116 The four vertices of a regular tetrahedron are $A, B, C, D$, with edge length $1, P \in AB, Q \in CD$, then the range of the distance between points $P, Q$ is $\qquad$ .
$116\left[\frac{\sqrt{2}}{2}, 1\right]$. It is easy to know that the maximum distance between points $P$ and $Q$ is 1, when $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$, the distance between points $P$ and $Q$ is the smallest. At this time, $PQ=\frac{\sqrt{2}}{2}$, so the range of values is $\left[\frac{\sqrt{2}}{2}, 1\right]$.