4. Determine the largest natural number $k(n)$ with the following property: There are $k(n)$ different subsets of a given set with $n$ elements, such that any two of them have a non-empty intersection.
Solution. Let's fix the element $a_{1}$ from the set $X=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ and consider the subsets of the set $X$ that contain the element $a_{1}$. There are as many such subsets as there are subsets of the set $\left\{a_{2}, a_{3}, \ldots, a_{n}\right\}$, which means $2^{n-1}$. Clearly, any two of these subsets have a non-empty intersection, so $k(n) \geq 2^{n-1}$.

Let's assume that more than $2^{n-1}$ subsets of the set $X$ are chosen. Let's divide all $2^{n}$ subsets of $X$ into $2^{n-1}$ pairs, such that in one pair belong a set and its complement. From the Pigeonhole Principle, it follows that at least two of the chosen subsets form such a pair, which means their intersection is empty. Therefore, it cannot be that $k(n)>2^{n-1}$.

Finally, from the previous considerations, it follows that $k(n)=2^{n-1}$.