Re: Nowhere continuous function

From: Artur (artur_at_opendf.com.br)
Date: 10/04/04 

Date: 4 Oct 2004 14:00:16 -0700

The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote in message news:<waderameyxiii-F4AC5D.19205502102004@news.supernews.com>...
> In article <93b1dfbf.0410021553.7ef07e52@posting.google.com>,
> artur@opendf.com.br (Artur) wrote:
>
> > Hello
> >
> > I have a question: Can there exist a function f:I->R, I an open
> > interval of R, which has a limit at every point of I but is nowhere
> > continuous?
>
> No, in fact the set of discontinuities is at most countable. Here's a
> thread where this was discussed:
>
> http://snipurl.com/9i2m
>
> One way to proceed is to let g(x) be the limit of f at x; so g : I -> R.
> Let D_n = {x in I : |f(x)-g(x)| > 1/n}. The set D of discontinuities of f
> is the union of the D_n's. If D were not countable, then some D_n would be
> uncountable, hence would have an accumulation point p somewhere in I. Now
> check that lim_{x->p} f(x) fails to exist, contradiction.

Let me try to finish up. g is continuous on I. Let {x_m} be any
sequence in D_n that converges to p and satisfies x_m<>p for every m.
Then, g(x_m) -> g(p) and |f(x_m) - g(x_m)| >1/n for every m. In
addition, |f(x_m) - g(p)|>= ||f(x_m)- g(x_m| - |g(p) - g(x_m)|| for
every m. If we choose m sufficiently large that |g(p) - g(x_m)|
<1/(2n), then |f(x_m) - g(p)|>1/(2n), and f(x_m) does not converge tp
g(p), a contradiction.

Actually, we have to show g is continuous. If x is in I and {x_m} is a
sequence in I that converges to x, then for every m there's y_m in y
such that |y_m - x_m| < 1/m, y_m <>x and |f(x_m) – g(x_m)| < 1/m..
Then, y_m – > x and and |f(x_m) – g(x_m)|-> 0. It follows that
f(y_m) -> g(x), which implies g(x_m) ->x, showing g is continuous.
Artur

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