The Mandlebrot set![]()
a) is a compact subset of![]()
b) is symmetric under reflection in the real axis
c) meets the real axis in the interval![]()
d) has no holes it it
Proof
For
each iterate of 0,
defines a polynomial function of![]()
![]()
![]()
![]()
and so on. In general
for![]()
To prove a) define the set
for n=1,2,3,... so that
and so on.
![]()
and
so that M is bounded .
Each set
is closed because it has complement
which is open. In fact if
for some
then this inequality must also hold for all
in some open disc with centre
by the continuity of the function
It follows that
must be closed because if
then
for some
and so some open disc with centre
must lie outside
and hence outside
This proves that
is closed and bounded, or compact.
To prove b), notice that because
is a polynomial in
with real coefficients,
for
but
if and only if
by symmetry of
in the real axis.
To prove c), note
is disconnected for
and
so
for
and![]()
To prove d) we can prove
is connected so that each pair of points in
can be joined by a path in
Consider the set
We can show that each point in
can be joined to a point of
by a path in![]()
Suppose in fact that
is a point of
which cannot be joined to
in this way. Define![]()
since
and
is open because if
can be joined to
then so can any point of any open disc in
with centre
and
is connected because pairs of points in
can be joined in
via
thus
is a subregion of
Since
cannot be joined in
to
and
is open we deduce![]()
Now use the Maximum Principle. If
then
since if it were we could enlarge
slightly. Thus
for![]()
By applying the Maximum Principle to each polynomial function
on
we obtain
for
and
contradicting that
so
is connected.