If $x$ is right-continuous with left limits, does $\{t>0:x(t)\ne\lim_{s\to t-}x(s)\}$ admit a minimum?

Question

Let $E$ be a normed $\mathbb R$-vector space and $x:[0,\infty)\to E$ be càdlàg with $x(0)=0$. Moreover, let $x(0-):=x(0)$, $$x(t-):=\lim_{s\to t-}x(s)\;\;\;\text{for }t>0$$ and $$\Delta x(t):=x(t)-x(t-)\;\;\;\text{for }t\ge0.$$

Now let $\tau_0:=0$ and $$\tau_n:=\inf\underbrace{\{t>\tau_{n-1}:\Delta x(t)\ne0\}}_{=:\:I_n}\;\;\;\text{for }n\in\mathbb N.$$

Can we show that either $I_n=\emptyset$ or $\tau_n\in I_n$ for all $n\in\mathbb N$?

If $I_n=\emptyset$, then $\tau_n=\infty$. Assume $I_n\ne\emptyset$. Then there is a $t_n\in I_n$. Let $$J:=\{t\in(\tau_{n-1},t_n]:\Delta x(t)\ne0\}.$$ Clearly, $$\tau_n=\inf J\in[\tau_{n-1},t_n].$$ We know that $J$ is countable, but is that sufficient to conclude $\Delta x(\tau_n)\ne0$?

Answer 1

Let $E = \mathbb{R}$ and define $x : [0, \infty) \to E$ as

$$x(t) = \begin{cases} 0 & t = 0 \\ \frac{1}{\left\lceil \frac{1}{t} \right\rceil} & t \neq 0 \end{cases}$$

Then $I_n = \left\{ \frac{1}{n} : n \in \mathbb{N} \right\}$ nad $\tau_n = 0$ for each $n$, so $\tau_n \notin I_n$.