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If $f:[0,1] \rightarrow \Bbb R$ is a strictly increasing discontinuous function, can $f([0,1])$ be a subset of $\Bbb Q$?

Attempt: Since $f$ is a discontinuous strictly increasing function, this means that :

$1.~ \forall a \in [0,1]: \lim_{x \rightarrow a^-} f(x)$ and $\lim_{x \rightarrow a^+} f(x)$ exist

$2.~$ There can be no removable discontinuity. So, if $f$ is discontinuous at a point $a :~\lim_{x \rightarrow a^-} f(x) \ne\lim_{x \rightarrow a^+} f(x)$

$3.~$ There can be only a countable number of discontinuities.

Since there can only be a countable number of discontinuities, and $[0,1]$ is uncountable, $f$ must be continuous on an uncountable number of points. Thus, $f$ must traverse an interval by being continuous at these points and hence, cannot be a subset of rationals because each interval contains uncountable irrationals.

Is my approach correct?

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    $\begingroup$ $f$ doesn't have to be discontinuous. As long as it is strictly increasing, $f([0,1])$ can't be a subset of $\Bbb Q$, by the argument in @Integrand's answer. (What is a "discontinuous function" anyway? Do you just mean that it is not everywhere continuous?) $\endgroup$
    – TonyK
    May 3, 2020 at 20:37

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This is a naive thought but here goes: since $f$ is strictly increasing on $[0,1]$, if $x<y$, $f(x)<f(y)$. In other words, $f$ is injective. But then we couldn't have $f([0,1])\subseteq \mathbb{Q}$ because $|\mathbb{Q}|<|[0,1]|$.

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    $\begingroup$ I think you meant: since f is strictly increasing $\endgroup$
    – BinyaminR
    May 3, 2020 at 20:31
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    $\begingroup$ A function can't be "strictly increasing at a point"! $\endgroup$
    – TonyK
    May 3, 2020 at 20:41
  • $\begingroup$ @TonyK I think a reasonable definition can be provided that $f$ is not strictly increasing at a point if it is not strictly increasing in any neighborhood of such point. $\endgroup$ May 3, 2020 at 21:13
  • $\begingroup$ @TonyK: there is a definition of increasing at a point. Let $f$ be defined in some neighborhood of $c$. Then $f$ is said to be increasing at $c$ if there is a neighborhood $I$ of $c$ such that $x\in I, x<c\implies f(x) \leq f(c) $ and $x\in I, x>c\implies f(x) \geq f(c) $. The strict inequality lead to definition of strictly increasing. $\endgroup$
    – Paramanand Singh
    May 4, 2020 at 7:09
  • $\begingroup$ @TonyK: the key result is that if a function is increasing at every point of an interval then it is increasing in entire interval (proof requires completeness of reals). $\endgroup$
    – Paramanand Singh
    May 4, 2020 at 7:13

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