Re: Question on "Complex Made Simple"

On Thu, 26 Feb 2009 01:06:35 -0800 (PST), Rotwang
<sg552@xxxxxxxxxxxxx> wrote:

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.

Right.

The _really_ right proof is this: If A and B are compact subsets
of the plane then A + B = {a + b : a in A, b in B} is compact.

Proof: AxB is compact, and if f(x,y) = x + y then f is continuous
on AxB, so A + B = f(AxB) is compact. QED.

Decided to give the proof I did because it didn't use anything
about compactness other than the characterization in terms
of convergent subsequences (seemed likely that for some
young readers the convergent subsequence thing will be
the most familiar characterization and/or the one that
makes the most sense). Went for simplicity, or what
seemed to me to be simplicity, over niceness, in many
places in the first half or so.

(And then of course the "simplest" way to type
\overline{D(0,\rho} is to simply cut&paste
\overline{D(z,\rho} from above... aargh.)



David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
.

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