Re: pointwise but not quasi-uniform

On Sat, 22 Apr 2006, eugene wrote:
On Sat, 22 Apr 2006, eugene wrote:

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 characteristic functions: if
for example

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

then f_n -> 0 pointwise, and put

No, f_n(x) -> 0, if x in [0,1)
f_n(1) -> 1

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)

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

Hi, I'm from Oregon, USA. You're from Russia. Moscow?

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?

No, (f_n)_n is uniformly convergent on {1}
Thus you add to your intervals the set {1}, and you've shown
quasi-uniformity. For not quasi-uniformity you may have to adjust
your definition as I hinted above. As this is getting all much
complex with quotings, comments and corrections, may I suggest you present
a rewritten and revised copy instead of further patchings and correctings?
That make additional discussion much easier for both of us.
.

Relevant Pages

  • Re: pointwise but not quasi-uniform
    ... countable union ... convergence is uniform. ... 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 ...
    (sci.math)
  • Re: pointwise but not quasi-uniform
    ... set of all strictly increasing functions mapping N to ... This is immediate from the definition of uniform ... We will give an example of a sequence of functions ... Hence S cannot be a countable union ...
    (sci.math)
  • Re: pointwise but not quasi-uniform
    ... which is not quasi-uniform. ... Ok, quit quibbling. ... This is immediate from the definition of uniform ... Hence S cannot be a countable union ...
    (sci.math)