4. If the inequality $|x-a|+|x|<2$ has no real solution, then the range of real number $a$ is $\qquad$ .
4. $a \leqslant-2$ or $a \geqslant 2$.

Transform the original inequality into $|x-a|0) \\ 2+x,(x<0)\end{array}\right.$ Draw the graphs of $y_{1}$ and $y_{2}$.
If the inequality has no solution, then it should be $y_{1} \geqslant y_{2}$, i.e., the graph of $y_{1}$ coincides with or is above the graph of $y_{2}$. Clearly, $a \leqslant-2$ or $a \geqslant 2$.