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Let $F : \mathbb{R}\to \mathbb{R}$ be a monotone function then Prove that $F$ has countably many discontinuities.

Edit: Monotone function is not only increasing.

My Attempt : I know that the set of discontinuities of a monotone function is at most countable. If I take greatest integer function $[x]$ which is monotone function that has countable set of discontinuities as $\Bbb Z $. But I'm unable to Prove it. Please help me.

Baljeet
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  • Note that the discontinuities are jumps, the lateral limits exist. Now look inside an interval $[n,n+1]$ for $n\in\mathbb{Z}$. The maximum and minimum of the function in that interval are $f(n+1)$ and $f(n)$. How large can be the sum of the jumps of $f$ that happen in the interval $[n,n+1]$? Note that it cannot be larger than $|f(n+1)-f(n)|$, in particular it is finite. Then note that the sum of uncountably many numbers (moreover positive numbers) cannot be finite. – MoonLightSyzygy Jan 12 '20 at 04:03
  • The result holds for any INTERVAL in $\mathbb{R}$ – SL_MathGuy Jan 12 '20 at 04:22
  • Are you asking why $[x]$ has countably many discontinuities? – Kavi Rama Murthy Jan 12 '20 at 04:51
  • @Kavi Rama Murthy : No , I wants a general proof. – Baljeet Jan 12 '20 at 04:58
  • This is a partial answer. – Baljeet Jan 12 '20 at 05:36
  • What is partial about it? If $f$ is monotone then either $f$ is increasing or $-f$ is increasing, and they both have the same set of discontinuities. –  Jan 12 '20 at 08:35

1 Answers1

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Let $D$ denote the set of discontinuity points. For each $x\in D$, the left and right limits differ, and are therefore the endpoints of an interval with non-empty interior. In this manner we obtain a collection of such intervals $$ \{I_d\colon d\in D\}. $$ By monotonicity the intervals are disjoint. Choosing one rational number from each interval therefore yields an injection from $D$ into $\mathbb Q$. Hence $D$ is countable.

EDIT: I see there is some confusion on the part of the question asker. The terminology "countable" normally means "has cardinality less than or equal to that of the natural numbers". In particular, that is the only interpretation of that word which makes your question correct. (It is easy to give an example of a monotone function with only a finite number - or even zero - discontinuities.) So if the question asker understands already why there are at most countably many discontinuities, that means there is actually no question left to answer: since "at most countably many" is a synonym for "countably many".

pre-kidney
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  • This question has appeared many many times on MSE. – Kavi Rama Murthy Jan 12 '20 at 05:23
  • @pre-kidney, Interval contains rationals as well as irrationals and tell me that when we choose an irrational from each interval then why one one is not possible from $D$ to $\Bbb Q^c$ ? – Baljeet Jan 12 '20 at 05:24
  • That too is possible - you can get an injection from $D$ to the irrationals $\mathbb Q^c$, but since there are uncountably many irrationals that is not a very interesting fact. – pre-kidney Jan 12 '20 at 07:12
  • @Baljeet After reading your comments and question again, I think you are confused over terminology relating to the word "countable". See my edit. – pre-kidney Jan 12 '20 at 08:04