By Robert B. Burckel

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Since single points are connected sets, (iv) follows from (iii). Thus C contains a neighborhood of each of its points and so is open. Being open, C contains a point with rational coordinates. As there are only countably many such points and components are disjoint, there can be only countably many components. 31 Let U be an open subset of C. Then there exist compact sets K , such that (1) (2) Kn c & + I * 0 K, = U. n-1 (3) Every bounded component of C\Kncontains a component of C\U. In particular, ifC\U has no bounded components, then each set C\Kn is connected.

Proof: Let W be the set of unbounded components of C\11. For eachJ the last theorem provides unbounded, closed sets K2,. Then C\R1 = (C\8) U Ka U ** * UK,,. (I) Every bounded component of C\Q different from C,lies in some Kj, which is connected and unbounded. Therefore every point of C\Q, not in C1lies in an 0 5. Connectivity of a Set 35 unbounded component of C\n,. 38, Q u C1u is open, so its complement U W is closed. U K,, U u - . u C,, V, which shows that C, is relatively open in C\Q,. Being also compact and connected, C, is therefore a component.

Remark: These same techniques show, with a little more work (see KRAFFT [1932]), that f ( D ( 0 ,r)) is open and f-' is given by a power series near co, facts which we will secure in Chapter V by other means. , for each a E C\y there is a power series about a which agrees with G, in some neighborhood of a. 14. 8 2. 6 Given a E @\y, choose R > 0 so that D(a, R) c @\y. Then for any 0 < r < R, ZE D(a, r) and ( E Y we have 1 - 1 1 (t - Z ) n + l - (6 - a y + 1 with convergence uniform in such z and 1.