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# [Solved]: Is $\{\Theta(f)|f:\mathbb{N}\rightarrow\mathbb{N}\}$ Dedekind-complete?

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Problem Detail:

Let $\Theta$ and $o$ be defined as usual (Landau-notation). For two equivalence classes defined by $\Theta$ we define $$\Theta(f) <_o \Theta(g) :\Leftrightarrow f \in o(g)\qquad.$$ Let $$\mathbb{F}:= \{\Theta(f)\mid f:\mathbb{N}\rightarrow\mathbb{N}\}$$

My question: Is $\langle \mathbb{F},<_o\rangle$ Dedekind-complete, i.e. given a set $F\subseteq \mathbb{F}$ with an upper bound $f_u,\forall f \in Ff\in o(f_u)$ are there $f_\inf$ and $f_\sup$ s.t. $$\forall \Theta(f)\in F: \Theta(f_\inf) <_o \Theta(f) \qquad{(1)}$$ $$\forall g: (g\text{ fulfills (1)} \Rightarrow g \in o(f_\inf))$$

($f_\sup$ is defined analogously)?

Note: We don't need to assume a lower bound, since there is no infinite strictly decreasing series of $n \in \mathbb{N}$, i.e. $\Theta(0)$ is always a lower bound.

Background: I'd like to consider $\inf \{f \mid L \in N/DTIME/SPACE(f)\}$ for some Language $L$, but that only makes sense if such a (class of) function(s) exist(s).

#### Answered By : Yuval Filmus

Let $F$ be the set of all functions $f(n)$ such that $\sum_n f(n)$ diverges. Then $F$ is bounded from below by $f(n) = 1/n^2$ (for example), but there is no infimum: for every divergent sequence, you can find a sequence which diverges more slowly. This is a classical result of Hausdorff.

Similarly, if $F$ is the set of all functions $f(n)$ such that $\sum_n 1/f(n)$ converges, then $F$ is bounded from above by $f(n) = n$, but there is no supremum since for every convergent sequence you can find a convergent sequence which grows much faster.

Edit: This paper (among others) provides a proof of these claims.