Proof
Observe first, that if A and B lie on opposite sides of L, then
all i-lines through A and B must meet L. Thus, if there is an
i-line through A and B which does not meet L, then A and B
must lie
on the same side of L.
We proceed by stages:
-
If a point P lies outside a circle C , then there is an
extended line through P which does not meet C.
Let Q be the centre of C.
We may choose the line through P perpendicular to PQ.
-
If a point P lies inside a circle C , then there is a
circle through P which does not meet C.
Let Q be the centre of C.
We may choose the circle on QP as diameter -
all of its points are less than |QP| from Q.
-
If C is a circle, and tεI(2), then t maps
all points in the exterior of C to points on one side of t(C), and
all points in the interior of C to points on the other side of t(C).
If P lies in the exterior of C, then, by (1), there is an i-line
(an extended line) M through P which does not meet C.
Then t(M) is an i-line through t(P) and t(∞) which does
not meet t(C). Thus t(P) lies on the same side of t(C) as t(∞).
Similarly, using (2), any for any Q in the interior of C, t(Q) lies
on
the same side of t(C) as t(C), where C is the centre of C.
Since t maps C to t(C), it maps the complement of C to the
complement of t(C). Thus, the images of the interior and
exterior must cover both sides of t(C), so the interior and
exterior must map to opposite sides.
-
Finally, suppose that L is an extended line. We can choose
an inversion h
which maps L to a circle C, so h-1(C) = L.
Applying (3) to C and h-1, we see that h-1 maps
the interior of C to one side of L. Thus, h maps this side
of L to the interior of C, and the other side to the exterior.
Now apply (3) to C and toh-1. This maps the interior of C to
one side of (toh-1)(C) = t(h-1(C)) = t(L),
and the exterior
to the other side.
Now, applying h then toh-1, we see that t = (toh-1)oh maps
the sides of L to the sides of t(L), as required.
|
|