By R.B. Burckel
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Additional info for An Introduction to Classical Complex Analysis: Vol. 1
37 (i) Show that an open subset of IC is simply-connected if either (a) its boundary has no bounded component or (b) it is bounded and has a connected boundary. Hint: Every component of the complement meets the boundary. Show that if the subset A of IC is simply-connected, then A contains every (ii) bounded subset S of IC whose boundary lies in A. Hints: oS = S\S c A implies S () (qA) = S () (qA), so that this set is relatively e10pen in qA. It therefore contains every component which it meets.
Then z E 1(,.. To prove (I), we set r" = lIn - I/(n + I) and show that D(z, rn) every z E K". Indeed, given such a z and WE D(z, r,,), we have Iwi Izl + r" < n + rn, since z E K" c D(O, n), ~ n + I, so WE D(O, n + I). But also for any a E qv we have ~ Iw - al ~ Iz - al - Iw - zl > ~- r" =n ! ~ I ii - Iw - zl, since z E C" I' since WE D(z, r,,), c K"+l for § 4. Components and Compact Exhaustions It follows that WE Cn + l and so finally WE 31 D(O, n + 1) n Cn + l = Kn+l' It remains to look after (3).
It therefore contains every component which it meets. As it is bounded and qA has no bounded components, S () (qA) is empty. 38 Let U be an open subset of IC such that q U has exactly one bounded component C. Then U U C is open and simply-connected. If in addition U is connected, then so is U U C. Proof: The components of q( U U C) are just the unbounded components (if any) of q U. Thus there are no bounded ones, so U U C is simply-connected. 34 there is a compact, relatively open subset V of qu which contains C.
An Introduction to Classical Complex Analysis: Vol. 1 by R.B. Burckel