Page 292 - Sensing, Intelligence, Motion : How Robots and Humans Move in an Unstructured World

EXERCISES 267

Proof of Theorem 5.8.4. The proof is analogous to that of Ref. 110, VI.16.2.
Let F be a component of ∂D. Suppose F contains at least two points, u and v.
Since D is uniformly locally connected, u and v are accessible (Theorem 5.9.13)
and can be joined by a cross-cut, L, such that (D − L) determines two regions,
D 1 and D 2 ,in D (Theorem 5.9.12). No points of F except u and v can belong
to the boundary of both of these regions, because if w were such a point, any
1
arc in D joining two points x and y of D 1 and D 2 within ρ(w, L) of w would
2
necessarily meet L and would therefore have diameter at least 1 ρ(w, L).The
2
region would not be locally connected at w. Thus, the sets
E i = (F ∩ D i ) − (u ∪ v), i = 1, 2
have no common points. Their union is (F − (u ∪ v)), and since

E i = (F − (u ∪ v)) ∩ D i
they are closed in (F − (u ∪ v)). Thus, (F − (u ∪ v)) is not connected because
it has partition E 1 |E 2 ,so F is a continuum whose connectivity is destroyed
by the removal of any two points. F is therefore a simple closed curve.
Q.E.D.

5.10 EXERCISES

1. Consider a planar RR (revolute–revolute) arm manipulator (see Figure 5.E.1).
The arm’s links l 1 and l 2 are line segments of the same length l.Thereis an

O l /4
l 2
l/4

J 1

l 1

J o
Figure 5.E.1