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
Foreach iterate of 0,
defines a polynomial function of
and so on. In generalfor
To prove a) define the setfor n=1,2,3,... so that
and so on.
andso that M is bounded .
Each setis 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 becauseis a polynomial in
with real coefficients,
for
but
if and only if
by symmetry of
in the real axis.
To prove c), noteis disconnected for
and
so
for
and
To prove d) we can proveis 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 thatis 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. Ifthen
since if it were we could enlarge
slightly. Thus
for
By applying the Maximum Principle to each polynomial functionon
we obtain
for
and
contradicting that
so
is connected.