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

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.

## closed as off-topic by user21820, Siong Thye Goh, B. Goddard, Did, Simply Beautiful ArtSep 2 '17 at 15:03

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• What don't you understand? Sit down and draw a picture of a monotone function (do you know what "monotone" means?) and understand why there's a problem when it peaks. – user296602 Aug 26 '17 at 22:36
• The $x_1$ is necessarily less than $1$ which implies that there are infinitely many points in the very small open interval $(x_1,1)$ whose image should be greater than $1$ (this supposing $g$ increasing and you can see the case in which g is decreasing). – Piquito Aug 26 '17 at 22:43
• As an aside, if $f : U\to \mathbb{R}^n$ for some open $U\subset \mathbb{R}^n$ is continuous and injective, then $f(U)$ is open. This is called Brouwer's invariance of domain theorem. Proof here. This implies that for continuous and injective $f : (0, 1)\to [0, 1]$, $\operatorname{im}(f)$ must be open in $\mathbb{R}$ and therefore can't be $[0, 1]$ (more generally, there is no continuous bijection from open $U$ to non-open $V$ for any $U, V\subset \mathbb{R}^n$). – Michael Lee Aug 27 '17 at 0:16
• If the question is about details of this specific proof (as opposed to asking for any proof of the fact in the title) you should mark this by using (proof-explanation) tag. – Martin Sleziak Aug 27 '17 at 5:31
• – Martin Sleziak Aug 27 '17 at 5:32

Suppose that such a function $g$ exists. Take $x_0\in(0,1)$ such that $g(x_0)=1$ and take $x_1<x_0$. Since $g$ is one-to-one and $g(x_0)=1$, $g(x_1)<1$. Now, take $x_2>x_0$. 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. $g(x_1)<g(x_2)$. Then, by the intermediate value theorem, there is a $y\in(x_1,x_0)$ such that $g(y)=g(x_2)$.
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.
• @Intuition Of course it is not the same $y$. The first one is smaller than $x_0$, whereas the second one in is greater. – José Carlos Santos Aug 27 '17 at 8:44