Re: sequence of continuous functions
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Fri, 30 Jun 2006 08:41:57 -0500
On 29 Jun 2006 13:21:17 -0700, "Amanda" <sca18@xxxxxxxxxxx> wrote:
Hello
I'd like a hint to prove this statement from Rudin Real and Complex
Analysis
Let {f_n} be a sequence of continuous complex valued funtions defined
on a set X ( a topological space) that converges on X to a function f.
Then, there exists an open subset V of X and a M>0 such that |f_n(x)| <
M for every x in X and every n =1,2,3.....That is, theres an open set V
where {f_n} is uniformly bounded.
This is not true - probably you're leaving out a hypothesis.
For example, let X = Q, the rationals. Enumerate the rationals
r_1, ... . For each n, let I_n be an open interval containing
r_n, such that I_n does not contain r_j for any j < n (we can
do that since we only have to exclude finitely many r_j.)
Now let f_n be a continuous function such that f_n(r_n) = n
and f_n(t) = 0 for t not in I_n.
Then if j is fixed the fact that f_n(r_j) = 0 for all n > j
shows that f_n(r_j) -> 0 as n -> infinity. But the fact
that f_n(r_n) = n shows that f_n is not uniformly bounded
on any interval.
I tried using the fact that the set of discontinuities of f is meager,
but couldnt come to a conclusion.
Thank you
Amanda
************************
David C. Ullrich
.
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