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I have a feeling that no such map exists, i.e., there is no non-decreasing surjective function from $\mathbb{R}$ to $\mathbb{Q}$. But I am just unable to write an argument. Any hint or sketch of proof will be helpful.

ogirkar
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    If you mean strictly monotonical then that would imply injectivity. – dxiv Mar 20 '23 at 17:52
  • non-decreasing function – ogirkar Mar 20 '23 at 17:57
  • I have sketched an argument.. will add it as an answer after sometime.. just check the idea.. function can't be continuous.. being monotonic, it can only have jump discontinuity.. so because of that jump, it will fail to be surjective – ogirkar Mar 20 '23 at 18:34
  • Please [edit] your post to provide additional context. – Xander Henderson Mar 20 '23 at 18:50
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    The question has been answered here before, e.g. https://math.stackexchange.com/q/3326686/ . – Xander Henderson Mar 20 '23 at 18:52
  • @Believer a jump discontinuity doesn't mean that the function is not surjective onto the the Rationals though – joseville Mar 20 '23 at 18:59
  • @joseville but function is monotonic.. so I am having a feeling that jump discontinuity along with monotonicity will do the job – ogirkar Mar 20 '23 at 19:01
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    @joseville It actually does. If there is a jump discontinuity at $c$, then $$ L^- = \lim_{x\uparrow c} f(x) < \lim_{x\downarrow c} f(x) = L^+. $$ The interval $(L^-,L^+)$ must contain some rational number, and the monotonicity of $f$ ensures that $f(x) \le L^1$ for all $x < c$ and that $f(x) \ge L^+$ for all $x > c$. Thus none of the rationals in that interval are in the image of $f$, so $f$ cannot be surjective. – Xander Henderson Mar 20 '23 at 19:05
  • @XanderHenderson But what about the function you mentioned in a comment to the answer, $f : \mathbb{R}\to\mathbb{N}: x \mapsto \lceil x \rceil$. It has has jump discontinues but is still surjective onto the naturals. – joseville Mar 20 '23 at 19:10
  • @joseville That is only a counterexample to the claim that an increasing function must be injective. The difference here is that $\mathbb{Q}$ is dense in the reals, whereas $\mathbb{Z}$ is not. I was addressing only that first paragraph, not the rest of the argument. – Xander Henderson Mar 20 '23 at 19:11
  • Thanks for that info, but if I've understood correctly, just the fact that a function has one or more jump discontinuities, by itself, doesn't imply that it is not surjective. It can still be surjective, depending on its codomain. – joseville Mar 20 '23 at 19:17
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    @joseville Yes.. here along with jump discontinuity, we are using two facts.. one is function is monotonic and second is that rationals are dense – ogirkar Mar 20 '23 at 19:20

1 Answers1

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Suppose there is a monotonically increasing onto map $f: \Bbb R \to \Bbb Q$.

Consider $$A= \{ x \in \Bbb R : f(x) < \sqrt 2\}$$ $$B= \{ x \in \Bbb R : f(x) > \sqrt 2\}$$

clearly $A \cup B = \Bbb R$ and $A \cap B = \emptyset$.

Since $f$ is onto, we have $f(A) \cup f(B) = \Bbb Q$. Since $f$ is increasing, we have $f(A) < \sqrt 2 < f(B)$ and so $f(A) \cap f(B) = \emptyset$.

Moreover, since $f$ is increasing, we have that $$\forall a \in A \quad \forall b \in B \quad a \le b$$

This means that $A$ and $B$ are separated by a real number $\xi \in \Bbb R$ such that $$\forall a \in A \quad \forall b \in B \quad a \le \xi \le b$$ which is equal to $\xi = \sup A = \inf B$.

Now, using monotinicuty of $f$ we have $$\forall a \in A \quad \forall b \in B \quad f(a) \le f(\xi) \le f(b)$$ Hence $$\sqrt 2 = \sup f(A) \le f( \xi) \le \inf f(B) = \sqrt 2$$ Thus $f(\xi)= \sqrt 2$, which contradicts $\sqrt 2 \notin \Bbb Q$.

We have a contradiction, hence $f$ cannot exist.

Crostul
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    The argument in your first paragraph fails (though that might be a result of edits to the question): the map $f : \mathbb{R}\to\mathbb{Z}: x \mapsto \lceil x \rceil$ is non-decreasing and surjective, but is not injective. Increasing doesn't mean strictly increasing. – Xander Henderson Mar 20 '23 at 18:45