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It is a fact from analysis that a continuous and open real-valued function of a real variable is strictly monotonic. The proof I know runs something like this: Suppose $f$ is an open and continuous map but is not strictly monotonic. Consequently, there exist three numbers $a < c < b$ such that either $$ f(a) \geq f(c) \leq f(b) \;\;\;\; (1) $$ or $$ f(a) \leq f(c) \geq f(b) \;\;\;\; (2) $$ If $(1)$ holds then the exteme value theorem guarantees that $f$ attains its infimum on $[a,b]$; but by assumption, the infimum is at least as small as $f(c)$ so in fact $f$ attains its infimum on $(a,b)$. Also by assumption, $f$ carries open intervals to open intervals. With this though we have a contradiction since an open interval cannot contain it's own infimum. A similar argument yields considering suprema yields an analagous contradiction. Therefore, $f$ is strictly monotonic.

My question is, Is there a more constructive way to prove this that doesn't involve contradiction? Although I think the proof given is nice, I don't think I could have come up with it own my own because the consequences of $f$ not being strictly monotonic as exhibited in (1) and (2) would not have occurred to me. So, it would be good to see a direct way of proving this.

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    Those consequences are practically the *definition* of not strictly monotonic.2012-02-29
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    @Alex Well, nonetheless, when I first encountered the fact I had to work up a proof (1) and (2) before I was convinced. So, on first encounter, the fact was not "obvious" to me and so it wouldn't have occurred to me to apply it2012-02-29
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    If $f$ is not strictly increasing, we have some $a such that $f(a)\geq f(c)$. If $f$ is not strictly decreasing, we have some $c such that $f(c)\leq f(d)$. Combine the two. I doubt you will find a more intuitive proof than that.2012-02-29
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    Or simply: if it’s not monotone, it has a bend in it. That can be either a $\vee$ bend or a $\wedge$ bend. Those are the only possibilities. I’m with Alex on this one: proof don’t come much more straightforward than this one. Rather than look for a simpler one $-$ which probably doesn’t exist $-$ you should spend some time trying to figure out why this one wouldn’t have occurred to you.2012-02-29
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    @BrianMScott, AlexBecker : Of course Alex's observation is obvious, but I found it not obvious either (nice username) how to assemble the non-monotonicity into the forms $(1)$ or $(2)$. [It's not simple as merely 'combine the two'](https://wj32.org/wp/2013/01/15/every-continuous-open-mapping-of-r-into-r-is-monotonic/) (references [here](http://math.stackexchange.com/questions/203061/existence-of-vee-or-wedge-for-non-monotonic-functions)), or at least it takes additional steps I regard as non-trivial, and one needs to be more careful.2016-09-02
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    I experienced something similar before with the 'no continuous real function is a 2-to-1 mapping'; everyone around me would say such is 'intuitively clear' and leave me behind in the dust, and while I thought it was *plausible*, it took me lots of beating it to death (in the form of casework) to have confidence that it would work, something they had either swept under the rug or truly did find intuitive. (In which case, I wonder what their brains are made of, and whether it is me who is defective.) I'm still wondering whether there's a cleaner/slicker demonstration of this one too.2016-09-02

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