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I need to prove that any monotonic function whose domain is an interval $[a;b]$ can have only finite or countable number of discontinuity points...

I don't seem to have any insightful ideas. It even raises more questions in my head. What happens if we remove a requirement for monotonicity? Can you tell me any function (whose domain is all real numbers for example) have more than countable number of discontinuity points?

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    $\begingroup$ Take a look at this Wikipedia article for functions discontinuous everywhere. Also, the statement you are trying to prove is apparently called Froda's theorem. $\endgroup$ – Aeolian Feb 12 '13 at 21:20
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    $\begingroup$ See math.stackexchange.com/questions/56831/… $\endgroup$ – Jonas Meyer Feb 12 '13 at 21:28
  • $\begingroup$ @Aeolian Interesting, never heard this name. The wiki proof makes things much more complicated than they are, I find. $\endgroup$ – Julien Feb 12 '13 at 21:35
  • $\begingroup$ @julien You do have a point. A bit of Googling returned this (problem #2), which is essentially the Wikipedia proof, but better explained. $\endgroup$ – Aeolian Feb 12 '13 at 22:54
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$f(x)=\begin {cases} 1&x \in \mathbb Q \\ 0 & x \not \in \mathbb Q \end {cases}$

is discontinuous at every point.

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  • $\begingroup$ Is it appropriate to say that for this (Dirichlet function?) a set of discontinuity points has cardinality of the continuum? $\endgroup$ – Kapitonas Feb 12 '13 at 21:54
  • $\begingroup$ @Kapitonas: yes it is. In fact it is the continuum. $\endgroup$ – Ross Millikan Feb 12 '13 at 21:55
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Assume $f$ is non-decreasing first.

At a discontinuity point $x_0$, you have $$\lim_{x\rightarrow x_0^-}f(x)< \lim_{x\rightarrow x_0^+} f(x).$$

Now try to find a one-to-one mapping from the set of discontinuity points into $\mathbb{Q}$.

The non-increasing case can be treated similarly.

Your second question has already been answered many times, so I will stop here.

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  • $\begingroup$ @GitGud Yes, of course, I just added it. Thanks. $\endgroup$ – Julien Feb 12 '13 at 21:30
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Sorry for that but I will raise even more questions in your head : there exists functions that are not continuous in any point. Moreover, these kind of "monstruous functions" are dense ...

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