Image of compact set under piecewise continuous function Let $a,b>0\in\mathbb{R}$. Let $U$ be an domain in $\mathbb{C}^n$. Let $f:[a,b]\longrightarrow U$ be a piecewise continuous map. Then is $f[a,b]$ compact? If not compact, will it be bounded?
Ok. This is in the following context. I am given a piecewise smooth path $\gamma:[a,b]\longrightarrow U$. Where $\gamma(a)=z$ and $\gamma(b)=w$, for given $z,w\in U$. We are also given a function $\alpha:U\times\mathbb{C}^n\longrightarrow \mathbb{R}$, which is upper semicontinuous. Now it is said that $t\in[a,b]\longrightarrow \alpha(\gamma(t),\gamma’(t))$ is bounded and measurable. I wanted to know why the function is bounded. I know that $\gamma[a,b]$ is compact. And $\gamma$ being upper semicontinuous will attain its maximum on a compact set. But I am not sure about $\gamma’$.
 A: Compact, not necessarily: On $[0,1]$ let $f(x) = x, 0\le x<1,$ $f(1)=2.$ Then $f([0,1]) = [0,1)\cup\{2\}.$
Bounded, yes: First, a lemma: If $f$ is continuous on $(a,b)$ and $f$ has finite limits at the end points, then $f(a,b)$ is bounded.
Proof: Suppose $\lim_{x\to a^+} f(x)=L,$ $\lim_{x\to b^-} f(x)=M.$ Let $\epsilon=1.$ Then there exists $\delta_a>0, \delta_a<(b-a)/3,$  such that $|f(x)-L|<1$ for $x\in (a,a+\delta_a).$ Thus for such $x,$
$$|f(x)| = |f(x)-L+L|\le |f(x)-L|+|L| <1+|L|.$$
Similarly, there exists $\delta_b>0,\delta_b<(b-a)/3,$ such that $|f(x)|<1+|M|$ for $x\in (b-\delta_b,b).$ It follows that $f$ is bounded on the set $(a,a+\delta_a)\cup (b-\delta_b,b).$
Since $f$ is continuous on the compact set $[a+\delta_a,b-\delta_b],$ $f([a+\delta_a,b-\delta_b])$ is compact, hence is bounded. It follows that $f(a,b)$ is bounded.
Now suppose $f$ is piecewise continuous on $[a,b].$ Then there exist points $a=x_0<x_1<\cdots <x_n=b$ such that $f$ is continuous on each $I_k=(x_{k-1},x_k)$ and has finite limits at the end points of $I_k.$ By the lemma, each $f(I_k)$ is bounded. The set $f(\{x_0,\dots x_n\})$ is also bounded. Therefore
$$f([a,b])=f(I_1)\cup \cdots \cup f(I_n)\cup f(\{x_0,\dots x_n\})$$
is bounded.
