Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Misunderstanding part of Spivak's proof of second hard theorem?

  1. Nov 2, 2013 #1
    Hello,

    First off, the theorem is "if f is continuous on [a,b] then f is bounded above on [a,b]. What is the name of this theorem? He doesn't say but I'm sure it has a popular name.

    Anyway, it's on page 115. Let me type what he wrote, then I'll say which part I don't understand.

    Proof: Let A= {x: a≤ x ≤ b and f is bounded above on [a,x]}

    Before I continue (because this is one of two things I don't understand here), why do we assume f is bounded above on [a,x]? What if f is not bounded above on [a,x]? Also, how do we KNOW f is bounded above on [a,x]? We are trying to prove a function is bounded above by assuming it is bounded above? Doesn't make sense to me, I must be missing something.

    Anyway, continuing on:

    Clearly, A≠ null set, because a is in A, and A is bounded above (by b), so A has a least upper bound, Ω (in the book it is alpha but it looks too similiar to the letter a so I'm gonna roll with Ω)

    Our first step is to prove that we actually have Ω=b. Suppose instead Ω < b. By theorem 1 (which I already proved on the previous page) there is δ>0 such that f is bounded on (Ω-σ, Ω+σ). Since Ω is sup A there is some Xo in A satisfying Ω-σ<Xo<Ω. This means x is bounded on [a,Xo]. But if X1 is any number with Ω<X1<Ω+δ, then f is also bounded on [X0,X1]. Therefore x is bounded on [a,X1] so X1 is in A, contradicting the fact that Ω is an upper bound of A.

    Ok, so I understand why what he just showed contradicts that Ω is an upper bound of A. I just don't understand what any of that has to do with Ω<b. What about when Ω=b, how does all of that fail to show a contradiction?

    Thanks in advance
     
    Last edited: Nov 2, 2013
  2. jcsd
  3. Nov 2, 2013 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    From the Weirstrauss definition of continuity we know that for every ##\epsilon>0##, there is a ##\delta >0## such that

    $$ | x -a | < \delta \longrightarrow | f(x) - f(a) | < \epsilon.$$

    Therefore there is always an ##x = a + \delta## such that ##f## is bounded, even if ##[a,x]## is just a small neighborhood of ##a##.

    The way I would understand this part is the following. We have shown that ##f## is bounded on ##[a,X_0]## and that this further implies that it is bounded on ##[a,X_1]##, ##X_1>X_0##. We can keep iterating this for ##X_2, \ldots ## until we get arbitrarily close to ##[a,b]##.

    Part 1 showed from continuity that ##f## is bounded in a small region ##[a,a+\delta]##. Part 2 shows that continuity further lets us show that the function is also bounded on ##[a+\delta, a+ 2\delta]##, etc, allowing us to piece these regions together into ##[a,b]##.
     
  4. Nov 3, 2013 #3

    pasmith

    User Avatar
    Homework Helper

    We're not. We're saying that if f is bounded above on [a,x] then x is in A (and also that if x is in A then f is bounded above on [a,x]). It might be that f is such that A is empty, but as we will see the condition that f be continuous excludes that possibility.

    The argument which works for [itex]\Omega < b[/itex] fails for [itex]\Omega = b[/itex] because if [itex]b < x_1 < b + \delta[/itex] then trivially [itex]x_1 > b[/itex], and so [itex]x_1 \notin A[/itex] (recall that A was defined to be a subset of [a,b]).
     
  5. Nov 3, 2013 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    We don't need continuity to exclude that possibility, for clearly ##a \in A## for any ##f## that is defined at ##a##.
     
  6. Nov 8, 2013 #5
    I understand that he's saying that IF f is bounded above...but how would we know that f is bounded above? Isn't the whole point of this theorem to prove that a function is bounded above? How would we know it is bounded above? Why not just assume the whole thing is bounded above and not even bother proving it? (I'm not trying to be irritating I just legimately don't understand how we are allowed to make that assumption)
     
  7. Nov 9, 2013 #6

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    Here is a topological way to think about it.

