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

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 $$F\subset C(X)$$?

• 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 $F$. – Yemon Choi Sep 27 '18 at 1:32
• @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. – Nik Weaver Sep 27 '18 at 1:45
• @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. – erz Sep 27 '18 at 2:07
• @NikWeaver Thank you! Apologies to the OP. – Yemon Choi Sep 27 '18 at 3:43
• 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). – YCor Sep 27 '18 at 9:52

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.
• 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 $\beta$. – erz Sep 27 '18 at 2:46
• 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 $a(n) = {\rm max}(f_1(n), \ldots, f_n(n))$. – Nik Weaver Sep 27 '18 at 3:01