4
$\begingroup$

I am wondering if the boundedness of growth can be characterized by sequences. I am not sure if I use the term "growth" correctly, or use the correct tags for this question. Here is what I mean.

Let $X$ be an uncountable set and let $F$ be a collection of real-valued functions on $X$ with the following property:

For every countable set $A\subset X$ there is a function $\alpha_{A}:A\to(0,+\infty) $ such that for every $f\in F$ the set $\left\{\frac{f(x)}{\alpha_{A}(x)}, x\in A\right\}$ is bounded.

Does there exist a function $\alpha:X\to(0,+\infty)$ such that for every $f\in F$ the set $\left\{\frac{f(x)}{\alpha(x)}, x\in X\right\}$ is bounded?

If the answer is "no", can we remedy the situation by assuming that $X$ is a topological space satisfying some mild conditions and $F\subset C(X)$?

Improve this question
$\endgroup$
5
  • $\begingroup$ If I do not misunderstand, then I think the answer is yes for easy reasons: applying your given condition in the special case where each A is a singleton, it follows that for each $x\in X$ we have $h(x):=\sup\{ f(x) \colon f\in F\}<\infty$. Then $h:X\to (0,\infty)$ is your required upper bound for the collection $F$. $\endgroup$
    – Yemon Choi
    Sep 27, 2018 at 1:32
  • 3
    $\begingroup$ @YemonChoi: I think you do misunderstand --- the sup over $F$ is not supposed to be finite, just the sup over $A$ for each $f$ separately. $\endgroup$
    – Nik Weaver
    Sep 27, 2018 at 1:45
  • $\begingroup$ @YemonChoi it is exactly as Nik Weaver says. In fact, $F$ is WLOG a vector space, and the condition should be viewed as existence of a weighted supremum seminorm. $\endgroup$
    – erz
    Sep 27, 2018 at 2:07
  • 1
    $\begingroup$ @NikWeaver Thank you! Apologies to the OP. $\endgroup$
    – Yemon Choi
    Sep 27, 2018 at 3:43
  • $\begingroup$ A remark: let $G$ be a (semi)group. Let $F$ be the set of subadditive non-negative functions on $X$. If $G$ satisfies the property that every countable subset is contained in a finitely generated semigroup, then $G$ satisfies the "countable boundedness property". (Several naturally defined uncountable (semi)groups are known to satisfy this; the first example is maybe the semigroup $E^E$ of self-maps of a set $E$, Sierpinski 1935.) In this context, I don't know if the conclusion always hold (yes for $E^E$, since every subadditive function on $E^E$ is bounded, following from Sierpinski's proof). $\endgroup$
    – YCor
    Sep 27, 2018 at 9:52

1 Answer 1

Reset to default
6
$\begingroup$

This was a fun question! The answer is no.

Let $\Omega$ be the set of all countable ordinals. For each limit ordinal $\alpha \in \Omega$ let $f_\alpha: \Omega \to [1,\infty)$ be a function which increases to infinity on $[0,\alpha)$ and is constantly zero on $[\alpha,\Omega)$.

For any limit ordinal $\alpha \in \Omega$ the set of $f_\beta$ with $\beta \leq \alpha$ a limit ordinal is countable, and therefore there is a function $a_\alpha: [0,\alpha) \to [1,\infty)$ such that $f_\beta \preceq a_\alpha$ for all $\beta < \alpha$ on $[0,\alpha)$, where $\preceq$ denotes "$\leq$ except at finitely many points". Since any $f_\beta$ with $\beta > \alpha$ is bounded on $[0,\alpha)$ we get the stated condition.

However, there is no $a: \Omega \to (0,\infty)$ that works. Given any such function $a$, there is an $n \in \mathbb{N}$ such that $a(\alpha) \leq n$ for uncountably many $\alpha$. Let $(\alpha_k)$ be an increasing sequence of such $\alpha$'s and let $\alpha$ be their supremum. Then the function $f_\alpha$ goes to infinity on $[0,\alpha)$ so its quotient with $a$ will still be unbounded.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Do I understand correctly that $a_{\alpha}$ from your second can be constructed in the following way? Form a sequence $\beta_n$ of the limit ordinals smaller than $\alpha$ and then say that $a_{\alpha}(\beta)=\max f_{\beta_k}(\beta)$, where $k\le m$, where $\beta_m$ is the the greatest limit ordinal smaller than $\beta$. $\endgroup$
    – erz
    Sep 27, 2018 at 2:46
  • 1
    $\begingroup$ I think so, yes. It's easier to say more generally that any countable family of functions on a countable set can be majorized in this way: wlog the functions are $f_n: \mathbb{N} \to (0,\infty)$, and then you let $a(n) = {\rm max}(f_1(n), \ldots, f_n(n))$. $\endgroup$
    – Nik Weaver
    Sep 27, 2018 at 3:01
  • $\begingroup$ @NikWeaver you're right, thanks (mistaken comments erased now) $\endgroup$
    – YCor
    Sep 27, 2018 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.