# How many points of discontinuity?

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?

• 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. – Aeolian Feb 12 '13 at 21:20
• – Jonas Meyer Feb 12 '13 at 21:28
• @Aeolian Interesting, never heard this name. The wiki proof makes things much more complicated than they are, I find. – Julien Feb 12 '13 at 21:35
• @julien You do have a point. A bit of Googling returned this (problem #2), which is essentially the Wikipedia proof, but better explained. – Aeolian Feb 12 '13 at 22:54

$f(x)=\begin {cases} 1&x \in \mathbb Q \\ 0 & x \not \in \mathbb Q \end {cases}$

is discontinuous at every point.

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

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.

• @GitGud Yes, of course, I just added it. Thanks. – Julien Feb 12 '13 at 21:30

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 ...