Let $X$ be a Banach space and consider the following definition.
Definition. $f:[a,b]\to X$ is regulated if it has one-sided limits at every point of $[a,b]$, i.e. for every $c\in [a,b)$ there is a value $f(c+)\in X$ such that $$\lim_{t\to c^+}\left\|f(t)-f(c+)\right\|=0$$ and if for every $c\in (a,b]$ there is a value $f(c-)\in X$ such that $$\lim_{t\to c^-}\left\|f(t)-f(c-)\right\|=0.$$
Question. If $f:[a,b]\to X$ is regulated, then how do we show that for every $\epsilon>0$, the set $$N=\{c\in[a,b):\left\|f(c+)-f(c)\right\|\geq \epsilon\}$$ is a finite set.
NOTE: If $f$ is continuous then the question is clear to me. So we have to assume that $f$ has discontinuities (in fact, the set of such discontinuities is at most countable). What bothered me actually is to show that $N\neq \varnothing$ whenever $f$ has discontinuities.
Tips or suggestions are very much appreciated.
If $\epsilon>0$, choose a step function $\tau$ such that $\sup_{x\in[a,b]}\|f(x)-\tau(x)\|<\frac\epsilon4$. Now, since $\tau$ is a step function, there is a partition $\{I_k\}_{k=1}^n$ of $[a,b]$ such that $\tau$ is constant on each interval $I_k$. Let $M$ denote the set of endpoints of the intervals $I_k$. This set is finite.
Now, suppose that $x\not \in M$. Then there is a $\delta>0$ such that $(x-\delta,x+\delta)\subset I_k$ for some $k$. Also, since $f$ is regulated, there are points $t_1\in (x-\delta,x), t_2\in (x,x+\delta)$ such that $\|f(t_1)-f(x-)\|<\frac\epsilon4$ and $\|f(t_2)-f(x+)\|<\frac\epsilon4$. Note that $t_1, t_2\in I_k$, so that $\tau(t_1)=\tau(t_2)$. Then $$\begin{align}\|f(x-)-f(x+)\|&\le\|f(x-)-f(t_1)\|+\|f(t_1)-\tau(t_1)\|\\ &+\|\tau(t_2)-f(t_2)\|+\|f(t_2)-f(x+)\|<\epsilon,\end{align}$$ so $x\not \in N$. Then $N\subset M$ and in particular $N$ is finite.