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.
Best Answer
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.