F is strictly increasing at each point in (a,b)

[Mr Davis 97]

Homework Statement

Let $(a,b)\subset \mathbb{R}$ be a bounded open interval. Prove that if $f:(a,b)\to\ \mathbb{R}$ is strictly increasing at each point in $(a,b)$, then $f$ is strictly increasing on $(a,b)$.

Homework Equations

Let $(a,b)\subset \mathbb{R}$ be a bounded open interval in $\mathbb{R}$. We say that $f:(a,b)\to \mathbb{R}$ is strictly increasing at $c\in(a,b)$ if there is a $\delta>0$ such that both the following hold:
1) for each $x\in(c-\delta,c)$ we have $f(x)<f(c)$
2) for each $x\in(c,c+\delta)$ we have $f(c)<f(x)$.

We say that $f:(a,b)\to \mathbb{R}$ is strictly increasing on $(a,b)$ if whenever $x,y\in(a,b)$ with $x<y$, we find that $f(x)<f(y)$.

The Attempt at a Solution

I am not sure where to start here. This is a challenge problem in the section on the least upper bound, so I suppose it relates somehow, but I am not seeing the connection. Hints would be nice.

EDIT: I think I solved it.

 

[Eclair_de_XII]

1) for each $x\in(c-\delta,c)$ we have $f(x)<f(c)$
2) for each $x\in(c,c+\delta)$ we have $f(c)<f(x)$

The way I did it, I combined these two statements, changed the second $x$ to a $y$ and chose a $\delta>0$ such that $(c-\delta,c+\delta)\subset (a,b)$.

 

[Mr Davis 97]

The way I did it, I combined these two statements, changed the second $x$ to a $y$ and chose a $\delta>0$ such that $(c-\delta,c+\delta)\subset (a,b)$.

I don't think you're allowed to choose $\delta$. It says only that "there exists."

 

[mathwonk]

did they prove the heine borel property in the section on lub's? (i.e. every open covering of a closed bounded interval by open intyervals has a finite sub covering.) given points c< d in the interval, the fact that f(c) < f(d) seems to follow from the existence of a finite cover of the closed bounded interval [c,d] by open intervals of the type mentioned.

or did they discuss greatest lower bounds? given c in (a,b) you want to show there are no points x >c where f(x) ≤ f(c). Assuming there are some, the set of such x is a bounded non empty set and has a greatest lower bound d. That means every x with f(x) ≤ f(c) has x ≥ d, and also that there are such x arbitrarily close to d and greater than d.

case 1) f(d) ≤ f(c). rule this out.

case 2) f(d) > f(c). Rule this out.

 

[Ray Vickson]

Homework Statement

Let $(a,b)\subset \mathbb{R}$ be a bounded open interval. Prove that if $f:(a,b)\to\ \mathbb{R}$ is strictly increasing at each point in $(a,b)$, then $f$ is strictly increasing on $(a,b)$.

Homework Equations

Let $(a,b)\subset \mathbb{R}$ be a bounded open interval in $\mathbb{R}$. We say that $f:(a,b)\to \mathbb{R}$ is strictly increasing at $c\in(a,b)$ if there is a $\delta>0$ such that both the following hold:
1) for each $x\in(c-\delta,c)$ we have $f(x)<f(c)$
2) for each $x\in(c,c+\delta)$ we have $f(c)<f(x)$.

We say that $f:(a,b)\to \mathbb{R}$ is strictly increasing on $(a,b)$ if whenever $x,y\in(a,b)$ with $x<y$, we find that $f(x)<f(y)$.

The Attempt at a Solution

I am not sure where to start here. This is a challenge problem in the section on the least upper bound, so I suppose it relates somehow, but I am not seeing the connection. Hints would be nice.

EDIT: I think I solved it.

What is the definition of the concept "is strictly increasing at $c \in (a,b)$"? Using some definitions the problem is trivial, but using others it presents a bit of a challenge.

 

[WWGD]

I think this may work: choose x>y. Then there is a chain of $\delta$'s from y to x, but maybe you can join them with a line that includes the endpoints and then use compactness to use only finitely-many $\delta$s.