Re: pointwise but not quasi-uniform

On Sat, 22 Apr 2006, eugene wrote:

Once again, i hope it would be more readable:

We'l say that a sequence f_n : X->R or C tends to
f(x)
quasi-uniformly if X we can represent as a
countable union
of sets on which f_n converges to f uniformly.
Prove that there exists

No, don't you mean
f_n|A -> f|A uniformly for each A in the union?

yes, i meant that f_n|A _> f|A uniformly on each component of the union of X.

sequences f_n:[0,1]->R, such that f_n->f=0 on
[0,1]pointwise but not
quasi-uniformly.

Could you please chare you ideas and check my
example
I tried some kind of characteruistic functions: if
for example

f_n= charactersitic function of [ 1-1/n ,1 ],

then f_n -> 0 pointwise, and put

X=union_{k=1}^{\infty} A_k.

It is readily that if f_n->0 on some A_k then f_n=0
on A_k

for all n>some N. So, we may notice that all the
closed
intervals [a,b] strictly in [0,1] are the sets of
uniform
convergence.

[0,1]=union_n [ 1-1/n, 1-1/(n+1) ], n=1,2,3,...

No,
\/_n [1 - 1/n, 1 - 1/(n+1)] = [0,1)

A space on both sides of = sign makes for major
components
of an equation to be easily discerned when quickly
scanning.

and on all of the [ 1-1/n, 1-1/(n+1) ] the
convergence is uniform. This
means that my example doesn't work.

No. To show not quasi-uniform you have to show for
each countable
union of sets, (open, closed or whatever), there is
out of each
union one set upon which (f_n)_n isn't uniformly
convergent to 0.

No, i write it very bad and you didn't understand me. I wanted to say that if it were possible for X to represent it as a union of sets on which the convergence is uniform
as
[0,1] = \/_n A_k,

then some of the A_k must contain 1 , but we may notice that it can't because for each f_n(1) = 1 and it can't uniform convergence on a set which contains 1.

Is it ok?
.

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