2・44 $c \geqslant 1$ is a fixed positive integer. For each non-empty subset $A$ of the set $\{1,2, \cdots, n\}$, a positive integer $\omega(A)$ is assigned from the set $\{1,2, \cdots, c\}$, satisfying
$$
\omega(A \cap B)=\min (\omega(A), \omega(B)),
$$

where $A, B$ are any two non-empty intersecting subsets of $\{1,2, \cdots, n\}$. If there are $a(n)$ such assignment methods, calculate
$$
\lim _{n \rightarrow \infty} \sqrt[n]{a(n)} .
$$
［Solution］When $c=1$, it is obvious that $a(n)=1$, thus we have
$$
\lim _{n \rightarrow \infty} \sqrt[n]{a(n)}=1 \text {. }
$$

Now consider the case when $c \geqslant 2$.
The set $\{1,2, \cdots, n\}$ has $2^{n}-1$ non-empty subsets, and these $2^{n}-1$ non-empty subsets form the domain of the function $\omega(A)$.
$$
\text { Let } \begin{aligned}
b_{k} & =\{1,2, \cdots, k-1, k+1, \cdots, n\} \\
& =\{1,2, \cdots, n\}-\{k\},(k=1,2, \cdots, n)
\end{aligned}
$$

Let $D=\{1,2, \cdots, n\}$.
First, we prove that if $\omega\left(b_{1}\right), \omega\left(b_{2}\right), \cdots, \omega\left(b_{n}\right), \omega(D)$ these $n+1$ values are determined, then for any non-empty subset $A$ of $D$, $\omega(A)$ is also determined.
We use mathematical induction.
By the given condition, when $A$ is an $(n-1)$-element subset of $D$, $\omega(A)$ is determined.
Assume that when $A$ is an $(n-k)$-element subset of $D$, $\omega(A)$ is determined, then when $A$ is an $(n-k-1)$-element subset of $D$, take an element $m$ in $D$ that does not belong to $A$, then $B=A \cup\{m\}$ is an $(n-k)$-element subset of $D$, by the induction hypothesis $\omega(B)$ is determined. Then by
$$
A=B \cap b_{m}
$$

we get
$$
\omega(A)=\min \left\{\omega(B), \omega\left(b_{m}\right)\right\} .
$$

Therefore, $\omega(A)$ is also determined.
By the principle of mathematical induction, the function $\omega$ is completely determined by the $n+1$ values $\omega\left(b_{1}\right), \omega\left(b_{2}\right), \cdots$, $\omega\left(b_{n}\right), \omega(D)$.

Notice that each of these $n+1$ values can take at most $C$ different values, therefore, the number of ways to choose the function $\omega$
$$
a(n) \leqslant C^{n+1} .
$$

On the other hand, for any of the $C^{n}$ ways to choose the $n$ values $\omega\left(b_{1}\right), \omega\left(b_{2}\right), \cdots, \omega\left(b_{n}\right)$, we can take
$$
\omega(D)=\max \left\{\omega\left(b_{1}\right), \omega\left(b_{2}\right), \cdots, \omega\left(b_{n}\right)\right\}
$$

which satisfies the given condition, thus providing one way to choose the function $\omega$. Therefore, the number of ways to choose the function $\omega$
$$
C^{n} \leqslant a(n) .
$$

Thus, we have
$$
C^{n} \leqslant a(n) \leqslant C^{n+1},
$$

taking the $n$-th root, we get
$$
C \leqslant \sqrt[n]{a(n)} \leqslant \mathrm{C}^{1+\frac{1}{n}}
$$

taking the limit,
$$
C \leqslant \lim _{n \rightarrow \infty} \sqrt[n]{a(n)} \leqslant \lim _{n \rightarrow \infty} C^{1+\frac{1}{n}},
$$

which gives $\lim _{n \rightarrow \infty} \sqrt[n]{a(n)}=C$.
It is clear that the above equation also holds when $C=1$.