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I'm aware that strict monotonicity implies invertibility, but I've never seen the opposite direction being stated formally. It seems to me invertibility also implies strict monotonicity if the function is continuous. Both Claude and ChatGPT kept telling me that this is not true until I pointed out some error in their counterexamples, so I'm half believing half doubting myself. Here's my reasoning:

If $f$ is a continuous, invertible function defined on $I$, then for any $x_1,x_2\in I$ such that $x_1 < x_2$, we have $f(x_1) \neq f(x_2)$. Suppose $f$ is not strictly monotone, then there exists $a, b, c \in I$ such that $a<c<b$ but $f(b)$ is not between $f(a)$ and $f(c)$. We may assume that $f(a) < f(c)$ but $f(c) \geq f(b)$. If $f(c) = f(b)$, we have a contradiction as $f$ is invertible. If $f(c) >f(b)$, by the Intermediate Value Theorem, there exists $s\in(a,c)$ and $t\in (c,b)$ such that $f(s)=f(t)$, a similar contradiction. Thus, $f$ is strictly monotone.

Are this reasoning and the proposed statement correct?

Thank you!

ten_to_tenth
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    Are you assuming something about the domain of $f$? – José Carlos Santos Aug 17 '24 at 18:01
  • Remember the IVT usually only applies on an interval. – Bcpicao Aug 17 '24 at 18:20
  • @JoséCarlosSantos, if we suppose that the domain $I$ of the function is connected, then the OP’s claim is correct. On the other hand, if the domain is not connected, the claim is false. – Angelo Aug 17 '24 at 18:23
  • @Angelo Sure. But I want to know what the OP thinks about it? – José Carlos Santos Aug 17 '24 at 18:24
  • @JoséCarlosSantos, in my opinion he is considering a function $f$ defined on an interval. For that reason he calls $I$ the domain of $f$, indeed $I$ stands for interval. – Angelo Aug 17 '24 at 18:29
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    By the way, AI bots are not reliable sources of information for mathematics beyond basic arithmetic. I would suggest searching the internet, finding a book, or speaking to a teacher instead. – Joe Aug 17 '24 at 18:43
  • @JoséCarlosSantos Hi, thank you for your answer! I'm a little confused about your question and Angelo's comment. Since $f$ is continuous, doesn't it imply that the domain is an interval? – ten_to_tenth Aug 18 '24 at 05:04
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    Are you claiming that the function$$\begin{array}{ccc}(-1,0)\cup(2,4)&\longrightarrow&\Bbb R\x&\mapsto&x\end{array}$$is discontinuous? If so, at which points it is discontinuous? – José Carlos Santos Aug 18 '24 at 08:39
  • @JoséCarlosSantos Hmm, that's certainly new to me. For some reason, I always thought a continuous function should look like a curve without any gaps or holes and one-sided limits at endpoints should only apply to the rightmost and leftmost points in the domain. I suppose in this case, we could say the function is continuous on each piece of its domain? – ten_to_tenth Aug 18 '24 at 08:46
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    I don't know what a piece of the domain of a function is. Anyway, we simply say that the function from my previous example is continuous, and nothing else. The idea that the graph of a continuous function should look like a curve without any gaps or holes applies to functions defined on intervals, but not in general. – José Carlos Santos Aug 18 '24 at 09:09
  • @JoséCarlosSantos Thank you very much! – ten_to_tenth Aug 18 '24 at 10:44

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