I do not understand the justification of why III is false, could anyone clarify this for me please?
Which of the following statements are true about the open interval $(0,1)$ and the closed interval $[0,1]$?
I. There is a continuous function from $(0,1)$ onto $[0,1]$.
II. There is a continuous function from $[0,1]$ onto $(0,1)$.
III. There is a continuous one-to-one function from $(0,1)$ onto $[0,1]$.
(A) none (B) I only (C) II only (D) I and III only (E) I, II, and III
Statement I is true. Consider $f(x):=|\sin(2\pi x)|$; $f(1/2)=0$, $f(1/4)=1$, and every value between follows from the intermediate value theorem.
Statement II is false. The image of a compact set under a continuous map is compact. It follows that $f([0,1])$ must be compact when $f$ is continuous. But the Heine–Borel theorem implies $f([0,1])$ must be closed and $(0,1)$ is open. Thus $f([0,1])\ne(0,1)$, if $f$ is continuous.
Statement III is false. Suppose for the sake of contradiction that $g:(0,1)\to[0,1]$ is one-to-one and onto. If $g$ is one-to-one, then it must be monotonic. Since $g$ is onto there exists an $x_1$ in $(0,1)$ such that $g(x_1)=1$. But this means $g$ must be increasing for values of $x$ less than $x_1$ and decreasing for values greater than $x_1$. This contradicts monotonicity.