Re: Nowhere continuous function

From: The World Wide Wade (waderameyxiii_at_comcast.remove13.net)
Date: 10/03/04 

Date: Sat, 02 Oct 2004 19:20:55 -0700

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.

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