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Elementary Real and Complex Analysis

  1. Jun 2, 2006 #1
    Hi.
    So I was reading through "Elementary Real and Complex Analysis" by Georgi E. Shilov (reading the first chapter on Real Numbers and all that "simple" stuff like the field axioms, a bit of set stuff, etc.).
    Anyways, so while I was reading, I ran into something I couldn't understand... the least upper bound axiom..
    It says "A set E (is a subset of) R is said to be bounded from above if there exists an element z (is an element of) R such that x (is less than or equal to) z for every x (is an element of) E, a fact expressed concisely by writing E (is less than or equal to) z. Every number z with the above property relative to a set E is called an upper bound of E. An upper bound zo of the set E is called the least upper bound of E if every other bound z of E is greater than or equal to zo."
    OK, so let me get this straight... does this imply anything along the lines of zo being the (numerically) greatest element of E as a subset of R such that every other element R has that E doesn't have is greater than E? (I'm guessing R means all reals here)

    So, I continued reading and ran into a similar problem with the greatest lower bound:
    "Suppose the set E is bounded from below. Then a lower bound zo of E is called the greatest lower bound of E if every other lower bound z of E is less than or equal to zo."
    So could this be interpreted as E having an element zo being the (numerically) smallest element such that elements z of R are all smaller than this zo?

    Also, what does it mean when a system of nested half open intervals have an empty intersection? Does it mean they don't share any of the same elements...?

    Thanks for any replies, I appreciate any pointers.
     
  2. jcsd
  3. Jun 2, 2006 #2

    TD

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    Your conclusions are a bit confusing to me, but perhaps I can clarify it a bit with some examples.
    We're working in the reals, R and I'll denote E as a subset again.

    Consider E = {1,2,3,4,5}.
    An element u of R is called an upperbound of E if u ≥ e, where e is an element of E, for all e.
    It is clear that u = 6 is such an upper bound, but u = 100, u = 5.001 and u = 5 are upper bounds as well. It's easy to see that 5 is the smallest. In fact, if a set E has a maximum, denoted max(E), then this is always the least upper bound, denoted lub(E) or sup(E) where sup stands for 'supremum' (same as least upper bound).

    Now consider E = {1-1/n | n element of N\{0}}. Then E = {0, 1/2, 2/3, 3/4, 4/5, ...}.
    Easy to see again, u = 2 is an upper bound. Actually, we always start with 1 and then substract a positive number, so our result can never be larger than 1, so also u = 1 is an upper bound. But does E have a maximum? No: we can get arbitrarily close to 1, but 1 isn't an element of E. With some intuition, you'll see that there is no smaller upper bound than 1, since we can get as close to 1 as we want (taking n large enough). Hence here, the least upper bound of E is 1. Conclusion: the least upper bound of E doesn't have to be an element of E.

    I already told you that if a set has a maximum, then max(E) = sup(E). Now it shouldn't be too hard to understand the analogues 'greatest lower bound', glb or also called infimum, inf(E). The same goes here: if a set has a minimum, then min(E) = inf(E). An example where this isn't the case:

    Consider E = {1/n | n element of N\{0}}, so E = {1, 1/2, 1/3, 1/4, ...}
    Here you can see that we'll get arbitrarily close to 0, but 0 isn't an element of E. Lower bounds are -100, -3, ... but the greatest lower bound is 0.

    Now, the important theorem that you'll book will probably cover is this:
    "Every non-empty set (of real numbers) which is (upper)/(lower) bounded has a (least upper)/(greatest lower) bound"

    This can be proven with using nested intervals.

    Hopefully this helps :smile:
     
    Last edited: Jun 2, 2006
  4. Jun 2, 2006 #3

    matt grime

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    ok so far.

    no, there is no reason to suppose that the least upper bound of E is an element of E. For instanve the least upper bound of the open interval (0,1) is 1.


    that doesn't make grammatical sense.

    again, ok

    no, again the greatest lower bound is not necessarily in E, and it is impossible for all elements of R to be less than any fixed number.


    it means that there is no element that is in all of the sets. That is what empty intersection means in general and has nothing to do with the nature of the sets being intersected.
     
  5. Jun 2, 2006 #4
    DEFINITELY.
    THANKS.

    Is there any chance you could refer me to an example?
    Thanks.
     
  6. Jun 2, 2006 #5

    matt grime

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    X=(0,1) Y=(1,2) XnY is empty.
     
  7. Jun 2, 2006 #6

    matt grime

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    If you wanted an intersection of nested half open intervals then let S_n=(0,1/n], and the intersection over all n in N is empty.
     
  8. Jun 2, 2006 #7
    Ah... after further inspection, now I understand. :)
    Thanks alot for the help guys.
    There's one other thing I don't get..
    Can someone explain how this notation works:
    E = sup{a:[a,b] (is an element of) Q}
    (or with inf instead of sup and b: instead of a:)
    Thanks.
     
    Last edited: Jun 2, 2006
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