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What does a delta symbol mean in mathematics?

  1. Dec 21, 2011 #1
    In Spivak's calculus book he provides a proof for:

    Theorem: If f is continuous on [a, b] and f(a) < 0 < f(b), then there is some number x in [a, b] such that f(x) = 0.

    In the proof he explicitly says, "...A has a least upper bound [itex]\alpha[/itex] and that a < [itex]\alpha[/itex] < b. We now wish to show that f([itex]\alpha[/itex]) = 0, by eliminating the possibilities f([itex]\alpha[/itex]) < 0 and f([itex]\alpha[/itex]) > 0."

    This doesn't seem to apply as intuitively to me. How can a least upper bound be in the middle of a set? If we describe x to be between values a and b, then shouldn't the least upper bound and greatest lower bound be b and a, respectively?

    Also, he continues to explain the first case, where f([itex]\alpha[/itex]) < 0. He says, "There is a [itex]\delta[/itex] > 0 such that f(x) < 0 for [itex]\alpha[/itex] - [itex]\delta[/itex] < x0 < [itex]\alpha[/itex]."

    I'm confused with the notation here. I'm unsure what [itex]\delta[/itex] represents, so if someone can offer an alternative explanation of this segment that would be greatly appreciated.

    Thank you!

    Also, the theorem is intuitive to me. I understand that if f is continuous and a < 0 and b > 0 then there has to be a point where it crosses the x-axis, but of course the proof is of more value here than the intuitive sense of the problem.
  2. jcsd
  3. Dec 21, 2011 #2
    δ is just probably a small non zero number.
  4. Dec 21, 2011 #3


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    in this context, [itex]\delta[/itex] just means a small positive number. [itex]\delta[/itex] is used to connect with the "epsilon-delta" notation used in the definition of continuity (or more generally any limit).

    the whole point of the theorem, is the essential nature of the hypothesis that f is continuous. it's not enough to have an intuitive idea of what continuous is, because the notion of continuity DEPENDS on the notion of a limit, and our intuition can lead us astray when it comes to limits (limits are a subtler concept than just numbers). if we are to be sure that we can assert the truth of this for ANY continuous function (including some rather bizarre functions that are extremely "erratic"), we need to involve the definition of continuity in an essential way.

    about the set A, which is the set of all x in [a,b] for which f < 0 on [a,x], it's clear that b cannot be in A, since f(b) > 0. so although b is an upper bound for A (since it's an upper bound for [a,b]), it's not the least upper bound of A (this is a consequence of the continuity of f...since f(b) > 0, we can let ε = f(b) > 0. then we know THERE IS a δ > 0, such that |x-b| < δ implies |f(x) - f(b)| < f(b). now we're only interested in the part of (b-δ,b+δ) that lies in [a,b], which is (b-δ,b]. for all the x in that interval, f(x) is positive. so the least upper bound of A, is at most b-δ, which is certainly LESS than b).

    similarly, since f(a) < 0, for some (possibly different) δ' > 0, f(x) for all x in [a,a+δ'), is negative (because...f is continuous). this eliminates the possiblility that the least upper bound of A is merely a.

    so now we know that sup(A) is somewhere in (a+δ',b-δ). we've not only eliminated a and b, but some small regions near the endpoints. that is, A contains the interval [a,a+δ'], and is likewise contained IN [a,b-δ]. in effect, we're putting the "squeeze" on sup(A).

    now if α = sup(A), and f(α) < 0, then the same argument we used above (the continuity of f), shows that in some (perhaps very small) interval surrounding α (and perhaps we should use δ", here), say (α-δ",α+δ"), f(x) is always negative (we can use ε" = -f(α)).

    but that would mean that [a,x1] for some x1 in (α,α+δ") is contained in A, in which case α isn't even an upper bound for A. (we know that [a,α) is in A because of our definition of A and α, and since we are assuming f(α) < 0, in fact [a,α] is contained in A).
    Last edited: Dec 21, 2011
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