Does compactness hold for a cadlag function on some closed, bounded set $[a,b]$

Question

Let $f$ be some cadlag function, left limits exist and right continuituous. Does compactness hold for a cadlag function on some closed, bounded set $[a,b]$. In other words, is the set $\{f(c):a\leq c\leq b\}$ compact?

My attempt:

Clearly, it is bounded. In my view, it need not be closed since our cadlag function $f$ could have some jump at $d$, and then constructing some monotone sequence from below to $d$ would result in $\lim_{e \nearrow d} f(e)\neq f(d)$. And since $$\text{ bounded and closed }\iff\text{ compact }$$ in euclidean spaces, $\{f(c):a\leq c\leq b\}$ need to be compact.

Answer 1

You are right that the image of a compact interval under a càdlàg function is bounded, but not necessarily closed. The conclusion is that the image is not necessarily compact.

A concrete example is $f:[0, 1] \to \Bbb R$, $f(x) = x + \lfloor x \rfloor$, with $f([0, 1]) = [0, 1) \cup \{ 2 \} $.

Answer 2

You could take

$$f(x) = \begin{cases} x^2+1 & 0 \le x \le 1 \\ -x^2 & -1 \le x < 0\end{cases}$$

This has left limits and is right continuous, but its image is not compact since its range is $[-1,0)\cup [1, 2]$.