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

  • #1
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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.
 
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Answers and Replies

  • #2
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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)##.
 
  • #3
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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."
 
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  • #4
mathwonk
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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.
 
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  • #5
Ray Vickson
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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.
 
  • #6
WWGD
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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.
 

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