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If f is continuous on [a,b], then f is bounded on [a,b].

  1. Jan 21, 2008 #1
    Dear friends,

    I just joined the forums, and I'm looking forward to being a part of this online community. This semester, I signed up for Analysis II. I'm a math major, so I should be able to understand pretty much everything you say (hopefully). However, I'd really appreciate it if you try not to be too arcane in your explanations.

    So, here's the theorem I'm stuck on.

    Question: Prove...

    Suppose f is continuous on [a,b] and S is the set such that x is in S if and only if x is in [a,b] and, (1) x = a or (2) f is bounded on the subinterval [a,x]. Then S is [a,b].

    This is what I have so far.

    Attempted solution:

    Let S = {x in [a,b] | x = a or f is bounded on [a,x]}
    Thus, by definition, S is a subinterval of [a,b].
    Because of this fact, S is bounded.
    Because S is bounded, S has a least upper bound, call it p.
    It follows that p > a must be true.
    Because b is an upper bound of S, p <= b.

    Now, we employ an indirect argument.
    Thus, assume p < b.
    By definition, b is the LUB of set [a,b].

    Other questions, and relevant background:

    From here on out, I'm confused about where to go and what to do. Does anyone have any pointers or suggestions? What approach do I need to take to complete this proof? The only thing I can think of is that it might have to do with f being continuous on an interval implying that f is bounded on that interval. However, I'm not sure how to prove this implication either.

    Thank you in advance for your help.
  2. jcsd
  3. Jan 21, 2008 #2
    It seems like you need to show there's an element of the set other than a. I mean, it's pretty obvious, but it's kind of needed, I think.

    Of course, I would personally just say that a continuous function maps compact sets to compact sets, then apply Heine-Borel. But that's a copout.
  4. Jan 21, 2008 #3
    Hmmm...true, but remember. Ultimately, I'm trying to show that S is [a,b]. The continuous function bit is a part of the problem I think, though it might not be.
  5. Jan 21, 2008 #4
    Well, if it's discontinuous it could be that it's not bounded on any such interval [a,x] (take f(x) = a for x = a and f(x) = 1/(x - a), for x > a.
  6. Jan 21, 2008 #5
    It seems to me you can do this much more efficiently if you merely use the fact that f is continuous on [a,b]. I know you ask this later in your OP, but you haven't clearly used it anywhere in your outline. Remember that f continuous on [a,b] means f achieves a max/min on [a,b].
  7. Jan 21, 2008 #6
    Yeah, it seems like you're going to have to use some form of the compact sets being mapped to compact sets thing. If you want to show there's an x =/= a in the set, I think you're gonna have to invoke the intermediate value theorem, which is a result of continuous functions taking on a maximum and minimum value on a closed interval, which is proved with the stated theorem and Heine-Borel.
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