# Continuity and strictly increasing functions

1. Apr 2, 2012

### kingstrick

1. The problem statement, all variables and given/known data

Let f:[0,1] →ℝ be a continuous function that does not take on any of its values twice and with f(0) < f(1), show that f is strictly increasing on [0,1].

2. Relevant equations

3. The attempt at a solution

Assume that f is not strictly increasing on [0,1]. Therefore there exists a,b,c between 0 and 1. where 0 < a < b < c < 1. such that f(0) < f(a) < f(b) and f(c) < f(b) < f(1). By the intermediate value theorem there exist a x between (a,b) and a y between (b,c) for there exists a k where f(a) < k < f(b) and f(c) < k < f(b) and f(x) = k and f(y) = k. Contradiction Therefore f is strictly increasing on [0,1].

Does this proof work or am I missing something again?

2. Apr 2, 2012

### micromass

Staff Emeritus
I don't think this is necessarily true. How do you conclude the existence of those a,b and c??

The idea of your proof is good though.

3. Apr 2, 2012

### kingstrick

If [0,1] is an interval then there exists a value between 0 and 1 that I will call a, and therefore there exists a value between a and 1 since it is a interval which I call b and then there is a value between b and 1 called c as [b,1] is also an interval.

4. Apr 2, 2012

### Joffan

micromass' question seemed to be asking how you can select a, b and c to give the relationships you state for f(a), f(b) and f(c). Certainly there are non-monotonic functions that do not allow such a selection: $x^2-0.5x$, for example.

5. Apr 2, 2012

### kingstrick

Am I not allowed to assume and/or define their relationship? I mean for it to be non-mono there must be a point in the middle (not literally) of the interval where f(x) is either the min/max or at least curve up and then back down (or down then back up). So why can't I pick that number to be the value of f(b)?

Last edited: Apr 2, 2012
6. Apr 2, 2012

### Joffan

Well, you have to define how you pick the points properly for a proof. And my example function $x^2-0.5x$ does not allow you to pick points in the way you defined them. Neither does $1.5x-x^2$. So you have to deal with those possibilities.

I think you are on the right track, but you're not quite nailling down the proof yet.

7. Apr 2, 2012

### jgens

You are allowed to define their relationship so long as their relationship holds for all non-monotonic functions in this case. Consider the example that Joffan posted. Making the selection you mentioned is not possible.

8. Apr 2, 2012

### Office_Shredder

Staff Emeritus
Essentially b is a local maximum in your proof. If a function goes down and then up, it has no local maximum, but the function isn't strictly increasing, So you can't assume there is a local maximum.

Instead of trying to pick three points, just pick two points that break the definition of increasing

9. Apr 2, 2012

### Joffan

Actually, scrub that. By all means look at those functions but only to see that there are alternative shapes you might need to think about in [0,1].

The basic problem as that you are picking three points, and you really only need to pick two, based on some interval where the function is not increasing.

[[ edit: as Office Shredder said ]]