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Isolated continuity point

  1. Aug 25, 2014 #1
    Hi
    The question is the following: is it possible for a (say) real function to be continuous at a certain point internal to its domain, and be discontinuous in some neighborhood of that point?
    I am not talking about a function defined at a single point or things like that, but of a function defined on the entire [itex]\mathbb{R}[/itex] (or some interval in it, whatever).

    Now, i have also came up with an answer: a function [itex]f[/itex] defined as [itex]f(0)=0[/itex], [itex]f(x)=x[/itex] for every rational [itex]x[/itex], and [itex]f(x) = 2x[/itex] for every irrational [itex]x[/itex]. Such a function would be (seems to me) continuous at [itex]x=0[/itex] and discontinuous for any other [itex]x[/itex]. I am not completely certain of this, though, and for that reason i would like some feedback on this.

    I am also asking this question because strangely enough I have never heard of the concept of an isolated continuity point, while for example the "opposite" (that of an isolated singularity) is quite common, and I would like to know if it's just me or if it is just a "useless" pathological concept.

    Thank you in advance.
    Bye
     
  2. jcsd
  3. Aug 25, 2014 #2
    Yes.

    In fact, there is a function defined on all of ##\mathbb{R}## which is continuous at a single point.
     
  4. Aug 25, 2014 #3

    micromass

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    Yes, that is a correct example.

    Now, can you come up with an example of a function defined on entire ##\mathbb{R}## that is differentiable only in one point? :tongue:
     
  5. Aug 25, 2014 #4

    pasmith

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    If [itex]f : \mathbb{R} \to \mathbb{R}[/itex] and [itex]g : \mathbb{R} \to \mathbb{R}[/itex] are continuous such that there exists a unique [itex]a \in \mathbb{R}[/itex] such that [itex]f(a) = g(a)[/itex], then the function [tex]
    h : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} f(x) & x \in \mathbb{Q} \\
    g(x) & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} [/tex] is discontinuous on [itex]\mathbb{R} \setminus \{a\}[/itex] and continuous at [itex]a[/itex].
     
  6. Aug 26, 2014 #5
    That is a very interesting example, thank you.

    That seems to be tricky! I did some research and stumbled upon this discussion of that matter, in which that question is very well explained.

    Now however I wonder if it is accidental that in all of these examples the functions are constructed using rational and irrational numbers. I think that the important point is to have one subset which is dense in the other. Is there some example of functions having this kind of "pathologies" NOT using rational/irrational numbers in the definition?
    Even better, is it possible to find a function of this kind NOT using at all dense subsets of the real numbers in the definition?

    Thanks
     
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