4. Let $A$ be any set of 11 real numbers, and let the set $B=\{u v \mid u, v \in A, u \neq v\}$. Find the minimum number of elements in $B$.

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Prove that $|B| \geqslant 17$. Since when all elements in $A$ are changed to their opposites, the set $B$ remains unchanged, we can assume without loss of generality that the number of positive numbers in $A$ is no less than the number of negative numbers.

Case one: $A$ contains no negative numbers. Let $a_{1}c_{1} b_{2}>\cdots>c_{1} b_{k}>c_{2} b_{k}>c_{3} b_{k}>\cdots>c_{l} b_{k}$ be $k+l-1=10$ elements in $B$, and they are non-positive; and $b_{2} b_{3}<b_{2} b_{4}<b_{2} b_{5}<b_{2} b_{6}<b_{3} b_{6}<b_{4} b_{6}<b_{5} b_{6}$ are 7 elements in $B$, and they are positive. In this case, $B$ has at least 17 elements.
In summary, $|B| \geqslant 17$. On the other hand, if $A=\left\{0, \pm 1, \pm 2, \pm 2^{2}, \pm 2^{3}, \pm 2^{4}\right\}$, then $B=\left\{0,-1, \pm 2, \pm 2^{2}, \pm 2^{3}, \pm 2^{4}, \pm 2^{5}, \pm 2^{6}, \pm 2^{7},-2^{8}\right\}$ is a 17-element set.
In summary, the minimum number of elements in $B$ is 17.