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I got this problem:

Prove that if $f:[a,b]\to\Bbb{R}$ is one to one and has the intermediate value property, then $f$ is strictly monotone.

That is, we must show that $\forall x,y\in[a,b], x<y \to f(x)<f(y)$ or $\forall x,y\in[a,b], x<y \to f(x)>f(y)$.

Start of my proof:
Suppose not.
That is suppose that $\exists x_1,y_1\in[a,b], x_1<y_1 \,\,and\,\, f(x_1)\geq f(y_1)$ and suppose that $\exists x_2,y_2\in[a,b], x_2<y_2 \,\,and\,\,\, f(x_2)\leq f(y_2)$.

Now since $f$ is one to one we get that $\exists x_1,y_1\in[a,b], x_1<y_1 \,\,\,and f(x_1)> f(y_1)$ and that $\exists x_2,y_2\in[a,b], x_2<y_2 \,\,and \,\,\,f(x_2)< f(y_2)$.

Now I am not sure how to split the proof into cases?

Thanks on any hints.

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    $\begingroup$ Note to posters: if $f$ is not strictly monotone, it means that there exist $x_1,x_2,x_3,x_4$ such that (i) $x_1 < x_2$ and $f(x_1) \le f(x_2)$; and (ii) $x_3 < x_4$ and $f(x_3) \ge f(x_4)$. It does not mean that there exist $x_1 < x_2 < x_3$ such that either (i) $f(x_1) \le f(x_2)$ and $f(x_2) \ge f(x_3)$ or (ii) $f(x_1) \ge f(x_2)$ and $f(x_2) \le f(x_3)$. This is something that has to proven! (EDITED TO ADD: Ewan Delanoy has now proven it.) $\endgroup$ – TonyK Sep 28 '14 at 14:34
  • $\begingroup$ @TonyK +1 well said. $\endgroup$ – MathNerd Sep 28 '14 at 14:40
  • $\begingroup$ possible duplicate of Injective functions with intermediate value property are continuous. Better proof? $\endgroup$ – Yiorgos S. Smyrlis Sep 29 '14 at 10:19
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Hint : if $f$ is not monotone, then there is a "saddle" point, i.e. $x_1<x_2<x_3$ such that $f(x_1)<f(x_2)>f(x_3)$ or $f(x_1)>f(x_2)<f(x_3)$.

Indeed, suppose for example that $f(a)<f(b)$ (replace $f$ with $-f$ otherwise). If $f$ is not increasing on $(a,b)$, there are $u<v$ in $[a,b]$ with $f(u)>f(v)$.

If $f(a)<f(u)$, we can take $x_1=a,x_2=u,x_3=v$.

If $f(b)>f(u)$, we can take $x_1=u,x_2=v,x_3=b$.

If none of those two inequalties hold, we have $f(a)>f(u)>f(v)>f(b)$ contradicting the initial hypothesis.

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  • $\begingroup$ This is true, but it's messy to prove $-$ there are so many cases. $\endgroup$ – TonyK Sep 28 '14 at 14:26
  • $\begingroup$ @TonyK I think it’s important to separate the combinatiorial, order-only aspect from the continuity aspect in the proof. $\endgroup$ – Ewan Delanoy Sep 28 '14 at 14:33
  • $\begingroup$ @TonyK I don’t think my proof is that "messy". $\endgroup$ – Ewan Delanoy Sep 28 '14 at 14:33
  • $\begingroup$ I agree $-$ you have done well to keep it so simple. +1. $\endgroup$ – TonyK Sep 28 '14 at 14:40
  • $\begingroup$ @EwanDelanoy +1 that's exactly what I needed. Thanks. $\endgroup$ – MathNerd Sep 28 '14 at 15:02
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There is no reason to ever consider $f(x_1)=f(x_2)$ for different $x$'s since the function is one-to-one. So only strict inequalities are needed!

If $f$ is not strictly monotone we may find $x_1<x_2<x_3$ (WLOG since otherwise we could just consider $g=-f$) such that $$ f(x_2)>f(x_1),f(x_3) $$ But then by the intermediate value property we must have $c_1\in[x_1,x_2]$ and $c_2\in[x_2,x_3]$ with $f(c_1)=f(c_2)$ which contradicts $f$ being one-to-one.

