4. $\left(0, \frac{\sqrt{2}}{2}\right)$.
$\arcsin x<\arccos x \Rightarrow x \in\left[-1, \frac{\sqrt{2}}{2}\right], \arccos x<\operatorname{arccot} x \Rightarrow x \in(0,+1]$, so the range of $x$ is $\left(0, \frac{\sqrt{2}}{2}\right)$.
4. Given $\arcsin x<\arccos x<\operatorname{arccot} x$, the range of real number $x$ is $\qquad$