Is c a bounded neighborhood of f?

[kingstrick]

Homework Statement

Let I:[a,b] and let f: I→ℝ be a function with the property where for all x in I, the function f is bounded on a neighborhood V_δx(x) of x. Prove that f is bounded on I.




2. The attempt at a solution

Let I:[a,b] and let f: I→ℝ be a function with the property where for all x in I, the function f is bounded on a neighborhood V_δx(x) of x. The interval I consists of real numbers a and b where a < b. According to the density theorem, there exists a c in I where a<c<b wher c is a cluster point of I. Also, following the density theorem there exists a x in I where a < x < c or c < x < b such that |a-x| = |b-x| = |c-x|. Now given an ε > 0, there exists a δ > 0 where |x - c | < δ ...

Is this right up to this point?
If it is, how do I proceed from here?

 

[Dick]

I really have know idea where you are going with that. Shouldn't you use that the interval [a,b] is compact?

 

[kingstrick]

I really have know idea where you are going with that. Shouldn't you use that the interval [a,b] is compact?

I don't really know where i am going with it either. I do not recall what compact means, is that the same thing as a closed bounded interval.

My text said to solve it like this limit theorem, If A is contained in R and f : A→ R has a limit at c in ℝ the f is bounded on some neighborhood of c.

I thought to have a neighborhood of c, i needed to have a cluster point, so i tried proving that [a,b] had a cluster point and then work towards bounding. I guess i am approaching it wrong.

 

[Dick]

I don't really know where i am going with it either. I do not recall what compact means, is that the same thing as a closed bounded interval.

My text said to solve it like this limit theorem, If A is contained in R and f : A→ R has a limit at c in ℝ the f is bounded on some neighborhood of c.

I thought to have a neighborhood of c, i needed to have a cluster point, so i tried proving that [a,b] had a cluster point and then work towards bounding. I guess i am approaching it wrong.

Yes, in your problem compact means closed and bounded. The property I'm looking for is that if [a,b] is closed and bounded and O is a set of open sets that cover [a,b], then there is a finite subset of O that also covers [a,b]. Do you know that?

 

[kingstrick]

Yes, in your problem compact means closed and bounded. The property I'm looking for is that if [a,b] is closed and bounded and O is a set of open sets that cover [a,b], then there is a finite subset of O that also covers [a,b]. Do you know that?

I did but I don't see how to use that here.

 

[Dick]

I did but I don't see how to use that here.

Ok, then every point in [a,b] has a neighborhood in which it is bounded. Call those neighborhoods your collection O. The theorem tells you that [a,b] is covered by a finite number of neighborhoods on which is it bounded. Can you see how to combine the bounds on each of the finite number of neighborhoods to get a bound on all of [a,b]?

 

[kingstrick]

Or do you know that if c_n is a sequence in [a,b], then c_n has a limit point?

That one i know.. let me think about that for a second.

 

[Dick]

That one i know.. let me think about that for a second.

You can use either that or the open set formulation. They are really the same thing. To do it that way, assume f is unbounded. Then pick c_n such that |f(c_n)|>n for all n. What happens at a limit point of that sequence?

 

[kingstrick]

You can use either that or the open set formulation. They are really the same thing. To do it that way, assume f is unbounded. Then pick c_n such that |f(c_n)|>n for all n. What happens at a limit point of that sequence?

how can i assume it is unbounded wshen i am told that it is bounded?

 

[kingstrick]

You can use either that or the open set formulation. They are really the same thing. To do it that way, assume f is unbounded. Then pick c_n such that |f(c_n)|>n for all n. What happens at a limit point of that sequence?

Does this work? Have i established the conclusion by the end?

Let I := [a,b] and let f: I→ℝ be a function with the property where for all x in I the function f is bounded on a neighborhood V_δx(x) of x. Since I is a closed set, I is also bounded. Therefore any sequence (Xn) in I must also be bounded. Thus by the Bolzano-Weirstrauss theorem, there exist a subsequence (Xn_k) of (xn) that converges to a number x. Since I is closed and the elements of the subsequence (Xn_k) belong to I it follows that x is in I. Then f(x) is continuous at x. so that (f(Xn_k) ) converges to f(x). Then (f(Xn_k) ) must be bounded. Since (Xn) are arbitrary sequences, f(x) is bounded on I.

