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How can I show that a bijective continuous function $f:\mathbb R\to \mathbb R$ is monotone ?

My attempts

Let $a<b$. Since $f$ is injective, $f(a)\neq f(b)$. Therefore $f(a)<f(b)$ or $f(a)>f(b)$. Suppose with out loss of generality that $f(a)<f(b)$ and let $x,y\in[a,b]$ such that $x<y$. By injectivity, $f(x)\notin \{f(a),f(b)\}$. Suppose that $f(x)\notin ]f(a),f(b)[$.

If $f(x)<f(a)$, by the intermediate value theorem (IVT), there is a $c\in[x,b]$ such that $f(x)=f(c)$ what contradict injectivity.

If $f(x)>f(b)$, by IVT we have a $d\in[a,x]$ such that $f(d)=f(b)$ what also contradict injectivity.

Then $f(x)\in [f(a),f(b)]$. We can show as well that $f(y)\in [f(a),f(b)]$ and that $f(x)<f(y)$. Therefore we have that $f(x)<f(y)$ if $x<y$.

The problem is that I'm not sure that my proof is finish. I just proved that for all $a,b\in\mathbb R$, s.t. $a<b$ the function $f$ is monotone on $[a,b]$. But may be it's enough to say that it's monotone on all $\mathbb R$. What do you think ?

hardmath
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Rick
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  • Since it works for all possible pairs of values, it works on the real line. – Terra Hyde Oct 07 '15 at 21:15
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    It's not duplicate since the OP is not asking how to prove the statement on $[a,b]$ but how to generalize the result on $\mathbb R$. – Surb Oct 07 '15 at 21:17
  • For each pair $x < y$ in $\mathbb {R}$, they are contained in a compact interval, apply your argument, (wolg) we get $f(x) < f(y)$. And we are done. – Xiao Oct 07 '15 at 21:23
  • Little correction: in the 2nd paragraph of your proof, "If $f(x) < f(a)$, ... there is $c \in [x, b]$ such that..." Here you have "$f(x) = f(c)$", but you mean/want $f(a) = f(c)$, or better still, $f(c) = f(a)$. (In the next paragraph when considering the other endpoint, you use that order, "we have $d \in [a, x]$ such that $f(d) = f(b)$" -- it reads better, I think). – BrianO Oct 07 '15 at 22:44
  • And no, I don't see that you're done -- so far you've shown that for all $a, b$, $f$ is monotone on $[a, b]$; however, it may be monotone increasing on $[0, 1]$ but decreasing on $[100, 101]$. So, next show that (*) for any $b' > b$, $f$ is increasing on $[a, b']$, and similarly that (**) for any $a' < a$, $f$ is increasing on $[a', b]$. Having shown (*), you can hand-wave through (**) "by symmetry". Then you're done. – BrianO Oct 07 '15 at 23:09

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