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I know that on $\mathbb{R}$ there are at most countably many disjoint unions of the open intervals because I can take one rational point on each of the open intervals (*). Since the open intervals are disjoint, the rational points that I have taken from each interval must be different. Let $A$ be a set of all these rational points I have taken; it must be a subset of all the rational number $\mathbb{Q}$. Since $\mathbb{Q}$ is countable, a subset of a countable set is at most countable, $A$ is countable. $A$ and the disjoint open intervals on $\mathbb{R}$ have a one-to-one correspondence; therefore, the disjoint open intervals on $\mathbb{R}$ must be at most countable.

I want to use this fact to prove the statement. Suppose $x$ is the discontinuity point on $\mathbb{R}$, by definition of continuity we know $f(x^-)\neq f(x^+)$, therefore $(f(x^-),f(x^+))$ is the open interval containing the discontinuous point. I think if I can prove for two discontinuous point $x$ and $y$, if $x\neq y$, $(f(x^-),f(x^+))\cap(f(y^-),f(y^+))=\emptyset$ I can utilize the statement I have put forward at the beginning to complete the proof. But I found myself stuck at this point.

I tried to suppose the function is monotone increasing, therefore $f(x_1)\leq f(x)\leq f(x_2)$ if $x_1<x<x_2$ but get myself confused. It seems I am unable to utilize the fact that the function is monotone properly.

This is not a homework question but is an exercise from the textbook I am using to learn real analysis myself. Apologize for asking this type of basic question, and I find myself unable to use more rigorous and concise maths language to write the first two paragraphs...

update:

(*) The statement as pointed in the comment is unclear. It should be the family of all pairwise disjoint non-empty open intervals is at most countable.

I found this post: How to show that a set of discontinuous points of an increasing function is at most countable most relevant to my question. But Showing that monotone functions have at most countable discontinuities. this question provides addresses my most concerned problem (how to show if $x,y$ are two discontinuity points $x< y\Rightarrow f(x^+)\leq f(y^-)$ assuming function $f$ is monotone increasing). Is there an everywhere discontinuous increasing function? the first answer of this question provides insights to why $f(x^+)$ and $f(x^-)$ must exist if I understand the answer correctly.

Andrés E. Caicedo
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JoZ
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  • Also: https://math.stackexchange.com/q/2793202/42969, https://math.stackexchange.com/q/3556482/42969, https://math.stackexchange.com/q/3031620/42969 – Martin R Oct 18 '20 at 17:20
  • To be clear, I think what you mean in your first paragraph is that any family of pairwise disjoint open intervals must be at most countable (strictly speaking you also need to assume the intervals are non-empty, since an uncountable family of empty interval are pairwise disjoint and open). – Jack M Oct 18 '20 at 17:44
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    @MartinR Those absolutely do not answer the question. OP is asking if their specific approach is correct. – Jack M Oct 18 '20 at 17:52
  • @JackM Thank you for making it clear and thanks for your clarification for me too! In fact, I found https://math.stackexchange.com/questions/84870/how-to-show-that-a-set-of-discontinuous-points-of-an-increasing-function-is-at-m this post as mentioned by MartinR quite useful. I think what concerns me is I don't know why $x_1<x_2$ will necessarily lead to $g(x_1^+)\leq g(x_2^-)$ as stated in that post. I find myself confused whenever I am dealing with limiting points... – JoZ Oct 18 '20 at 19:00

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