Theorem 5.1

  is path connected.

Proof

} Let M be with the unit circle shrunk to a point P. Again M is topologically a sphere, so we may give it a bounded metric d. Let be the set of sequences which converge to P and define a metric on by

Let the group of permutations of act on by permuting the coordinates. Define a metric D on the quotient space by letting

Here denote the projections of to . (That if and only if requires the convergence of x,y.) The set of zeros of a power series

inside forms a sequence in M converging to P (by Proposition 2.1) or else is finite, in which case we append an infinite sequence of P's. This defines a map

By the same Rouché's theorem argument used in the proof of Theorem 4.1, this map is continuous. The conditions of Lemma 4.1 hold for the same reason as before, so the image of f is path connected.

Suppose . Let be a path from the image under f of a 0,1 power series vanishing at to .

Fix , and let be with the annulus shrunk to a point Q. Define on by letting . By Proposition 2.1 there is an upper bound n on the number of zeros of a 0,1 power series inside . The path induces a path

(Apply the projection to each element of , and throw away infinitely many Q's to get .)

Pick such that . We define inductively a sequence of paths

each extending the one before. First apply Lemma 5.1 to lift to a path . Since some coordinate of is and since all coordinates of are Q, we get a path from to Q in . Let be the smallest such that . Then by restriction to we get a path in since can be identified with . Finally, since is always a coordinate of , for all .

By the same process, we inductively find for each a path such that , let be the smallest such that

and obtain a path

which we append to to obtain

such that is always a coordinate of .

Let . Piecing together the 's gives a continuous map

such that is a coordinate of for all . The set of limit points of as is a closed interval I. Let . If is distinct from then (since ) and by continuity of , r also differs by some from all coordinates of for t in a neighborhood of , so r cannot be a limit point of as . Thus but so I must be a single point. Since for , .

Case 1: 1 is the only limit point of as . Then extends to a path from to 1.

Case 2: There is a limit point , , of as . By Theorem 3.1, there is an open disc centered at contained in . For some , is in this disc, so we can replace the tail end of on by a straight line from to in .

In either case we can connect to a point on the unit circle via a path in . The same is true if , , since is closed under . Since contains the unit circle, this proves that is path connected.