Let $a,b\in\mathbb R$ with $a<b$, $(E,d)$ be a metric space and $x:[a,b]\to E$. Assume that, for all $t\in[a,b]$, $x$ has a left-sided limit $x(t-)$ and a right-sided limit $x(t+)$, where $x(a-):=x(a)$ and $x(b+):=x(b)$.
We can easily show$^1$ that $x$ is bounded.
How can we show that
- $x$ even attains its infimum and supremum (maybe assuming that $x$ is either right- or left-continuous)?
- If $x$ is right-continuous, then it is uniformly right-continuous?
- If $x$ is right-continuous, then $$x^-(t):=x(t-)\;\;\;\text{for }t\in[a,b]$$ is left-continuous?
My finial goal is to show (3.), but I think that we need (2.) for that and it seems to me that we need (1.) to show (2.).
I've actually no idea how we can show (1.), since the analogue statement for continuous functions on compact spaces depends on the fact that the image space is compact as well. This can obviously not applied here ... However, I guess we need to assume that $x$ is at least either right- or left-continuous, but even then the former result cannot be applied. So, how do we need to argue?
$^1$ Assuming the contrary, there is a $(t_n)_{n\in\mathbb N_0}\subseteq[a,b]$ with $$d(x(t_0),x(t_n))\ge n\;\;\;\text{for all }n\in\mathbb N.$$ Since $[a,b]$ is sequentially compact, $t_{n_k}\xrightarrow{k\to\infty}t$ for some increasing $(n_k)_{k\in\mathbb N}$ for some $t\in[a,b]$. By passing to another subsequence, if necessary, we can assume that $(t_{n_k})_{k\in\mathbb N}$ is monotonic. Assume that it is nondecreasing. Then, $$d(x(t_{n_k}),x(t-))\xrightarrow{k\to\infty}0$$ and hence $$\infty\xleftarrow{k\to\infty}n_k\le d(x(t_0),x(t_{n_k}))\le d(x(t_0),x(t-))+d(x(t_{n_k},x(t-))\xrightarrow{k\to\infty}d(x(t_0),x(t-));$$ which is impossible.