions/1393/relative-sizes-of-sets-of-integers-and-rationals-revisited-how-do-i-make-sense/203828#203828. – fmkatz Nov 03 '12 at 20:25
  • @fmkatz, the answer also depends on the set axiom system: according to Herrlich's Axiom of Choice, in ZF+AD there are no non-principal (he calls them "free") ultrafilters on $\mathbb N$ – alancalvitti Jan 14 '13 at 19:41

1 Answers1

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If I correctly understand your notation, the set $[kn]_X$ in requirement 2 is just $\{kx:x\in X\}$, so you are asking for the ultrafilter $F$ to be invariant under the operation $\mathbb N\to\mathbb N$ of multiplication by $k$ (for all $k$). There is no ultrafilter that accomplishes this, even for a single $k>1$. The reason is the theorem that, if an ultrafilter is mapped to itself by an operation $f$ on its underlying set, then the ultrafilter must contain the set of fixed-points of $f$.

For the particular operation you asked about, the (proof of) the general theorem can be specialized to the following. Partition $\mathbb N$ into the set $X$ of those numbers whose binary expansion ends with an even number of 0's (i.e., odd numbers, numbers congruent to 4 mod 8, etc.) and the set $Y$ of numbers whose binary expansion end with an odd number of 0's (congruent to 2 mod 4, or to 8 mod 16, etc.) An ultrafilter $F$ must contain exactly one of these two sets. But $[2n]_X=Y$ and $[2n]_Y=X$, so $F$ cannot satisfy your requirement 2.

Andreas Blass
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