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.



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



Bounded growth of functions vs bounded growth of functions on countable sets
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.


