Example 5

For $x$ real tending to $+\infty$ the set $\mathscr{E}$ of functions of the form $x^\alpha (\log x)^\beta$ (defined for $x > 1$), where $\alpha$ and $\beta$ are arbitrary real numbers, is a comparison scale. Indeed, here again the quotient of two functions in $\mathscr{E}$ is a function in $\mathscr{E}$; it is enough to show that if $\alpha$ and $\beta$ are not both zero then $x^\alpha (\log x)^\beta$ tends to 0 or $+\infty$; this is clear if $\alpha = 0, \beta \neq 0$; if $\alpha > 0$ one has $(\log x)^{-\beta} \ll x^\alpha$, and if $\alpha < 0$ one has $(\log x)^\beta \ll x^{-\alpha}$ for any $\beta$, whence the proposition.

Note that this last comparison scale is a totally ordered set (for the relation "f $\ll$ g" or $f = g$") whose order structure is isomorphic to the lexicographic order on $\mathbf{R}^2$ (Set Theory, III, p. 157); recall that in this structure the relation $(\alpha, \beta) < (\gamma, \delta)$ means "$\alpha < \gamma$, or $\alpha = \gamma$ and $\beta < \delta$").

Similarly, the scale formed by the functions $\exp(p(x))$, where $p$ runs through the set $P_0$ of polynomials with no constant term, has order structure isomorphic to the order structure of $P_0$, in which the relation $p < q$ implies that the dominant term of the polynomial $q - p$ has a coefficient $> 0$ (cf. Alg., VI. 19, Example 2).

Let $\varphi$ be a map from a set F into E, such that $\varphi^{-1}(\mathfrak{F})$ is a filter base on F. If $\mathscr{E}$ is a comparison scale on E (for the filter base $\mathfrak{F}$) then the functions $f \circ \varphi$, as $f$ runs through $\mathscr{E}$, form a comparison scale on F (for the filter base $\varphi^{-1}(\mathfrak{F})$).