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?
Best Answer
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 $|\mathbb{Q}|<|[0,1]|$.