Bounded Function on Closed Interval: Proving Boundedness

[Icebreaker]

If f is defined on [a,b] and for every x in [a,b] there is a d_x such that if is bounded on [x-d_x, x+d_x]. Prove that f is bounded on [a,b].

This question seems very odd. If every point, and indeed the neighbourhood of every point is bounded, then of course the function itself must be bounded. Of course I doubt this can be passed as a proof, so any suggestions would be helpful.

 

[EnumaElish]

The statement "[x-d_x, x+d_x] is bounded" is a tautology. Is it meant to say f is bounded on [x-d_x, x+d_x]?

 

[Icebreaker]

Yes, I believe so. I've made the modification but on the question sheet it was written exactly as I had copied it. My professor has a habit of handing out handwritten assignments.

 

[EnumaElish]

Suppose that f was not bounded on [a,b]. Then there must exist a point c in [a,b] such that f(c) > M for all real M.

Does that sound like the right beginning?

Proceed to show a contradiction with boundedness on [x-d_x,x+d_x] for all x in [a,b].

 

[Muzza]

If every point, and indeed the neighbourhood of every point is bounded, then of course the function itself must be bounded.

Consider 1/x on (0, 1) then.

If we assume that all d_x are strictly greater than 0, then I believe you can use the compactness of [a, b] to solve this problem.

 

[Icebreaker]

Consider 1/x on (0, 1) then.

The interval must be closed.

 

[Muzza]

The interval must be closed.

You missed my point. I was objecting to your implication that the statement was rather trivial (or obvious) just because the function was bounded on every neighbourhood in its domain.

 

[Icebreaker]

True. However, I have found a proof using Bolzano-Weierstrass theorem. Thanks for the inputs.

 

[EnumaElish]

True. However, I have found a proof using Bolzano-Weierstrass theorem. Thanks for the inputs.

When you have a chance can you post a sketch of that proof?

 

[Icebreaker]

This will be sketchy, but the proof itself is not mine:

Assume the contrary, that f is unbounded. By Weierstrass, we can find a sequence x_n that converges to some number c in I = [a,b] and f(x_n)>n for every n. For this c, let d_c be as in the hypothesis. Then, for n large enough, x_n is in (c-d_c, c+d_c). This contradicts our hypothesis that f is bounded on (c-d_c, c+d_c).