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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 thatand so on.

andso that M is bounded .

Each setis closed because it has complementwhich is open. In fact iffor somethen this inequality must also hold for allin some open disc with centreby the continuity of the functionIt follows thatmust be closed because ifthenfor someand so some open disc with centremust lie outsideand hence outsideThis proves thatis closed and bounded, or compact.

To prove b), notice that becauseis a polynomial inwith real coefficients,forbutif and only ifby symmetry ofin the real axis.

To prove c), noteis disconnected forandsoforand

To prove d) we can proveis connected so that each pair of points incan be joined by a path inConsider the setWe can show that each point incan be joined to a point ofby a path in

Suppose in fact thatis a point ofwhich cannot be joined toin this way. Define

sinceandis open because ifcan be joined tothen so can any point of any open disc inwith centreandis connected because pairs of points incan be joined inviathusis a subregion ofSincecannot be joined intoandis open we deduce

Now use the Maximum Principle. Ifthensince if it were we could enlarge slightly. Thusfor

By applying the Maximum Principle to each polynomial functiononwe obtainforandcontradicting thatsois connected.