 

[SteveL27]

how can i assume it is unbounded wshen i am told that it is bounded?

We already know that for each x there's a little nbd of x where f is bounded. Taken all together, one nbd for each x, that's a lot of neighborhoods. But what if we knew something about the domain that let us dramatically reduce the number of neighborhoods we need to deal with?

If you review the definition of compactness, this will be very helpful.

 

[kingstrick]

We already know that for each x there's a little nbd of x where f is bounded. Taken all together, one nbd for each x, that's a lot of neighborhoods. But what if we knew something about the domain that let us dramatically reduce the number of neighborhoods we need to deal with?

If you review the definition of compactness, this will be very helpful.

We haven't learned compactness yet. Our text has that two chapters down and we are not permitted to use any information in the text on HW if we haven't covered that section yet

 

[bwood01]

http://www.infoocean.info/avatar1.jpg I really have know idea where you are going with that.

 

[kingstrick]

http://www.infoocean.info/avatar1.jpg I really have know idea where you are going with that.

I don't know how to solve this then. everybody keeps saying use compactness but i can't...

 

[Dick]

how can i assume it is unbounded wshen i am told that it is bounded?

The idea here is to use a proof by contradiction. You assume f is unbounded and try to show that leads to a contradiction with the assumption that it has a bounded neighborhood of every point. That would prove it must be bounded. If you choose c_n such that |f(c_n)|>n and c_n has a limit point c, can c have a neighborhood where it is bounded?

 

[kingstrick]

The idea here is to use a proof by contradiction. You assume f is unbounded and try to show that leads to a contradiction with the assumption that it has a bounded neighborhood of every point. That would prove it must be bounded. If you choose c_n such that |f(c_n)|>n and c_n has a limit point c, can c have a neighborhood where it is bounded?

ohhhh.

 

[Dick]

I don't know how to solve this then. everybody keeps saying use compactness but i can't...

And using that every sequence has a limit point is, in fact, using 'compactness'. You might not learn the abstract definition for a few chapters, but you are using it. That's why everyone is saying that.

 

[kingstrick]

is it because c can't be in the set I if it is a cluster point for F

 

[Dick]

is it because c can't be in the set I if it is a cluster point for F

What are F and c and what do you mean by that? Can you explain in full?

 

[kingstrick]

The idea here is to use a proof by contradiction. You assume f is unbounded and try to show that leads to a contradiction with the assumption that it has a bounded neighborhood of every point. That would prove it must be bounded. If you choose c_n such that |f(c_n)|>n and c_n has a limit point c, can c have a neighborhood where it is bounded?

I meant to reference your quote. So are we saying that c is a cluster point of f(x) and so can't be bounded as that would imply values greater than c and thus beyond the bound?

 

[Dick]

I meant to reference your quote. So are we saying that c is a cluster point of f(x) and so can't be bounded as that would imply values greater than c and thus beyond the bound?

You probably have the right idea, but the details you give are all wrong. c was defined as a cluster point of the c_n. It's between a and b. It's not unbounded. What IS unbounded?

 

[kingstrick]

the f(Xn) is unbounded

 

[Dick]

the f(Xn) is unbounded

The f(c_n) is unbounded. What does that tell you about the possibility of c having a bounded neighborhood?

 

[kingstrick]

Since f(c_n) is divergent, then c does not have a bounded neighborhood, as c_n goes to c, f(c_n) won't go to L?

 

[Dick]

Since f(c_n) is divergent, then c does not have a bounded neighborhood, as c_n goes to c, f(c_n) won't go to L?

There is no limit L in the problem. You just know c is supposed to have a bounded neighborhood. You don't know that it has a limit. What you do know is that any neighborhood of c contains an infinite number of c_n because it's a limit point.