Let $ \mathbb{Q}$ and $ \mathbb{R}$ be the set of rational numbers and the set of real numbers, respectively, and let $ f : \mathbb{Q} \rightarrow \mathbb{R}$ be a function with the following property. For every $ h \in \mathbb{Q} , \;x_0 \in \mathbb{R}$, \[ f(x\plus{}h)\minus{}f(x) \rightarrow 0\] as $ x \in \mathbb{Q}$ tends to $ x_0$. Does it follow that $ f$ is bounded on some interval? 

[i]M. Laczkovich[/i]
To determine whether the function \( f : \mathbb{Q} \rightarrow \mathbb{R} \) is bounded on some interval given the property that for every \( h \in \mathbb{Q} \) and \( x_0 \in \mathbb{R} \), \[ f(x+h) - f(x) \rightarrow 0 \text{ as } x \in \mathbb{Q} \text{ tends to } x_0, \] we will construct a counterexample to show that \( f \) may be unbounded on any interval.

1. **Enumeration of Primes and Definition of \( q_n \)**:
   Let \( p_1, p_2, \ldots \) be an enumeration of the prime numbers. Define \( q_n = (p_1 p_2 \ldots p_n)^n \) for \( n \in \mathbb{N} \).

2. **Definition of Sets \( A_n \)**:
   Define the sets \( A_n \) as follows:
   \[
   A_n = \left\{ \frac{m}{q_n} \mid m \in \mathbb{Z} \right\}, \quad n = 1, 2, \ldots
   \]
   These sets have the following properties:
   - \( A_n \subset A_{n+1} \)
   - \( \bigcup_{n \in \mathbb{N}} A_n = \mathbb{Q} \)

3. **Behavior of \( A_n \) with respect to \( h \in \mathbb{Q} \)**:
   For every fixed \( h \in \mathbb{Q} \), for all sufficiently large \( n \), if \( x \in A_n \), then \( x + h \in A_n \).

4. **Definition of Sets \( B_n \)**:
   Define the sets \( B_n \) as follows:
   \[
   B_1 = A_1, \quad B_n = A_n \setminus A_{n-1}, \quad n = 2, 3, \ldots
   \]
   These sets have the following properties:
   - \( \bigcup_{n \in \mathbb{N}} B_n = \mathbb{Q} \)
   - \( B_n \cap B_m = \emptyset \) for \( m \neq n \)

5. **Behavior of \( B_n \) with respect to \( h \in \mathbb{Q} \)**:
   For every fixed \( h \in \mathbb{Q} \), for all sufficiently large \( n \), if \( x \in B_n \), then \( x + h \in B_n \).

6. **Definition of the Function \( f \)**:
   Define the function \( f \) as follows:
   \[
   f(x) = n \quad \text{if } x \in B_n
   \]
   This function is clearly unbounded on any interval since \( f(x) \) can take arbitrarily large values.

7. **Verification of the Given Property**:
   Fix \( h \in \mathbb{Q} \) and \( x_0 \in \mathbb{R} \). According to the properties of \( B_n \), we can choose \( N \in \mathbb{N} \) such that for all \( n > N \), \( x \in B_n \implies x + h \in B_n \). Consider a neighborhood \( U \) of \( x_0 \) that does not contain any members of \( B_1, B_2, \ldots, B_N \), except possibly \( x_0 \) itself. Thus, for all \( x \in \mathbb{Q} \) and \( x \in U \setminus \{x_0\} \), we have \( f(x) = f(x + h) \). Therefore,
   \[
   \lim_{x \to x_0, x \ne x_0} (f(x + h) - f(x)) = 0.
   \]

This construction shows that \( f \) satisfies the given property but is unbounded on any interval.

The final answer is \( \boxed{ f } \) may be unbounded on any interval.