By R.B. Burckel

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**Extra info for An Introduction to Classical Complex Analysis: Vol. 1**

**Sample text**

Vc V for some n. Then the set W = Xn () V = X,. , contains X () V => K and so is non-void. Since X" is connected, it follows that X,. = W. But then X" is compact, a contradiction. 5. In this case n c (D"o) U X"o' a union of compact sets, so n is bounded. Finally notice that §5 8n = n Xn (exercisette). Connectivity of a Set The theorems of this section are technical results which will be needed later (in Chapter X). They require only material presented so far and it is a good idea to dispose of them now.

Since H(C) = h(C) lies in q{O}, H-l(q{O}) is an open neighborhood of C. The compact sets Ak = {x E IRR: d(x, A) ~ Ilk} satisfy n'" k=l Ak x [0, I] =A x [0,1] c C c H-l(q{O}). Therefore by compactness there is some k such that Ak x [0, 1] C H-l(q{O}). Then for every x E B the point (x, min{l, kd(x, A)}) lies in Ak x [0, I] u B x {I} c H-l(q{O}) and so F(x) = H(x, min{l, kd(x, A)}) is the desired continuous extension off. Remark: This is called Borsuk's homotopy extension theorem. The latter notion will be defined and explored in Chapter IV.

And so _ C(x) - I ~ (_1)kx2k . + N~co hm L, k=l ~ [(_1)2"-lX4n-2 + ~~ "~1 Observe that for 0 ix) 2kX 2k k~O (2k)! n! 2 k=O = 2 [1 + (_1)n] -=2:inxn co i = L, _ - 1 + E( - E(ix) [X2 + 4! - 1] 2! x 4-2 (2k)! + (-1)2"X4n ] (4n)! ]X4n - 2. 1 < 1 ~--=-: (4n - 2)! co [ x 2 + n~2 1] (4n)! - (4n _ 2)! x 4n-2 Therefore C(V3) < -t. Since C(O) = Re E(O) = 1, we have that C[O, V3] is a connected subset of IR which contains positives and negatives and hence zero; that is, the compact set [0, V3] r. C-l(O) is not void.