4. Let $n$ be a positive integer, try to find the largest integer $k$, such that in an $n$-element set, $k$ subsets can be taken, where the intersection of any two subsets is non-empty.
4. $k_{\max }=2^{n-1}$.

First, construct $2^{n-1}$ subsets of an $n$-element set such that the intersection of any two subsets is non-empty. Let the $n$-element set be $\{1,2, \cdots, n\}$, then the subsets of the $(n-1)$-element set $\{2,3, \cdots, n\}$, each combined with the set $\{1\}$, yield $2^{n-1}$ subsets, all of which have the element 1 in common.

Next, prove that $k \leqslant 2^{n-1}$. Use proof by contradiction, assume $k \geqslant 2^{n-1}+1$, divide the subsets of the set $A=\{1,2, \cdots, n\}$ into unordered pairs, such that a subset $A_{0}$ of $A$ and its complement $A \setminus A_{0}$ form a pair $\left(A_{0}, A \setminus A_{0}\right)$. Since $A$ has $2^{n}$ subsets, there are $2^{n-1}$ such pairs, and the intersection of the two sets in the same pair is the empty set, and $(A, \varnothing)$ is one of these pairs. Because the subsets satisfying the problem's condition are non-empty, at least $k-1 \geqslant 2^{n-1}$ of the $k$ subsets are distributed among the remaining $2^{n-1}-1$ pairs. Since $2^{n-1}>2^{n-1}-1$, by the pigeonhole principle, there must be two subsets in the same pair, and their intersection is the empty set, which is a contradiction.
$$
\text { In conclusion, } k_{\max }=2^{n-1} \text {. }
$$