Bounded function on an interval

In summary, we are trying to prove that a function f defined on the interval I=[a,b] is bounded, given that for every point x in I, f is bounded in a neighborhood of x. This can be done by using the concept of compactness and open covers. By picking an open neighborhood of x and using the fact that f is bounded on that neighborhood, we can create an open cover for I. Since I is compact, this cover has a finite subcover, which means there are only a finite number of these open neighborhoods. Using these neighborhoods, we can define an upper bound for f and show that it is bounded on I. This can also be proven without using compactness by considering the sequence of points (d
  • #1
kbfrob

Homework Statement


Let I=[a,b] and let f:I->R be a function (not necessarily continuous) with the property that for every x in I, f is bounded in a neighborhood of x.
Prove that f is bounded on I

The Attempt at a Solution


I have no idea
 
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  • #2
Hi kbfrob,

If you're familiar with the notion of compactness in terms of open covers, let x be a point in I and pick an open neighbourhood of x; then f is bounded on that open neighbourhood by your hypothesis. The collection of such open neighbourhoods (each corresponding to a point in I) provide an open cover for I, which is compact, so this collection has a finite open subcover of I.

I'll let you try and use these ideas to define an upper bound for f. Show us your work if you get stuck.
 
  • #3
that is exactly what i tried to do, but I'm not familiar with the idea of compactness. Essentially what i was thinking about doing was starting at a, which is bounded in a neighborhood of (d1). then go to a+(d1), which is bounded in a neighborhood of d2. then go to a+(d1)+(d2)...etc. what i can't figure out is how to show that there will be a finite number of di's that cover [a,b]. Is there a way to show this without using compactness?
 
  • #4
kbfrob said:
that is exactly what i tried to do, but I'm not familiar with the idea of compactness. Essentially what i was thinking about doing was starting at a, which is bounded in a neighborhood of (d1). then go to a+(d1), which is bounded in a neighborhood of d2. then go to a+(d1)+(d2)...etc. what i can't figure out is how to show that there will be a finite number of di's that cover [a,b]. Is there a way to show this without using compactness?

If you infinitely many d_i's, they would form a bounded sequences (bounded because they are in [a,b]) and by Bolzano-Weierstrass such a sequence must have a accumulation point d. (Think about what happens if it has more than one, or maybe infinitely many accumulation points!). You then know that f is bounded in a neighbourhood of d and this neighbourhood covers all but finitely many of your d_i's (if d is the only accumulation point).
 

FAQ: Bounded function on an interval

What is a bounded function on an interval?

A bounded function on an interval is a mathematical function that has a finite range of values within a given interval or set of numbers. This means that the function's output or values are limited and do not tend towards infinity.

How do you determine if a function is bounded on an interval?

To determine if a function is bounded on an interval, you need to evaluate the function at the endpoints of the interval. If the function's output at the endpoints is finite, then the function is bounded on that interval. If the endpoints have infinite output, the function is not bounded on that interval.

What is the difference between a bounded function and an unbounded function?

A bounded function has a finite range of values within a given interval, while an unbounded function does not have a finite range and its values can tend towards infinity. In other words, an unbounded function has no limit on its output, while a bounded function has a limit.

Why is it important to study bounded functions on an interval?

Studying bounded functions on an interval is important because it helps us understand the behavior of functions and their limits. Bounded functions have specific characteristics and properties that can be used to make predictions and solve problems in various fields of mathematics and science.

Can a function be bounded on one interval and unbounded on another?

Yes, a function can be bounded on one interval and unbounded on another. The boundedness of a function depends on the specific interval or set of numbers it is being evaluated on. A function may be bounded on one interval, but not on another, depending on its behavior and values within those intervals.

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