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I'm an autodidact, and so, I really doubt sometimes if my solutions, especially in Analysis, are correct or not. Here is the theorem I have to prove:

If $f$ is a one-to-one continuous function on an interval $I$. Then $f$ is strictly increasing or strictly decreasing

Proof: Let $f : [a,b] \to \mathcal{R}$. Consider $x_1, x_2 \in [a,b]$ such that $x_2> x_1$. $f(x_1) = f(x_2) \implies x_1 = x_2$ by the virtue of one-to-one, therefore it is not possible. The only two remaining possibilities are: $f(x_1) > f(x_2)$, or $f(x_2) > f(x_1)$ As $x_1 ~\text{and}~x_2$ were arbitrary, therefore the above working shows, that either $f$ is strictly increasing or strictly decreasing.

Was I rigorous enough in my proof?

P.S.: Let me know how many questions am I allowed per week (or other duration) to post in "solution verification" tag, because these types questions shall not be contributing to the real aim of the site.

Delta Psi
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    It does not prove then result. You just proved that for fixed $x_1$ and $x_2$ either $f(x_1)<f(x_2)$ or $f(x_1)>f(x_2)$. This is the definition of a monotone function. – geetha290krm Nov 03 '22 at 09:56
  • See https://math.stackexchange.com/q/170147/42969 for better proofs – Martin R Nov 03 '22 at 09:57

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No, it isn't.

You must also prove that the direction of the inequality is always the same.

Hint: Use two intervals $[x_1,x_2]$ and $[x_2,x_3]$ and intermediate value theorem for continuous functions.

ajotatxe
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  • If we assume CASE 1: that for $x_1 \lt x_2$ we have $f(x_1) \lt f(x_2)$. Then, by IVT for every $x$ in $x_1 \lt x \lt x_2 $ we have $f(x_1) \lt f(x) \lt f(x_2)$. But does that imply for every $x \lt x_2$ we have $f(x) \lt f(x_2)$? – Delta Psi Nov 03 '22 at 12:25