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Show that if $f: \mathbb{R} \to \mathbb{R}$ is a continuous injective map, then it is strictly monotonic.


Could someone give me a proof for this? I have the intuition for why it's true - I'm just having trouble expressing that intuition in a rigorous manner. Basically consider two points $x_1, x_2 \in \mathbb{R}$. By the problem statement, $f$ is continuous on $[x_1, \, x_2]$. WLOG, assume that $f$ is strictly increasing. It there exists a point where it is not increasing, then $f$ hits a value twice, and it's not injective.

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  • $\begingroup$ The intermediate value theorem should come in handy. Assume $f$ is not strictly monotonic. Then there are $x_1 < x_2 < x_3$ with $f(x_1) \geq f(x_2)$ and $f(x_2) \leq f(x_3)$. If one if the values are equal, you're done - $f$ isn't monotonic. So assume $f(x_1) > f(x_2)$ and $f(x_2) < f(x_3)$, and also $f(x_1) \neq f(x_3)$ (again, if that doesn't hold, you're done). Now use the intermediate value theorem on either $x_1,x_2$ or $x_2,x_3$, depending on whether $f(x_1) < f(x_3)$ or not. $\endgroup$ – fgp Apr 13 '14 at 18:05
  • $\begingroup$ "WLOG, assume that f is strictly increasing" is confusing the hypothesis and the conclusion. $\endgroup$ – Did Apr 16 '14 at 6:10
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Let $M:=\{(x,y)|x<y\}$. Define $g\colon M\to \mathbb R$ by $g(x,y):=f(x)-f(y)$. Clearly $g$ is continuos and $M$ is connected. Hence $g(M)$ is connected, i.e., an interval of $\mathbb R$ which does not contain $0$ as $f$ is injective. That is, $g$ is either strictly negative or strictly positive, hence $f$ is strictly increasing resp. decreasing.

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Let $x,y,z$ be real numbers with $x<y<z$. If $f(y)=\max\{f(x),f(y),f(z)\}$, then by injectivity $f(y)>f(x)$ and $f(y)>f(z)$. Let $\eta=\frac{f(y)+\max\{f(x),f(z)\}}2$. Then $f(x)<\eta<f(y)$, hence by the IVT there exists $\xi_1\in(x,y)$ with $f(\xi_1)=\eta$. Also, $f(y)>\eta>f(z)$, hence by the IVT, there exists $\xi_2\in(y,z)$ with $f(\xi_2)=\eta$. Clearly,m $\xi_1\ne \xi_2$, hence we have a contradiction to injectivity of $f$. We conclude that $f(y)\ne\max\{f(x),f(y),f(z)\}$. Similarly, $f(y)\ne\min\{f(x),f(y),f(z)\}$. Thus we either have $f(x)<f(y)<f(z)$ or $f(x)>f(y)>f(z)$.

Then for any finite sequence $x_1<x_2<\ldots <x_n$ we have either $f(x_1)<f(x_2)<\ldots < f(x_n)$ or $f(x_1)>f(x_2)>\ldots > f(x_n)$. Now for any two numbers $x_1<x_2$, sort $\{0,1,x_1,x_2\}$ into ascending order and thus conclude that $f(x_1)<f(x_2)\iff f(0)<f(1)$.

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WLOG we have $f(0)<f(1)$. We can show by induction that $f(n)<f(n+1)$ for all $n\in\mathbb N$. If $f(n)<f(n+1)$ but $f(n+1)\ge f(n+2)$ then either $f(n+2)\ge f(n)$, and by the IVT we have some $c\in [n,n+1)$ such that $f(c)=f(n+2)$, or $f(n+2)<f(n)$ and by the IVT we have some $d\in (n+1,n+2)$ such that $f(d)=f(n)$. These contradict injectivity, so $f(n+1)>f(n+2)$ and by induction $f(n+1)>f(n)$ for all $n\in\mathbb N$. The same argument shows that $f(n+1)>f(n)$ for negative $n$, so we have $f(n+1)>f(n)$ for all $n\in\mathbb Z$.

Now let $x<y\in\mathbb R$. If $n=\lfloor x\rfloor=\lfloor y\rfloor$ and $f(x)\ge f(y)$, then either $f(y)\ge f(n)$, and by the IVT we have some $c\in [n,x)$ such that $f(c)=f(y)$, or $f(y)<f(n)$. In this case, either $f(y)\ge f(n-1)$, and by the IVT we have some $d\in [n-1,n)$ such that $f(d)=f(y)$, or $f(y)<f(n-1)$ so by the IVT we have some $e\in (n,y)$ such that $f(e)=f(n-1)$. All $3$ possibilities give us a contradiction, thus $f(x)<f(y)$ so $f$ is strictly increasing on the intervals $[n,n+1]$. If $\lfloor x\rfloor<\lfloor y\rfloor$ then $f(x)<f(\lfloor y\rfloor)\le f(y)$. Thus $f$ is strictly increasing.

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