I am self studying point set topology with MAT327 from Toronto University. In the topic of countability (chapter 4), I am asked to: "given a fixed set of functions $f_n:\mathbb{N}\rightarrow\mathbb{N}$, $n\in\mathbb{N}$ construct a function $g: \mathbb{N}\rightarrow\mathbb{N}$ that grows faster than all $f_n$"

this is: $\lim_{k\rightarrow\infty}\frac{g(k)}{f_n(k)}= \infty$

In the course is assumed that $0\not\in\mathbb{N}$. If the set of functions $f_n$ were finite $g(k)=\Pi_nf(k)^{f(k)}$ would be a solution, but I am not sure I can use this trick for an infinite set of functions.

How can I construct a function than grows faster than a given (infinite) set of functions?

  • 1
    $\begingroup$ What definition of "growth" do you use? As in big-O? Also your $g$ is not necessarily growing strictly faster than all $f_n$ in finite case, e.g. if all $f_n$ are constant, then so is your $g$. $\endgroup$
    – freakish
    Jul 5 at 12:32
  • $\begingroup$ @freakish Edited the question to define growth. Thanks for noticing the caveat for constant f_n. I take your comment as a solution. Thx $\endgroup$ Jul 5 at 12:41

1 Answer 1


Your $g(k)=\prod_{n} f_n(k)^{f_n(k)}$ does not necessarily work even in finite case. For example when all $f_n$ are constant then so is $g$.

What works for both finite and infinite case is

$$g(k)=k\cdot\max_{j\leq k}\{f_j(k)\}$$

Indeed, for any fixed $n$ we have

$$\lim_{k\to\infty}\frac{g(k)}{f_n(k)}=\lim_{k\to\infty}\frac{k\cdot\max_{j\leq k}\{f_j(k)\}}{f_n(k)}$$

Now given $k\geq n$ we have $$\frac{k\cdot\max_{j\leq k}\{f_j(k)\}}{f_n(k)}\geq\frac{k\cdot f_n(k)}{f_n(k)}=k\to\infty$$


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