    The interval, [a,b] is compact and in general,the continuous image of any compact set is compact. The compact subsets of the real numbers are closed and bounded.
     
  8. Nov 10, 2013 #7
    I have no idea what any of that means unforunately.

    I still don't understand how we can assume f is bounded on [a,x] to assume it is bounded on [a,b] without doing something like this:

    Theorem: If a function is continuous on a closed interval, it is bounded on that interval

    Proof: Assume said function is bounded on said interval. QED
     
  9. Nov 10, 2013 #8

    jgens

    User Avatar
    Gold Member

    We only assume that f is bounded on [a,x] if x is an element of A. But this is exactly what it means for the element x to be in A! Your confusion makes no sense.
     
  10. Nov 10, 2013 #9

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    They do not assume that f is bounded on [a,x]. Rather they consider the set of all x for which it is true. A priori this set may be empty. It is their job to show that b in in the set. That is how the proof goes.
     
  11. Nov 10, 2013 #10

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I mention it because much of the gnarly detail of the proof boils down to these simple topological ideas.

    But here is the heart of it without using topological words - though in my opinion they are worth learning.

    On a closed bounded interval an infinite sequence of points must have a subsequence that converges inside the interval. One can not pack an infinite number of points in a closed interval without having some point where they bunch up to a limit point.

    If f were unbounded then there would be a sequence on which the values of f increase(or decrease) without bound. A limit point of such a sequence would be a point where the function is infinite. This is impossible.
     
    Last edited: Nov 10, 2013
  12. Nov 10, 2013 #11
    So why not assume f is bounded on [a,b] and just include b as an element of A?

    Obviously something is not clicking for me...I know it is my fault though
     
  13. Nov 10, 2013 #12
    I know....like I mentioned in the first place it is very easy to understand intuitively...the problem for me is in the proof
     
  14. Nov 10, 2013 #13

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    read my second to last post above.
     
  15. Nov 11, 2013 #14

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    The argument about an unbounded sub-sequence applies here to show the existence of the δ>0. One would get a sequence of points converging to Ω on which the values of f increase without bound.
     
  16. Nov 12, 2013 #15
    I think this may just be a misunderstanding of notation. Spivak never assumes that f is bounded. CuriousBanker, can you explain using plain English what you read in your mind when you see the notation "Let A= {x: a≤ x ≤ b and f is bounded above on [a,x]}" ? Only that line, not anything surrounding it.
     
  17. Dec 21, 2013 #16
    I know I am 6 weeks late, but I just was insanely busy with work over last 6 weeks and literally did not have time to check this...sorry for not responding.

    What I read in my mind: A is a set of all x with value between a and b. f(x) is bounded on that entire interval.

    Doesn't he say f is bounded? "f is bounded above on [a,x]"

    How would you possibly know that f is bounded above on [a,x]? How could you prove it?
     
    Last edited: Dec 21, 2013
  18. Dec 21, 2013 #17
    "Our first step is to prove that we actually have Ω=b. Suppose instead Ω < b. By theorem 1 (which I already proved on the previous page) there is δ>0 such that f is bounded on (Ω-σ, Ω+σ). Since Ω is sup A there is some Xo in A satisfying Ω-σ<Xo<Ω. This means x is bounded on [a,Xo]. But if X1 is any number with Ω<X1<Ω+δ, then f is also bounded on [X0,X1]. Therefore x is bounded on [a,X1] so X1 is in A, contradicting the fact that Ω is an upper bound of A."

    If Ω=b, I don't see how this fails to show contradiction.
    How does Ω=b fail to show x is bounded on [a,x1]
    Also, how does x being bounded on [a,x1] show that x1 is in a? We have already shown that Ω<x1<Ω+σ, meaning x1 is greater than Ω, meaning that although is it an upper bound of A, it is not the least upper bound
     
  19. Dec 24, 2013 #18

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    2250: Elementary proofs of big theorems

    The first theoretical result is the Intermediate Value Theorem (IVT) for continuous functions on an interval.

