Talk:Darboux function

I've created this redirect, since I have feeling that this term is used often and it was in fact defined in the entry Darboux function. I've created also the redirect from Darboux property to the same entry. --Kompik 06:06, 31 May 2006 (UTC)


 * The above comment should have been in Talk:Intermediate value property, i gave it here by mistake. --Kompik 14:03, 7 June 2006 (UTC)

One can use transfinite induction on Ω, or a construction involving Hamel bases. Why use such powerfull tools? Isn't $$\sin\left( \frac{1}{x}\right)$$, for $$x\ne 0$$ and 0 for x=0 a discontinuous Darboux function?

Furthermore, why mention the ways to construct it without even mentioning how (even if only heuristically).

Unless by discontinuous you mean discontinuous everywhere or some similar notion, but in that case the article needs rewriting.

cvalente 00:23, 14 July 2006 (UTC)

Discontinuities of the second kind

 * Every discontinuity of a Darboux function is of the second kind, that is, at any point of discontinuity, neither the left hand limit nor the right hand limit is well-defined.

Is this correct? Surely $$ f(x) = \begin{cases} 0 & x \le 0 \\ \sin\left(\frac{1}{x}\right) & x > 1 \\ \end{cases} $$ is Darboux? –EdC 20:51, 24 February 2007 (UTC)
 * Indeed. Fixed. Algebraist 21:58, 26 April 2008 (UTC)

Example function correct?
The given example function f(x) = x^2 * sin(1/x) is not differentiable at 0. f(0)=0 not needed? —Preceding unsigned comment added by 93.92.56.76 (talk) 16:46, 9 January 2009 (UTC)


 * Actually it is. $$f'(0)=lim){h\rightarrow0}(f(h)-f(0))/h=lim_{h\rightarrow0}h^2sin(1/h)/h=0$$ and you can also see there that $$f(0)=0$$ was indeed important.  franklin   23:33, 9 December 2009 (UTC)