Re: locally increasing everywhere implies globally increasing
- From: "Dave L. Renfro" <renfr1dl@xxxxxxxxx>
- Date: 23 Nov 2006 09:25:08 -0800
Dave L. Renfro wrote (in part):
I think there's a more elegant way of doing
this using some type of "induction over sets"
("continuous induction", "Cousin's lemma", etc.),
David C. Ullrich wrote:
Seems to me the relevant notion of "continuous
induction" is the fact that (a,b) is _connected_.
The connectedness is going to be used, at least
implicitly, in sorting out the details in the argument
by compactness that you suggest - in fact
connectedness by itself does it:
Fix c in (x,b). Say A is the set of all x > c
such that f is increasing on [c,x]. Then A is
nonempty, it's clear tha A is closed, and the
fact that f is "locally increasing" shows that
A is open. So A = [c,b).
Something like that, I guess. I was referring to
the methods given in papers such as
R. M. F. Moss and G. T. Roberts, "A creeping lemma",
American Mathematical Monthly 75 (1968), 649-652.
I have a list of about 15 references (some back to the
early 1900's) on this general notion of set-induction
that I'll probably post sometime. I thought I had already
posted my list, but when I googled my usenet posts yesterday
in order to cite the list for anyone interested, I couldn't
find such a post. Many of the papers are re-discoveries of
the idea, and one or two that I managed to uncover don't
seem to be known to those who have written about this
technique in recent years.
Dave L. Renfro
.
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