If f: R-->R is continuous and open, show that f is strictly monotone.?

Question

A function f: M-->N is open if for any open subset A of M, f(A) is an open subset of N.

Thanks to both of you for writing the proofs. Both of them are so good that is is hard for me to decide on the best answer ;p

Answer 1

First we show that f is injective: suppose ∃x, y∈R such that x≠y but f(x)=f(y). WLOG assume that x f(x), then it assumes this value at some specific point t∈(x, y), so {f(t): t∈(x, y)} also has a maximum, and thus cannot be an open set, contradicting the fact that f is open. Similarly, if min {f(t): t∈[x, y]} < f(x), then {f(t): t∈(x, y)} has a minimum, and thus cannot be an open set. Finally, if max {f(t): t∈[x, y]} = min {f(t): t∈[x, y]} = f(x) = f(y), then the image of [x, y] is just the singleton {f(x)}, and so the image of (x, y) is {f(x)} (since (x, y) is nonempty, as x≠y), which is not open, again contradicting the openness of f. Therefore, it cannot be the case that f(x)=f(y) unless x=y, which means f is injective.

Now to prove f is strictly monotone: suppose that f is not strictly monotone. Then ∃x, y, z∈R such that xf(z) or f(x)>f(y) and f(y)f(z) (the other case is similar). Now either f(x)>f(z) or f(x)f(z), then f(y)>f(x)>f(z), so by the intermediate value theorem ∃t∈(y, z) such that f(t)=f(x). But because xf(x), then we can find t∈(x, y) such that f(t)=f(z) by the intermediate value theorem, also contradicting the injectivity of f. Using the same method of proof in the case f(x)>f(y) and f(y)

Answer 2

This proof uses the intermediate value theorem and some topological facts about continuous functions. You could also use the extreme value theorem, which is what the poster above used. I'm fairly sure that works too, but I had forgotten about that when writing this proof.

Let f: R-->R be continuous and open.

Assume f is NOT strictly monotone.

This means that there exists some triple (a, b, c) such that

a < b < c

and one of the following holds

(i) f(a) ≤ f(b) and f(c) ≤ f(b)

(ii) f(b) ≤ f(a) and f(b) ≤ f(c)

We'll consider the implications of each case toward the end of the proof.

Since f is continuous and a function from the real numbers to the real numbers, by the Intermediate Value Theorem we know that the image of a connected set is connected, so since [a, c] is connected, f([a, c]) is also connected.

We also know thta since [a, c] is a compact set, f([a ,c]) must be a compact set (once again from the continuity of real numbers).

A connected compact set (in the real numbers) is always a closed interval, so f([a, c]) = [x_1, x_2], for some x_1, x_2 in R.

Now, let's recall the possible cases given above.

Assume (i) is true:

f(a) ≤ f(b) and f(c) ≤ f(b) so either f(b) = x_2 or we have f(a) < x_2 and f(c) < x_2.

Either way, this means that f((a, c)) = (x_1, x_2] or [x_1, x_2]. In either case we have contradicted the given that f is an open function, as neither of those two intervals are open, even though (a, c) is open.

But what if instead, (ii) was true?

f(b) ≤ f(a) and f(b) ≤ f(c) so either f(b) = x_1 or we have f(a) > x_1 and f(c) > x_1.

Either way, this means that f((a, c)) = [x_1, x_2) or [x_1, x_2]. In either case we have contradicted the given that f is an open function, as neither of those two intervals are open, even though (a, c) is open.

And we're done, as we've considered both of these cases!

Therefore, (if f is a continuous real to real function) we've shown that if f is not monotone then f cannot be open, so if f is open then f is monotone.

QED

If you have any questions, please feel free to ask in the form of an edit to your question (or by sending me a message). I need to brush up on my analysis/topology anyway.