［Solution］ When $c=1$, it is obvious that $a(n)=1$, thus we have
$$\lim _{n \rightarrow \infty} \sqrt[n]{a(n)}=1$$

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.
From the given conditions, we know that 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, take an element $m$ from $D$ 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, so the number of ways to choose the function $\omega$ is
$$a(n) \leqslant C^{n+1}$$

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 conditions, thus providing a way to choose the function $\omega$. Therefore, the number of ways to choose the function $\omega$ is
$$C^{n} \leqslant a(n)$$

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

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

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

which gives
$$\lim _{n \rightarrow \infty} \sqrt[n]{a(n)}=C$$

Naturally, the above equation also holds when $C=1$.