Consider the points $1,\frac12,\frac13,...$ on the real axis. Find the smallest value $k \in \mathbb{N}_0$ for which all points above can be covered with 5 [b]closed[/b] intervals of length $\frac1k$.
1. We need to find the smallest value \( k \in \mathbb{N}_0 \) such that the points \( 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \) on the real axis can be covered with 5 closed intervals of length \( \frac{1}{k} \).

2. Assume \( k < \frac{1}{10} \). This means each interval has length less than \( \frac{1}{10} \).

3. Let the 5 closed intervals be \([a_1, b_1], [a_2, b_2], [a_3, b_3], [a_4, b_4], [a_5, b_5]\), each of length \( \frac{1}{k} \).

4. If all \( a_i > 0 \), we can find a natural number \( N \) such that \( N > \frac{1}{a_i} \) for all \( i \). This implies \( \frac{1}{N} < a_i \) for all \( i \), meaning \( \frac{1}{N} \) is not covered by any interval, contradicting the assumption that all points \( 1, \frac{1}{2}, \frac{1}{3}, \ldots \) are covered.

5. Therefore, at least one \( a_i \leq 0 \), meaning the interval \([a_i, b_i]\) covers the point 0.

6. Consider the 6 points \( 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{5}, \frac{1}{10}, 0 \). Each of these points must lie in one of the 5 intervals.

7. By the pigeonhole principle, at least two of these 6 points must lie in the same interval. However, the pairwise differences between these points are all \( \geq \frac{1}{10} \), which is greater than \( k \) since \( k < \frac{1}{10} \).

8. This contradiction shows that \( k < \frac{1}{10} \) is not possible. Therefore, the smallest \( k \) must be \( \frac{1}{10} \).

The final answer is \( \boxed{ k = 10 } \).