Re: help with proof of Linderlöf's theorem
- From: "Stephen J. Herschkorn" <sjherschko@xxxxxxxxxxxx>
- Date: Thu, 05 Apr 2007 21:53:06 -0400
AK wrote:
I'm trying to learn a bit of analysis (real & functional) by myself,
and I'm having some trouble understanding a step in a proof of
"Lindelöf's Theorem" as presented in the book "Applied Function
Analysis: A First Course for Students of Mechanical and Engineering
Science" by J. Tinsley Oden (Theorem 3.4.1 from the book).
I'll quote the theorem and the beginning steps of the proof as stated
in the book, with (*) indicating the step that I find problematic:
(please note that I'm using C to indicate "subset of")
****
Theorem:
Let H denote an open covering of a set A C R (the real line). There
exists a countable subclass of H that also covers A.
Proof:
Consider the rationals Q C R, and suppose that G is the class of all
open intervals with rational centres q and radii r; i.e.
G = {I(q,r)}; I(q,r) = {x in R : |x - q| < r; r,q in Q}
If a is a point in R, and if A contains a, then at least one
neighborhood in G contains a and is contained in A (*);
we can verify this from previous observations on nested intervals.
Thus suppose I(k,p) in G, and a in I(k,p) C A.
...
****
My problem is that I'm not sure why the statement labelled (*) is
supposed to be true.
You are correct; the statement is incorrect. Even simpler counterexample: Let A be the set of all intergers.
As a simple counter-example, consider the closed
set A = [sqrt(2) 2], and let a = sqrt(2). Then any point in A at a
"rational distance" from a is irrational. Hence it isn't possible to
find an open interval I(k,p) with k,p in Q, such that I(k,p) C A (a in
this example being a boundary point of A, the only way I(k,p) C
A is true is if r is exactly the distance |k-a|, which is
irrational and contrary to our assumption of r being rational).
Am I doing something stupid here, or did the author simply make a
mistake?
As far as the rest of the proof goes, it seems to me that the
assumption I(k,p) be a subset of A is unimportant, as the countable
subcovering can be a superset of A.
--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan
.
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