3

Given a function from naturals to reals, can I pick countable many functions $$f_1,f_2,f_3, \ldots$$ Such that $$O(f_1) \subseteq O(f_2) \subseteq O(f_3)\subseteq\ldots$$ and every function is in $O(f_n)$ for some $n$?

If not countable, what is the smallest cardinal in which we can? I initially thought we cannot, but then I figured why not define it recursively with $f_2$ is chosen to be a function such that $O(f_1\circ f_1 \circ \ldots \circ f_1) \subseteq O(f_2)$ holds for arbitrary compositions. If we take $f_1(x) = x^k$ then $f_2(x) = \mathrm{exp}(x)$ works, and I am sure one could construct an $f_3$, however I feel as if this construction is too simplistic to exhaust all possible functions.

Scezory
  • 33

1 Answers1

7

Any countably many functions can be eventually exceeded (in jargon: dominated) by a single one: let $$g(x)=1+\sum_{i=1}^xf_x(x).$$ A similar trick shows that for any countably many functions $f_i$ ($i\in\mathbb{N}$) there is a function $g\not\in\bigcup_{i\in\mathbb{N}}O(f_i)$. In general, it's easier to analyze "$g$ dominates $f$" than the looser condition "$g\in O(f)$" (or "$g = O(f)$") and the set-theoretic questions about the two (to the best of my knowledge) all wind up being equivalent, so I'll focus on domination for the rest of this answer.

The question of exactly how large a set of functions must be so that each function from naturals to naturals (you ask about naturals to reals, but making the codmain $\mathbb{N}$ instead of $\mathbb{R}$ doesn't change anything - just think about applying $\lceil .\rceil$ after the fact) is dominated by an element of the set leads to the topic of cardinal characteristics of the continuum (see also this old answer of mine). Briefly, this cardinal - called "$\mathfrak{d}$" - is provably at least $\aleph_1$ and at most $2^{\aleph_0}$, but that's essentially all that can be said; e.g. it's consistent that $\mathfrak{d}=\aleph_{\omega^2+13}$ and $2^{\aleph_0}=\aleph_{\omega_1}$.

Noah Schweber
  • 260,658