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Infinite intersection

  1. Apr 5, 2014 #1

    Atomised

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    1. The problem statement, all variables and given/known data

    Let [itex]\Lambda[/itex] = N and set A[itex]_{j}[/itex] = [j, [itex]\infty[/itex]) for j[itex]\in[/itex] N Then

    j=1 to [itex]\infty[/itex] [itex]\bigcap[/itex] A[itex]_{j}[/itex] = empty set

    Explanation: x[itex]\in[/itex] j=1 to [itex]\infty[/itex] [itex]\bigcap[/itex] provided that x belongs to every A[itex]_{j}[/itex].

    This means that x satisfies j <= x <= j+1, [itex]\forall[/itex] j[itex]\in[/itex]N. But clearly this fails whenever j is a natural number strictly greater than x. In other words there are no real numbers which satisfy this criterion.



    2. Relevant equations

    I understand the importance of demonstrating that x belongs to Aj for all j

    3. The attempt at a solution

    Why not just choose x = j+1, thus it will belong to Aj

    1. The problem statement, all variables and given/known data

    I know this contradicts the x <= j+1 condition but I do not understand this condition, why can't x exceed j+1?

    Apologies if formatting unclear.
     
    Last edited: Apr 5, 2014
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  3. Apr 5, 2014 #2

    Zondrina

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    I think I can make out what you wrote there. Write a few terms of your intersection out...

    ##A_1 \cap A_2 \cap A_3 \cap ... ##
    ##= [1,∞) \cap [2,∞) \cap [3,∞) \cap ...##

    Notice as your intersection proceeds, the ##(j+1)^{th}## interval does not contain the previous ##j##. So when you intersect ##[1,∞) \cap [2,∞)##, the ##1## will not be included in the intersection and then the ##2## and so on...

    So eventually you should wind up with nothing at all.

    EDIT: Choosing ##x > j## ensures ##x \in A_j##.
     
    Last edited: Apr 5, 2014
  4. Apr 5, 2014 #3

    Atomised

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    But if you choose x to be j+1 it will always belong to Aj. ??
     
  5. Apr 5, 2014 #4

    statdad

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    You need to choose a fixed x, not one that changes with j. Your comment simply means that you can find, for any set, a number in THAT set - that doesn't imply that it is in all sets.

    Even more specific. If you choose x to be 51 then yes, it is in A50 and A51, but it won't be in A52.
     
  6. Apr 5, 2014 #5

    HallsofIvy

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    Yes, that x will belong to Aj but it won't belong to A(j+1) or any A with a larger subscript. The point is that j can take on any integer value- there is no x that is in all Aj.
     
  7. Apr 6, 2014 #6

    Atomised

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    Here is my naïve response:

    Is it possible that statdad's answer (which I'm sure is technically correct) simply exposes a flaw in this approach?

    It seems to me that the delta epsilon approach to limits is of the nature of a convergent sequence, but this is of the nature of a divergent sequence and not susceptible to this approach. Of course j will eventually exceed any given x, but it is not as if any Aj will ever run out of elements.

    Another garbled idea: R & N have different orders of infinitude, but all Aj are countably infinite.

    Is the given answer not just a convention rather than a truth?

    -----------------------------------------------------------------------------------------

    If I do accept the conventional argument would it not be more correct to say

    j <= x < j + k, k some arbitrary k in Z?

    I mean the j+1 is arbitrary is it not?

    This example was drawn from an excellent set of notes for an introductory course in real analysis.
     
    Last edited: Apr 6, 2014
  8. Apr 6, 2014 #7

    statdad

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    Again, the idea is: in order for the intersection of these (or any collection) of sets to be non-empty, there has to be at least one fixed number in each one. By setting x = j but then continuing to discuss sets Aj you are not giving a fixed value: every time you select a new set you select a new x.

    To attempt to save that line: suppose you fix a value j, and let x = j. Then this x is not in any Ak for k > j, so this x is not in the intersection.
     
  9. Apr 6, 2014 #8

    Atomised

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    Statdad: thanks your answer helps me to understand what is required. I still have a doubt and I will revisit the topic once I have more fully studied sequences.




    I have proven that the problem I am currently working on is non-understandable at n, for all n.
     
