the Euclidean distance between them is greater than $1$ unit. And also, both of them are neighbors with $a$, which means the Euclidean distance between them is less than $2$ units. If we embed $b$, $a$, $c$ onto a straight line respectively, then $a$ must be between $b$ and $c$.

Wherever we place $d$, within one unit distance from $a$, it must be a neighbor of either $b$ or $c$. If $d$ is neighbors with neither, then the Euclidean distance between $d$ and $a$ must be greater than one unit, contradicting the unit interval property.

Every unit interval graph is net-free.

Suppose that $N(a, b, c; x, y, z)$ is an induced net, as shown in Figure \[fig:claw\]. It is clear that $a$, $b$ and $c$ must be inside an interval of at most $1$ unit. Note that $a,x$ and $c,z$ are pairwise disjoint neighbors, and moreover neither $x$ nor $z$ is a neighbor of $b$. This means that $b$ must be between $a$ and $c$.

Wherever we place $y$, within one unit distance from $b$, it must be a neighbor of either $a$ or $c$ (also, at least one of $x$, $z$). If $y$ is a neighbor with neither, 