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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. –  Alex Becker Feb 29 '12 at 3:21
@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 it –  ItsNotObvious Feb 29 '12 at 3:29
If $f$ is not strictly increasing, we have some $a<b$ such that $f(a)\geq f(c)$. If $f$ is not strictly decreasing, we have some $c<d$ such that $f(c)\leq f(d)$. Combine the two. I doubt you will find a more intuitive proof than that. –  Alex Becker Feb 29 '12 at 3:32
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. –  Brian M. Scott Feb 29 '12 at 5:58

1 Answer 1

f is open map and continuous, let f:R->R Let the inverse of f is F.so F is continuous from f(R) to R. Now lets assume there exists distinct x,y in R such that f(x)=f(y). WLG lets assume f(x)=f(y)=c for some c in R. F(x) is not continuous at c as x,y are distinct and hence the contradiction.

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You don't explain why $f$ can be inverse of an other function.<br> You don't use that the function is open. –  Hoseyn Heydari Oct 9 '14 at 10:29
yeah its open that's why inverse of f will be continuous and I write inverse of f as F –  user182061 Oct 9 '14 at 10:31

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