On a property of sets of infinite sequences Consider some finite set of symbols $\Sigma$, and any two sets $P$ and $Q$ of infinite sequences constructed from the symbols in $\Sigma$.
Also consider the property $s$, defined with respect to any such infinite-sequence set $A$ such that
$$s(A) = \lim_{n\to\infty} \sqrt[n]{f_n(A)}$$
where $f_n(A)$ is equal to the number of distinct symbol sequences of length $n$ present at the beginning of at least one sequence in $A$. (Heuristically, $f_n(A)$ counts the number of distinct "prefixes" of length $n$ present in elements of $A$.)
Given these definitions, how would one show that:
$$s(P\cup Q) = \max \{s(P),s(Q)\}$$
I attempted to prove this by, assuming $s(P) \geq s(Q)$, "sandwiching" $\sqrt[n]{f_n(P) + f_n(Q)}$ between $\sqrt[n]{f_n(P)}$ and $\sqrt[n]{f_n(P) + f_n(P)}$ and showing both of these converge to the same limit. However, I can't use this trick since I have no statement that $f_n(P) \geq f_n(Q)$ for all $n$, just about the limit of these expressions at $n \to \infty$.
 A: Let me write $\Sigma^*$ for the set of all infinite sequences of elements of $\Sigma$. For $P \subseteq \Sigma^*$, let $p_n(P)$ be the set of sequences of length $n$ that are prefixes of some sequence in $P$, so that $f_n(P) = |p_n(P)|$. Clearly if $P, Q \subseteq \Sigma^*$, we have:
$$
p_n(P \cup Q) = p_n(P) \cup p_n(Q)
$$
whence by the inclusion-exclusion principle:
$$
|p_n(P \cup Q)| = |p_n(P)| + |p_n(Q)| - |p_n(P) \cap p_n(Q)|
$$
Assume $\lim_{n \to\infty}f_n(P) \ge \lim_{n \to \infty}f_n(Q)$ and let $T = \{n : \Bbb{N} \mid f_n(P) \ge f_n(Q)\}$.
We can assume w.l.o.g. that $T$ is infinite, for, if not, we have that $f_n(Q) > f_n(P)$ for infinitely many $n$, but then $\lim_{n \to \infty}f_n(Q) = \lim_{n\to \infty}f_n(P)$ and, after swapping $P$ and $Q$, the resulting set $T$ will be infinite, say $T = \{ t_1, t_2, \ldots\}$ with $t_1< t_2 < \ldots$.
Now, for $n \in T$, if we divide our equation for $|p_n(P \cup Q)|$ through by $|p_n(P)|$ and take $n$-th roots, we get
$$
1 \le \frac{\sqrt[n]{|p_n(P \cup Q)|}}{\sqrt[n]{|p_n(P)|}} = \sqrt[n]{1 + \frac{|p_n(Q)|}{|p_n(P|} - \frac{|p_n(P) \cap p_n(Q)|}{|p_n(P)|}} \le \sqrt[n]{2}
$$
By the squeeze theorem, we get:
$$
\lim_{n \to \infty}\left( \frac{\sqrt[t_n]{f_{t_n}(P \cup Q)}}{\sqrt[t_n]{f_{t_n}(P)}}\right) = \lim_{n \to \infty}\left( \frac{\sqrt[t_n]{|p_{t_n}(P \cup Q)|}}{\sqrt[t_n]{|p_{t_n}(P)|}}\right) = 1
$$
which gives your claim.
