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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 $F\subset C(X)$?

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  • $\begingroup$ 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 '18 at 1:32
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    $\begingroup$ @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 '18 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 '18 at 2:07
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    $\begingroup$ @NikWeaver Thank you! Apologies to the OP. $\endgroup$ – Yemon Choi Sep 27 '18 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 '18 at 9:52
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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.

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  • $\begingroup$ 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 '18 at 2:46
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    $\begingroup$ 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 '18 at 3:01
  • $\begingroup$ @NikWeaver you're right, thanks (mistaken comments erased now) $\endgroup$ – YCor Sep 27 '18 at 22:28

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