$f:[0,1] \to [0,1]$ is increasing. Does $0 \leq f(x) \leq x$ for all $x \in [0,1]$ imply that $f$ is continuous near $0$? As in the question. I'd really appreciate your help with this.
(1) $f:[0,1] \to [0,1]$ is increasing, that is, $x \leq y \implies f(x) \leq f(y)$
(2) $0 \leq f(x) \leq x$ for all $x \in [0,1]$
Does (1)+(2) imply that there exists $\epsilon>0$ such that $f$ is continuous on $[0,\epsilon)$?
If no: Can f have infinitely many jumps on $[0,\epsilon)$ for any $\epsilon>0$?
NOTE: Clearly $f$ is continuous at $0$. The question is whether we can conclude that it is also continuous in some (small) neighborhood "to the right of $0$" when it is increasing/monotone.
My (probably wrong) intuition is that if we have infinitely many jumps then the "slope" cannot be below $1$ as it must be close enough to $0$.
 A: No, you cannot infer continuity, nor indeed limit the number of jumps. You can say that the jumps must be countable (although this is true for any monotone function), and you can show that the total cumulative distance that the graph "jumps" is finite (i.e. $\sum_{x_0 \in [0, 1]} \left[\lim_{x \to x_0^+} f(x) - \lim_{x \to x_0^-} f(x)\right]$ converges), but there still could be infinitely many jumps.
As an example, let
$$f(x) = \begin{cases}\left\lceil x^{-1} \right\rceil^{-1} & \text{if }x \in (0, 1] \\ 0 & \text{if }x = 0,\end{cases}$$
where $\lceil \cdot \rceil$ is the ceiling function, i.e. $\lceil x \rceil$ is the least integer greater than or equal to $x$. We have
$$\left\lceil x^{-1} \right\rceil \ge x^{-1} > 0 \implies 0 < \left\lceil x^{-1} \right\rceil^{-1} \le x.$$
Since $\lceil \cdot \rceil$ is monotone increasing, and $x \mapsto x^{-1}$ is monotone decreasing, we see that $\left\lceil x^{-1} \right\rceil$ is decreasing, so $f$ is once again increasing (note $f(0) \le f(x)$ for $x > 0$).
The "jumps" occur whenever $x = \frac{1}{n}$ for some $n \in \Bbb{N}$, i.e. for countably many $n$. These points also accumulate around $0$, so $f$ is not continuous on any neighbourhood of $0$.