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  • $\begingroup$ Your second paragraph requires proof, I would say. $\endgroup$ – TonyK Sep 28 '14 at 14:28
  • $\begingroup$ how do you know that if $f$ is not monotone then there exists $x_1<x_2<x_3$ such that $f(x_2)>f(x_1),f(x_1)$ or such that $f(x_2)<f(x_1),f(x_1)$? $\endgroup$ – MathNerd Sep 28 '14 at 14:39
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I'll show that if $f$ is one to one and not monotone in $[a,b]$ then there exist $x_1,x_2,x_3\in[a,b]$ such that $x_1<x_2<x_3$ and $f(x_2)<f(x_1),f(x_3)$ or $f(x_1),f(x_3)<f(x_2)$:

(Note: I tried to write the proof similiar to nested if statements in computer programming for better flow)


Since $f$ is one to one we get that $f(a)\neq f(b)$ and hence there are two cases $f(a)<f(b)$ or $f(a)>f(b)$

if $f(a)<f(b)$ then:

Since $f$ is not monotone, we get that in particualr $f$ is not decreasing and so there exists $u,v\in[a,b]$ such that $f(u)>f(v)$ (because $f$ is one to one, there cannot be equalty).

Now there two cases $u=a$ or $u\neq a$:

if $u=a$ then:

Since $u<v$ we get that $a<v$ and that $f(a)>f(v)$

Now we'll prove that $v\neq b$: If $v = b$ we get that $f(a)>f(b)$ which contradicts the fact that $f(a)<f(b)$, And so $v\neq b$ and hence we get that $v\in(a,b)$. Now since $f(a)>f(v)$ and since $f(b)>f(a)$ we get that $f(v)<f(a),f(b)$.

Now set $x_1 = a, x_2 = v, x_3=b$ and we get that $x_1<x_2<x_3$ and that $f(x_2)<f(x_1),f(x_3)$.

if $u\neq a$ then:

Now we'll show that $u\neq b$ because if $u = b$ we get that $b=u<v$ and so $v\notin [a,b]$ which is a contradiction and so $u\in(a,b)$.

Now there are two cases: $v=b$ or $v\neq b$

if $v=b$ then:

We get that $f(u)>f(b)$ and since $f(b)>f(a)$ we get that $f(a),f(b)<f(u)$ and since $u\in(a,b)$ we get that $a<u<b$.

Now set $x_1 = a, x_2 = u, x_3=b$ and we get that $x_1<x_2<x_3$ and that $f(x_1),f(x_3)<f(x_2)$.

if $v\neq b$ then:

We'll show that $v\in(a,b)$:
Since $v\in[a,b]$ we get that $v\leq b$ but since $v\neq b$ we got that $v<b$. Now since $a<u<v$ we get that that $a<u$ and so $v\in(a,b)$.
and we got $a<u<v<b$.

Now there are two cases: $f(a)<f(u)$ or $f(b)>f(v)$

Because if $f(a)>f(u)$ and $f(b)<f(v)$ and since $f(u)>f(v)$ we get that $f(a)>f(u)>f(v)>f(b)$ and so $f(a)>f(b)$ which is a contradiction.

if $f(a)<f(u)$ then:
Take $x_1=a,x_2=u,x_3=v$ and we get that $x_1<x_2<x_3$ and since $f(u)>f(a),f(v)$ we get $f(x_1),f(x_3)<f(x_2)$.

if $f(b)>f(v)$ then:
Take $x_1 = u, x_2=v,x_3=b$ and we get that $x_1<x_2<x_3$ and since $f(b),f(u)>f(v)$ we get $f(x_2)<f(x_1),f(x_3)$.


Similarly we prove for the case $f(a)>f(b)$.

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Edit: Suppose your function is not monotone, then there are four points such that $y_1<y_2$ and $y_3<y_4$ and $f(y_1)<f(y_2)$ and $f(y_4)<f(y_3)$. So we can take three of these four points and organize then in a increasing order $x_1<x_2<x_3$ such that $f(x_1)< f(x_3)<f(x_2)$, $f(x_2)< f(x_1)<f(x_3)$, $f(x_2)< f(x_3)<f(x_1)$, or $f(x_3)< f(x_1)<f(x_2)$, since $f$ is one to one, the values of $f(x_i)\,\,\,i=1,2,3$ can not be the same, suppose the first case, the others are similar. Since your function has the property of the intermidiate value, we have that exists $x_4\in (x_1,x_2)$ such that $f(x_4)=f(x_3)$ what is a contradiction, since the function is one to one. The same argument is valid for the others cases.

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  • $\begingroup$ Your first sentence requires proof, I would say. $\endgroup$ – TonyK Sep 28 '14 at 14:30
  • $\begingroup$ why? the function is one to one, as I said, so it can not have the same values for different points, and suppose it is not monotone give me the for possibilities above. why I should prove that, if it is clear by the hypothesis of the quention. $\endgroup$ – math_man Sep 28 '14 at 14:34
  • $\begingroup$ See my comment to the main question. $\endgroup$ – TonyK Sep 28 '14 at 14:35
  • $\begingroup$ You're right, thanks for comment. $\endgroup$ – math_man Sep 28 '14 at 14:49

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