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If $f$ is an increasing function defined on $[a,b]$ . Then show that the set of discontinuities of $f$ is countable and that $f$ has points of continuity in every open subinterval of $[a,b]$

Attempt: I have trouble understanding why the number of discontinuities of $f$ should be countable. For example, if we define a function which is increasing but different at each point in the interval $[0,1]$, then, there can be uncountable number of discontinuities.

I think if we prove that the number of discontinuities is countable, then we can prove that the function has points of continuity in every open interval of $[a,b]$.

Please help me understand intuitively why $f$ must necessarily have countable number of discontinuities. Suppose, $f$ has an uncountable number of discontinuities, then what can happen?

Thank you for your help.

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    $\begingroup$ Show that, for any $x$, $\lim_{t\to x^-}f(t)$ and $\lim_{t\to x^+}f(t)$ exist, and therefore any discontinuity is a jump discontinuity, and the interval $(\alpha,\beta)=(\lim_{t\to x^-}f(t),\lim_{t\to x^+}f(t))$ witnesses this. For different points of discontinuity, these intervals are disjoint. $\endgroup$ – Andrés E. Caicedo Oct 9 '14 at 7:10
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An $x$ at which $f$ is not continuous is in one of the sets $$ A_n = \left\{ x: \lim_{x+}f - \lim_{x-}f > 1/n \right\} $$

On the other hand, $$ f(b) - f(a) <\infty $$

As a countable union of finite sets is countable, you are done.

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  • $\begingroup$ Thank you for your answer. $x$ could be possibly discontinuous when $\lim_{x+}f = \lim_{x-}f $ but $f(x)$ might be missing. Is there a reason we have not included that case? $\endgroup$ – MathMan Oct 9 '14 at 11:20
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    $\begingroup$ Points of continuity must only be searched into the domain of the function, since the definition of continuity requires the value of the function to be computed in the point. If you decide that a function is not continuous where it is not defined, then the function $f(x) = \sqrt(x)$ would be discontinuous for every $x<0$ which is a not-countable set. $\endgroup$ – Emanuele Paolini Oct 9 '14 at 13:09
  • $\begingroup$ @EmanuelePaolini Thank you very much for your comment. I get it now. $\endgroup$ – MathMan Oct 9 '14 at 13:52

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