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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.

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

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  • 2
    $\begingroup$ 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. $\endgroup$ – user296602 Aug 26 '17 at 22:36
  • $\begingroup$ 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). $\endgroup$ – Piquito Aug 26 '17 at 22:43
  • $\begingroup$ 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 $U, V\subset \mathbb{R}^n$). $\endgroup$ – Michael Lee Aug 27 '17 at 0:16
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    $\begingroup$ 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. $\endgroup$ – Martin Sleziak Aug 27 '17 at 5:31
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    $\begingroup$ See also: Continuous bijection from $(0,1)$ to $[0,1]$. $\endgroup$ – Martin Sleziak Aug 27 '17 at 5:32
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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.

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  • $\begingroup$ why we assumed that the y in the first possibility is the same y in the second possibility. $\endgroup$ – Intuition Aug 27 '17 at 1:47
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    $\begingroup$ @Intuition: I don't think we did. They just happen to be written as the same letter. $\endgroup$ – Kevin Aug 27 '17 at 5:27
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    $\begingroup$ @Intuition Of course it is not the same $y$. The first one is smaller than $x_0$, whereas the second one in is greater. $\endgroup$ – José Carlos Santos Aug 27 '17 at 8:44
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    $\begingroup$ @Intuition: How do you know that both are excellent answers? You can't just accept something as correct just because some user on Math SE said so! $\endgroup$ – user21820 Aug 27 '17 at 9:04
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    $\begingroup$ @Intuition: You have read both, but it seems you do not know enough to evaluate their correctness. Therefore in my opinion you are not qualified to judge whether they are excellent or not. By doing so, you are giving the (false) impression that you know they are correct. The fact is that one is wrong. $\endgroup$ – user21820 Aug 27 '17 at 9:36

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