Talk:Nowhere continuous function

It Would be great if someone could elaborate this a bit more. I have read it a few times now and i still do not get it.

I expanded some on the example, which hopefully will help some. Kevinatilusa

codomain of indicator function[edit]

The codomain of the indicator function is defined to be {0,1} and not R. 194.199.97.94 23:01, 20 May 2005 (UTC)[reply]

Codomain ≠ range. --Kinu ^t/_c 17:36, 23 March 2008 (UTC)[reply]

The following sentence from the article has a repetitive phase:

More general definitions of this kind of function can be obtained, by replacing the absolute value by the distance function in a metric space, or the continuity definition by the definition of continuity in a topological space.

Should it read?:

More general definitions of this kind of function can be obtained, by replacing the absolute value by the distance function in a metric space, or by the definition of continuity in a topological space. Alephtwo (talk) 14:00, 5 May 2009 (UTC)[reply]

Dirichalet function[edit]

Should the Dirichalet (pardon misspelling) function have a separate page from the Nowhere continuous function page? Tazerdadog (talk) 04:38, 31 May 2012 (UTC)[reply]

I think there should be a page only for Dirichlet function. 'Nowhere continous' is just one aspect of this function. Doyoon1995 (talk) 19:34, 1 July 2015 (UTC)[reply]

More fancy nowhere continous fcts?[edit]

Are there (real-valued) functions which are nowhere continous and can't be changed on a measure 0 set to be a.e. continous?77.21.74.232 (talk) 22:25, 29 October 2012 (UTC)[reply]

Dirichlet function[edit]

I don‘t understand this: When y is a rational number f(y) is 1. Then how can f(x)-f(y)≥ ε > 0 , when f(x) can only be either 0 or 1 ? Oliver Carstens (talk) 19:26, 22 April 2020 (UTC)[reply]

The point is that if there were any two (three?) consecutive real numbers somewhere on the number line that were either all rational or all irrational, then the indicator function would be continuous (and constant) there. Since there are no rationals with adjacent rationals, however, the function is "nowhere continuous". -Bryan Rutherford (talk) 13:21, 15 June 2020 (UTC)[reply]