Re: simple analysis exercise (uniform continuity)
- From: José Carlos Santos <jcsantos@xxxxxxxx>
- Date: Fri, 24 Aug 2007 15:28:14 +0100
On 24-08-2007 14:59, Kiuhnm wrote:
Hint:
You are given that "given epsilon>0, there exists a
delta=delta(epsilon)
such that..."
You may need to choose a different delta for the extended function,
e.g. delta = delta(epsilon/3)/3 or the like
I already know that, but either I am missing something or it isn't as simple as it seems to be.
Let e>0 be given. There is a d>0 s.t.
d(x',y')<d => d(f(x'),f(y'))<e/3.
You should have added: for all x',y' in S.
Now, take _x_ and _y_ in cl(S) such that d(x,y) < d. Take x',y' in S
such that d(x,x'),d(y,y') < d. Deduce that
d(g(x),g(x')),d(g(y),g(y')) <= e/3.
Then
d(g(x),g(y)) <= d(g(x),g(x')) + d(g(x'),g(y')) + d(g(y'),g(y))
< e/3 + e/3 + e/3
= e.
Best regards,
Jose Carlos Santos
.
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