Find all non empty subset $ S$ of $ \mathbb{N}: \equal{} \{0,1,2,\ldots\}$ such that $ 0 \in S$ and exist two function $ h(\cdot): S \times S \to S$ and $ k(\cdot): S \to S$ which respect the following rules:
i) $ k(x) \equal{} h(0,x)$ for all $ x \in S$
ii) $ k(0) \equal{} 0$
iii) $ h(k(x_1),x_2) \equal{} x_1$ for all $ x_1,x_2 \in S$.
[i](Pierfrancesco Carlucci)[/i]
1. We start by analyzing the given conditions. We have a set \( S \subseteq \mathbb{N} \) such that \( 0 \in S \). There are two functions \( h: S \times S \to S \) and \( k: S \to S \) that satisfy the following rules:
   - \( k(x) = h(0, x) \) for all \( x \in S \)
   - \( k(0) = 0 \)
   - \( h(k(x_1), x_2) = x_1 \) for all \( x_1, x_2 \in S \)

2. Substitute \( x_1 = 0 \) into the third condition:
   \[
   h(k(0), x_2) = 0 \quad \text{for all } x_2 \in S
   \]
   Given \( k(0) = 0 \) from the second condition, we have:
   \[
   h(0, x_2) = 0 \quad \text{for all } x_2 \in S
   \]

3. From the first condition, \( k(x) = h(0, x) \) for all \( x \in S \). Since we have established that \( h(0, x) = 0 \) for all \( x \in S \), it follows that:
   \[
   k(x) = 0 \quad \text{for all } x \in S
   \]

4. Substitute \( k(x_1) = 0 \) into the third condition:
   \[
   h(0, x_2) = x_1 \quad \text{for all } x_1, x_2 \in S
   \]
   Since we have already established that \( h(0, x_2) = 0 \) for all \( x_2 \in S \), it follows that:
   \[
   0 = x_1 \quad \text{for all } x_1 \in S
   \]
   This implies that the only element in \( S \) is \( 0 \).

5. Therefore, the only possible set \( S \) is \( \{0\} \).

6. Verify that \( S = \{0\} \) satisfies all the conditions:
   - \( k(0) = 0 \)
   - \( h(0, 0) = 0 \)
   - \( h(k(0), 0) = h(0, 0) = 0 \)

Since all conditions are satisfied, we conclude that \( S = \{0\} \) is the only solution.

The final answer is \( \boxed{ \{0\} } \).