MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there an strictly increasing function $f : \Bbb{R}\longrightarrow \Bbb{R}$ such that its image is algebraically independent (over $\Bbb{Q}$) ?

share|cite|improve this question

As usual, take some sequence $a_q>0$ enumerated by rationals with $\sum_q a_q<+\infty$ and put $f(x)=\sum_{q<x}a_q$. Now employ the standard trick to ensure independence. Enumerate all polynomials of several variables with rational coefficients and all rationals by the same index $n$. Now, assuming that $a_1,\dots,a_n$ are already chosen so that they are algebraically independent (which implies that any set of their non-empty sums with no common terms is algebraically independent too), impose the restriction $a_k\le 2^{n-k}\delta_n$ for $k>n$ where $\delta_n$ is so small that all polynomials $p_j$ ($j\le n$) do not vanish on any $\delta_n$-perturbations of any set of non-empty disjoint sums of numbers $a_1,\dots,a_n$ to which they can be applied. This still allows to choose $a_{n+1}>0$ algebraically independent from $a_1,\dots,a_n$. Now we can easily see that if $x_1<x_2<\dots<x_p$, then $f(x_1), f(x_2)-f(x_1),\dots, f(x_p)-f(x_{p-1})$ are algebraically independent.

This would be an excellent question for AoPS but it is a bit below MO standards (as a challenge question). So, don't get surprised if I'll vote to close without comments next time.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.