Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.