$k$ is a fixed positive integer. Let $a_n$ be the number of maps $f$ from the subsets of $\{1, 2, ... , n\}$ to $\{1, 2, ... , k\}$ such that for all subsets $A, B$ of $\{1, 2, ... , n\}$ we have $f(A \cap B) = \min (f(A), f(B))$. Find $\lim_{n \to \infty} \sqrt[n]{a_n}$.
1. **Understanding the Problem:**
   We need to find the limit of the $n$-th root of $a_n$ as $n$ approaches infinity, where $a_n$ is the number of maps $f$ from the subsets of $\{1, 2, \ldots, n\}$ to $\{1, 2, \ldots, k\}$ such that for all subsets $A, B$ of $\{1, 2, \ldots, n\}$, we have $f(A \cap B) = \min(f(A), f(B))$.

2. **Analyzing the Function $f$:**
   The function $f$ must satisfy the condition $f(A \cap B) = \min(f(A), f(B))$. This implies that $f$ is determined by its values on sets of size $(n-1)$ and $n$. The value of $f$ on the set of size $n$ must be the largest value of $f$.

3. **Counting the Number of Valid Functions:**
   The number of possible values of $f$ on these sets is given by $k^n + (k-1)^n + \cdots + 1^n$. This is because we can assign the value of $f$ on the largest set and then independently choose values at most this large for the other sets.

4. **Extending $f$ to All Sets:**
   The formula for $f$ extends to intersections of any number of subsets, and there is a unique way to write any subset of size $\leq (n-1)$ as an intersection of sets of size $(n-1)$. Using the intersection formula minimum gives a unique and consistent extension of $f$ to all sets. Therefore, the number of maps is equal to the number of assignments of $f$ to the sets of size $\geq (n-1)$.

5. **Calculating the Limit:**
   We need to find $\lim_{n \to \infty} \sqrt[n]{a_n}$. Given that $a_n = k^n + (k-1)^n + \cdots + 1^n$, we can approximate $a_n$ for large $n$ by noting that the term $k^n$ will dominate the sum. Therefore, $a_n \approx k^n$ for large $n$.

6. **Taking the $n$-th Root:**
   \[
   \sqrt[n]{a_n} \approx \sqrt[n]{k^n} = k
   \]

7. **Conclusion:**
   Therefore, the limit is $k$.

The final answer is $\boxed{k}$.