    Theorem: If f is continuous on then interval I, then the set of values f assumes on I is also an interval. I.e. if a,b are points in I, then any number between f(a) and f(b) is also a value of f, taken at some point between a and b.

    proof: We assume f(a) < 0 and f(b) >0, and try to find c with f(c) = 0. We assume that every infinite decimal represents a real number.

    Lemma: If f is continuous at c and f changes sign on every interval containing c, then f(c) = 0.
    proof: This is the contrapositive of the fact that if f is positive (or negative) at c and continuous at c, then f is positive (or negative) on some interval containing c. This is immediate from the definition of continuity, and is an important exercise in understanding that definition. QED.

    Thus it suffices to find a real number c such that f changes sign on every interval containing c. Assume [a,b] = [0,1].
    Since f(0) < 0 and f(1) > 0, then f changes sign on some interval of form [r/10, (r+1)/10]. let c start out as the decimal .r.
    Then since f(r/10) < 0 and f((r+1)/10) > 0, f changes sign on some interval of form [(10r+s)/100, (10r+s+1)/100]. Then c continues as the decimal .rs.
    Continuing in this way forever, we obtain an infinite decimal, i.e. a real number
    c = .rs....., in the interval [0,1], such that f changes sign on every interval containing c. Hence f(c) = 0. QED Thm.

    We know the continuous image of an open bounded interval may be neither open nor bounded. But the next big theorem says that the continuous image of a closed bounded interval is also closed and bounded. We do it in two steps.

    Theorem: If f is a function which is continuous everywhere on a closed bounded interval [a,b], then f is bounded there.

    proof: We prove it by contradiction, i.e. assuming f is unbounded leads to finding a point where f is not continuous.

    Lemma: A function which is continuous at c, is also bounded on some interval containing c.
    proof: This is immediate from the definition of continuity. E.g. if e = 1, by continuity of f at c, there is an interval I containing c where the values of f lie between f(c)-1 and f(c)+1. Thus f is bounded on I. QED.

    Hence it suffices to show that if f is unbounded on [a,b], then there is a point c of [a,b] such that f is unbounded on every interval containing c.
    Assume [a,b] = [0,1], and that f is unbounded on [0,1]. Then there is some interval of form [r/10, (r+1)/10] where f is unbounded. Start out the decimal c as .r.
    Then there is some interval of form [(10r+s)/100, (10r+s+1)/100] where f is unbounded. Continue the expansion of c as the decimal .rs.
    Continuing forever, we construct an infinite decimal c = .rs......, in the interval [0,1], such that f is unbounded on every interval containing c. Thus f is not continuous at c. QED Thm.

    Theorem: If f is continuous on the closed bounded interval [a,b], then f assumes a maximum value there.
    proof: We know the set of values f takes on [a,b] is a bounded interval. If not closed it has form (c,d) or [c,d) or (c,d]. If of form [c,d) say, then the continuous function 1/(f(x)-d) is unbounded on [a,b], contradiction. QED.
     
  20. Dec 28, 2013 #19
    Ah, I see the exact slip in semantics. There is no period between the two statements: the "and" is a logical "and", not a vernacular "and". This notation being used is a prescription for the construction of a set, and as such, no statement within has a pre-assigned truth value. It may be that every statement is false, in which case we have merely constructed the empty set. In particular, we should read this statement as "A is the set of all x such that x is between a and b (inclusive) and f is bounded above on [a, x]." To be clearer, the statement after the "and" is still under the "such that" prescription. So that we can also read this as "A is the set of all x such that f is bounded above on [a, x] and such that x is between a and b (inclusive).", where I switched the order of the statements merely to emphasize the point that the order is not of any importance in an "and" statement.
    Without consideration of any other axioms or theorems, it may be that we have merely constructed the empty set, which is fine. For example, there may be no x values for which f is bounded above on [a, x], so our set would be empty then. However, we know that a is an element of our set, since a satisfies both requirements: a is between a and b (inclusive), and f(x) is bounded on the "interval" [a, a] by any value greater than f(a), so our set is not empty. Whether it includes more elements than a is then covered by the rest of the argument.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Misunderstanding part of Spivak's proof of second hard theorem?
  1. Part of proof (Replies: 0)

Loading...