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Prove that there does not exist a continuous, bijective function $f:[0,1)\to \mathbb{R}.$

By contradiction I can assume a function exists, so that function is surjective, onto and continuous. And I know I need to use the intermediate value theorem but I can't create such a contradiction.

user104235
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4 Answers4

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HINT: Show that a continuous injection is either order preserving or order reversing.

Asaf Karagila
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  • Hi Asaf, im not sure what you mean by order preserving. But is it okay if I do this? suppose there exists such a function. Let $x\in [0,1)$ such that $f(x) <f(0)$ thus $f(0) \in [f(x),f(1)]$ therefore there exists a $y$ such that $f(y) = f(0)$ which is a contradiction to one to one? Is that okay or can I not assume there exists $f(x) <f(a)$? – user104235 Nov 20 '13 at 02:07
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    To expand a bit on Asaf's hint, a continuous one-one function is either strictly increasing or strictly decreasing, hence either $f(0)$ is a minimum or a maximum of $f$ and thus range of $f$ is not $\mathbb{R}$, but an interval with one end point included. – Paramanand Singh Nov 20 '13 at 05:30
  • @ParamanandSingh So I cant do it that way since its not strictly increasing or decreasing. – user104235 Nov 20 '13 at 05:47
  • @ParamanandSingh So I have this. $f([0,1/2])$ is equal to either $[f(0),f(1/2)]$ or $[f(1/2),f(0)],$ so assume it equals $[f(0),f(1/2)].$ What can I say about $f(3/4)$? It has to be greater than $f(1/2)$ right? – user104235 Nov 20 '13 at 06:19
  • @user104235: Suppose $x<y$ and both $f(0)<f(x)$ and $f(y)<f(x)$. Show that the function is not injective (that would be the right place to use IVT). – Asaf Karagila Nov 20 '13 at 06:31
  • Yup, that confused me even more :(. I give up on this question. Thanks anyways! – user104235 Nov 20 '13 at 07:00
  • @user104235: C'mon, it's not very hard. You're just letting the details get the best of you. The idea is that between two intervals a continuous injection cannot change the order. That is, either $x<y$ implies $f(x)<f(y)$ or $x<y$ implies $f(x)>f(y)$. Do you understand why proving that would solve the problem? – Asaf Karagila Nov 20 '13 at 07:03
  • So if I let $x\in (a,b)$ and $f(x)<f(a)$ then by IVT $f([x,b])$ contains all points between $f(x)$ and $f(b)$ and contains $f(a)$ contradiction that $f$ is one-to-one? Is that right? – user104235 Nov 20 '13 at 07:12
  • @user104235: Assuming that you mean $f(x)<f(b)$ there, still no. It might be the case that $f(b)<f(a)$. But you can show there is $c\in(a,x)$ and $d\in(x,b)$ such that $f(c)=f(d)$. And then it contradicts injectivity. – Asaf Karagila Nov 20 '13 at 07:15
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    After Asaf's comments I think you must have got a clear idea of what needs to be done. What Asaf has shown in his comments is that "if a continuous function changes its direction of growth (like the graph rises and falls) then it will not be one-one". This is the crux of the solution. – Paramanand Singh Nov 20 '13 at 07:59
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From the interaction in comments I think OP needs bit more elaboration on the hint by Asaf. I will however refrain from providing a complete solution.

Assume that there is a function $f$ defined on $[0, 1)$ which is continuous and a bijection from $[0, 1)$ to $\mathbb{R}$. It means that $f$ is one-one function in particular i.e $f(a) = f(b)$ implies $a = b$.

Next let's consider $f(0)$ and $f(1/2)$. Because $f$ is one-one we must have $f(0) \neq f(1/2)$.

Let's assume that $f(0) < f(1/2)$ (the case $f(0) > f(1/2)$ can be handled similarly). Now we need to prove that if $x \in (0, 1)$ then $f(0) < f(x)$. This is where you need to use IVT for continuous functions. You should be able to do this by following Asaf's comments. And then we know that $f(0)$ is the minimum value of $f(x)$ and hence the part $(-\infty, f(0))$ of $\mathbb{R}$ is not mapped by this function $f$ and thus $f$ is not onto $\mathbb{R}$.

The proof of $f(0) < f(x)$ for all $x \in (0, 1)$ proceeds as follows. Clearly $f$ is one-one so $f(0) \neq f(x)$. If $x = 1/2$ then we already know that $f(0) < f(1/2) = f(x)$. So let $x \neq 1/2$. If $f(0) > f(x)$ then $f(x) < f(0) < f(1/2)$ so that $f(0)$ lies between $f(x)$ and $f(1/2)$ and hence by IVT we have ..... (I hope OP will be able to complete the dots)

If $f(0) > f(1/2)$ then we can show that $f(x) < f(0)$ for all $x \in (0, 1)$ so that $f(0)$ is the maximum value of $f$ and again $f$ is not onto $\mathbb{R}$.

Another thing to note. There is nothing special in $1/2$ we have chosen above. It can be replaced by any number lying in $(0, 1)$.

Paramanand Singh
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Let $f : [0,1) \rightarrow \mathbb{R}$ be a bijection. Then the restriction of $f$ to $(0,1)$ is a continuous bijection with range $\mathbb{R} - {f(0)}$ which is clearly disconnected but (0,1) is connected.

GA316
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Since there are continuous bijections (even homeomorphisms) from $(0,1)$ to $\mathbb R$, it somehow must have to do with that $0$. Now, how do we use it?

One observation would be that such a map $f$ restricts to a continuous bijection from $(0,1)$ to $\mathbb R\setminus\{f(0)\}$, the real line without one point. Why can such a map not exist?

A different version of the same argument would be that for such a map $f$ there would have to be an $a\in(0,1)$ with $f(a)<f(0)$ and a $b\in(0,1)$ with $f(b)>f(0)$. Now, what would the intermediate value theorem tell us then, and how would that lead to a contradiction?

Carsten S
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