  10. Apr 7, 2014 #9

    Atomised

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    Can somebody tell me which law of mathematics is contradicted by defining x = j+1 and rather than assuming that one must always choose x and test it first, assume instead that the mathematician must choose a (high) value for j, in order to 'escape' from x. Of course with this set up x will always belong to Aj. Is it not just a game where whoever goes first loses?
     
  11. Apr 7, 2014 #10

    PeroK

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    [tex]Let \ A_1 = [1, 1.9] \ and \ A_2 = [2, 2.9][/tex]
    [tex]And \ consider \ A_1 \cap A_2[/tex]
    [tex]Let \ x = j + 0.5[/tex]
    [tex] If \ j = 1, \ then \ x = 1.5 \in A_1 \ and \ if \ j = 2, \ then \ x = 2.5 \in A_2.[/tex]
    [tex]Hence \ x \in A_1 \cap A_2[/tex]

    Hence the intersection is not empty.
     
  12. Apr 7, 2014 #11

    Atomised

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    Thanks Perok, point taken. How about this scheme:

    Choose [itex]x_0[/itex] and when [itex]j<=x_0<=j+1[/itex]
    Choose [itex]x_1[/itex] and when [itex]j<=x_1<=j+1[/itex]
    Choose [itex]x_2[/itex] and when [itex]j<=x_2<=j+1[/itex]
    etc ad infinitum

    Then [itex]x_i[/itex][itex]\in[/itex][itex]A_j[/itex],[itex]\forall[/itex]j

    Is this not legitimate mathematically?
     
  13. Apr 7, 2014 #12

    PeroK

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    Mathematically, the intersection is non-empty if you can find an x that is in every set. You can define x however you like, but it must be a single x. You can't choose a different x for different sets. All you're showing then is that every A_j is non-empty.

    This is an important point of logic and the definition of intersection.

    You mentioned a game where who goes first loses. This is quite a good analogy.

    Intersection game: you specify x and I see if I can find a set that does not contain x. If the intersection is non-empty, you win; if it's empty I win.

    Every set is non-empty game: you specify a set and I see whether it contains an element. If every set is non-empty I win; if at least one of the sets is empty you win.

    If you're playing the intersection game, you have to stick by the rules of that game!
     
  14. Apr 7, 2014 #13

    Atomised

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    Although perhaps only relevant for this particular example with the interval [j,[itex]\infty[/itex]) my point is precisely that [itex]x_i[/itex] belongs to all [itex]A_j[/itex] if j<i, and that i can be set arbitrarily large - so that it is not just a case of [itex]A_j[/itex] being non-empty, but [itex]A_j[/itex] containing the same element.

    How about an Aussie Rules Intersection game:

    You specify subset [itex]A_j[/itex] and I give you an element that is in this and all prior [itex]A_j[/itex]?

    Is that illogical?
     
  15. Apr 7, 2014 #14

    PeroK

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    That proves that every finite sub-family of sets A_j has a non-empty intersection.
     
  16. Apr 7, 2014 #15
    It's not illogical, but it loses track of the original proposition, which was that the intersection of ALL [itex]A_j[/itex] is empty. If the intersection of all [itex]A_j[/itex] were not empty, it would be possible to specify at least one real number that belongs to all of them.

    BTW, I agree with your reluctance (in the original post) to accept the condition x <= j+1 as a condition of x's membership in [itex]A_j[/itex]. The only condition of x's membership in a particular [itex]A_j[/itex] is j <= x.
     
  17. Apr 7, 2014 #16

    Atomised

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    Perok, az_lender,

    Is there a mathematical double standard here, in the following theorem

    a = b iff | a - b | < ε, [itex]\forall[/itex] ε > 0

    it is accepted that the difference between a and b being arbitrarily small stands for equality, why then is [itex]x[/itex]'s ability to be arbitrarily large insufficient to secure membership in any A_j?
     
  18. Apr 7, 2014 #17

    SammyS

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    "Is there a mathematical double standard here" ?

    No.

    According to the Law of Trichotomy, exactly one of the following must be true.
    (i) a-b > 0
    (ii) a-b = 0
    (iii) a-b < 0​
    If either (i) or (iii) is true, then you can find ε > 0 such that | a-b | > ε .

    Otherwise, it must be true that a-b = 0, in which case a = b .
     
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