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I am trying to solve:

Let $f:[a,b] \to [c,d]$ be a continuous bijection. Suppose $f(a) < f(b)$

(i) Let $a < x_1 < b$. Prove $f(a) < f(x_1) < f(b)$

(ii) Let $a < x_1 < x_2 < b$. Prove $f(x_1) < f(x_2)$

I think I need to use the IVT for this. For (i) I have stated that for all y between c and d there exists $x_1 \in [a,b]$ such that $f(x_1) = y$. As f is continuous, $f(x_1)$ is such that $f(a) < f(x_1) < f(b)$. I'm not sure this is right? I'm lost for part (ii), but assume I need to manipulate the definition of IVT to fit?

Thanks for your help, it is much appreciated.

user127700
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  • http://www.proofwiki.org/wiki/Continuous_Function_on_Closed_Interval_is_Bijective_iff_Strictly_Monotone – Gabriel Romon Feb 21 '14 at 17:05
  • Did you mean to have $f(a) < f(b)$ instead? – copper.hat Feb 21 '14 at 17:08
  • Yes, thank you. I have edited it. – user127700 Feb 21 '14 at 17:09
  • What might help for (ii). Essentially, you wish to show that $f$ is an increasing continuous function. Is it possible to apply $(i)$ to show $(ii)$? Note that between every real number there is another real number. – Quickbeam2k1 Feb 21 '14 at 17:12
  • For (ii): By (i) a< $x_1$ <b implies f(a) < f($x_1$) < f(b). Similarly a< $x_2$ <b implies f(a) < f($x_2$) < f(b). Left to prove $x_1$ < $x_2$ implies f($x_1$) < f($x_2$). Let c be such that $x_1$ < c < $x_2$. c > $x_1$ so f(c) > f($x_1$) as strictly increasing. c < $x_2$ so f(c) < f($x_2$) as strictly increasing. So f( $x_1$) < f($x_2$). Does this work? Thanks! – user127700 Feb 21 '14 at 17:18

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Let us proof the following proposition and check that your assertion follows.

Proposition: Let $a<b\in \mathbb{R}$ and $f: [a,b]\rightarrow \mathbb{R}$ where $f$ is a continuous injection and $f(a)<f(b)$. We claim that $f$ is strictly monotonic increasing.

Proof: We shall show that for all $c\in (a,b)$ then $f(a)<f(c)<f(b)$. We argue by contradiction, suppose that there is some $c\in (a,b)$ such that either $f(c)<f(a)$ or $f(b)<f(c)$. Each case is similar so only we will show the contradiction in one. For the former, $f(c)<f(a)<f(b)$ since $f$ is continuous on the subinterval $[c,b]$ (why?) by the IVT there is some $x\in (c,b)$ such that $f(x)=f(a)$, which contradict that $f$ is $1$-$1$. To conclude suppose for sake of contradiction that $f$ is not strictly increasing on $[a,b]$, then exists $x_1<x_2\in [a,b]$ such that $f(x_1)\ge f(x_2)$. Then we have$f(a)<f(x_2)\le f(x_1)$. Contradiction.

Jose Antonio
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