Let $\mathcal F$ be the set of all functions $\mathbb R^+\to\mathbb R$ that are:

  • real analytic on $\mathbb R^+$,
  • monotonic on $\mathbb R^+$, and
  • having derivatives of any order that are also monotonic on $\mathbb R^+$.

Apparently, $\mathcal F$ has a cardinality of continuum $\mathfrak{c}$.

For $f,g\in\mathcal F$ let $f\prec g$ denote the eventual domination (i.e. $f(x)<g(x)$ for all sufficiently large $x$).


Does $\prec$ define a linear order on $\mathcal F?$ Denote its order type as $\phi$. What can be said about $\phi$? Apparently, the order type of reals $\lambda$ can be embedded into $\phi$. Are they equivalent? What is the height of $\phi$ (i.e. the smallest ordinal not embeddable into it)? What's the cofinality of $\phi$?


(1) You can construct continuum many pairwise incomparable (under eventual domination) functions of the form $f(x) = \sum_{n \geq 0} \frac{a_n x^n}{n!}$ where each $a_n \in [0, 1]$.

(2) $a \mapsto e^{ax}$, $a > 0$ embeds reals into $\phi$.

(3) For any increasing sequence of numbers $b_n$, there is an analytic function $f(x) = \sum_{n \geq 1} a_n x^n$ with each $a_n \geq 0$ such that for every $n$, $f(n) > b_n$. This means that the height of $\phi$ is same as that of $(\omega^{\omega}, \leq^{\star})$.

  • $\begingroup$ Is there a simple example of two pairwise incomparable functions of this form? Does "incomparable" mean that their difference will change sign infinitely often? $\endgroup$ – TauMu Jan 11 '15 at 4:31
  • 1
    $\begingroup$ OK, here's a simple one: $10 e^x$, $10 e^x + \sin{x}$. $\endgroup$ – user203787 Jan 11 '15 at 4:41

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