Re: Question on "Complex Made Simple"

On 26 Feb, 08:40, Rotwang <sg...@xxxxxxxxxxxxx> wrote:
On 26 Feb, 08:25, Steve Dalton <sdal...@xxxxxxxxxxxxxxx> wrote:

I recently bought "Complex Made Simple" by Prof. Ullrich, and am having a slightly difficult time comprehending Lemma 3.4.  I believe that it is correct, but the proof doesn't convince me.  Since I am quite tired, and don't feel like typing up the entire Lemma and proof, I will direct this question to either Prof. Ullrich or other owners of the book:
What if K = {1}, rho = 1/2 in the statement of the lemma.
The sequence (z_n) I choose is constant: 1/2.  Then by definition of K, alpha_n = 1, but the only possible beta_n (-1/2) lies outside the closed disk?  It's very possible I am missing something here, but if someone could kindly help me to understand, I would be most grateful.

I think that every incidence of Dbar(z,rho) in the proof is meant to
be Dbar(0,rho). Then the statement "for every n there exists alpha_n
in K and beta_n in Dbar(0,rho) such that z_n = alpha_n + beta_n" makes
sense, since every z_n is in Dbar(alpha_n,rho) for some alpha_n, and
z_n is in Dbar(alpha_n,rho) iff beta_n is in Dbar(0,rho), where beta_n
is defined to be z_n - alpha_n.

P.S. Here is (I think) another way to prove the lemma, which I mention
for no reason other than that I think it's quite nice: since K' is
obviously bounded, in order to prove it's compact we need only show
it's closed. However, the fact that K is compact means that, for every
point z in C, there is a point alpha in K such that |z - alpha| = d
(z,K) (this is proved by considering a sequence alpha_n in K such that
|z - alpha_n| -> d(z,K) and using sequential compactness). But then it
follows that K' is closed, since if z is not in K' then d(z,K) > rho,
so that D(z,epsilon) does not intersect K', where epsilon = d(z,K) -
rho.
.

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