I'm having some trouble with this problem:
Suppose $\mathcal{F} \subseteq \{f| f: \mathbb{Z}^+ \to \mathbb{R}\}$ and $\mathcal{F}$ is countable. Prove that there is a function $g: \mathbb{Z}^+ \to \mathbb{R}$ such that $\mathcal{F} \subseteq O(g)$.
The hint for the solutions is
First note that if $\mathcal{F}=\emptyset$ then $g$ can be any function. If $\mathcal{F} \neq \emptyset$, then since $\mathcal{F}$ is countable, we can write its elements in a list: $\mathcal{F} = \{f_1, f_2, \ldots\}$. Now define $g:\mathbb{Z}^+ \to \mathbb{R}$ by the formula $g(n) = \max \{|f_1(n)|,|f_2(n)|,\ldots,|f_n(n)|\}$.
Now I can see how this $g$ would work if $\mathcal{F}$ is finite, but what about if it is denumerable (countable and infinite)?
If $\mathcal{F}$ was defined as $\{f_k|k\in \mathbb{Z^+}, f(n)=k\cdot n\}$, then I can not see how $\max \{|f_1(n)|,|f_2(n)|,\ldots,|f_n(n)|\}$ can exist for any $n$. For $n=1$ this is the set of all integers, which is not even bounded from above and therefore can not have a maximum.
What am I missing here?
The point is that any finitely many integers have a maximum. You begin by enumerating your $\cal F$, and it doesn't really matter how, and at the $n$th step you just ensure that your value is greater than the first $n$ functions in the enumeration.
Your definition of $\cal F$ makes little sense, but it is also not a problem, because you first need to enumerate your $\cal F$, and then you can define $g$.
Of course, $g$ may depend on the enumeration, but it dominated over $\cal F$, that's the important point.