real analysis - Why are there no continuous one-to-one functions from (0, 1) onto [0, 1]?

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

Solution

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.

Answer

Suppose that such a function $g$ exists. Take $x_0\in(0,1)$ such that $g(x_0)=1$ and take $x_1x_0$. Again, since$g$is one-to-one and$g(x_0)=1$,$g(x_2)<1$. And, since$g$is one-to-one,$g(x_1)\neq g(x_2)$. There are then two possibilities:

1. $g(x_1)
   
2. $g(x_1)>g(x_2)$. Then, by the intermediate value theorem, there is a $y\in(x_0,x_2)$ such that $g(y)=g(x_1)$.

In both cases, this contradicts that $g$ is one-to-one.