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Proof checking.

  1. Jul 4, 2006 #1
    i need to prove that:
    1) every infinite set and bounded has at least one accumulation point.
    basically i proved it by using the method of proving the theorem of bolzano for infinite sequences, basically if A is an infinite set and bounded, then for every x in A there's a,b such that a<=x<=b we can divide [a,b] into two parts [a,(a+b)/2] [(a+b)/2,b] at least one of them has an accumulation point, and that way we continue the same way as in bolzano theorem.

    2) if E is a bounded set, then the set of accumulation points, E' is also bounded.
    basically what i did is as follows:
    let E be bounded, then for every[tex]x \in E[/tex] there exist a,b reals such that a<=x<=b and because E' is accumulation points set, for every x' in E' there exists x0 in E such that x'!=x0 and for every e, x0 in (x'-e,x'+e).
    if x'>x0, then x'-e<x0<x'+e x'<e+x0, a<=x0<x'<e+x0.
    if x'<x0, then x0-e<x'<x0<=b.
    either way, x' is bounded.

    are either of my proofs are correct?
    thanks in advance.
     
  2. jcsd
  3. Jul 4, 2006 #2

    matt grime

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    you do not define a or b, and what does the fact that those two closed intervals have accumulation points have to do with the original set?

    I'm not entirely sure I understand the question. The set of points 1/n for n a natural number is infinite and bounded and the set itself contains no accumulation points - 0 is not part of the set. So presumably the accumulation point is not supposed to be part of the set...

    you have your quantifiers in the wrong order.

    You can't do that. Again you have your quantifiers in the wrong order (you do not, and indeed cannot, pick an x0 for all e).
     
  4. Jul 4, 2006 #3
    the definition of accumulation point, is for example x is accumulation point if for a neighbourhood e of x, there's at least one point in a set E which is different, it doesnt say that all of the points should be different than x.

    a and b are real numbers.
    how should i place the quantifiers.

    thanks in advance.
     
  5. Jul 4, 2006 #4
    then it should be, for every e and every x' in E' there exists x0 in E.
     
  6. Jul 4, 2006 #5

    matt grime

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    Why don't you, for the first one, just use Bolzano Weierstrass properly? Pick some sequence from A and you're practically done.
     
  7. Jul 4, 2006 #6

    matt grime

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    so, say A is the interval [0,3], then I'll let a and b be 1 and 2 shall I?


    Now, as to your quantifiers, for *any* set A then for all x in A there are a and b with a<x<b, that is trivial (x-1, x+1 will do). You will have to pick a and b before quantifying x if you go down that path.
     
  8. Jul 5, 2006 #7
    but the theorem i know of bolzano and weierstrass, is that in the end you prove that a subsequence of an infinite and bounded sequnce converges. and here i need to prove that for an infinite and bounded set there exist at least one accumulation point, so i thought to use the same method of proving the former, but with one different thing, that for every domain we choose there will be at least one accumulation point, is this assumption not right?

    no, it should be the maximum and minimum of that set.

    but it's a general proof, so besides saying that a and b are the max and min, anything else you should point that was wrong in my proof.
    let me rephrase my proof:
    let E be bounded, then for any x in E there exist a,b reals (which are the maximum and minimum) such that a<=x<=b and because E' is accumulation points set, for any x' in E' there exists x0 in E such that x'!=x0 and there exists e such that x0 in (x'-e,x'+e).
    if x'>x0, then x'-e<x0<x'+e x'<e+x0, a<=x0<x'<e+x0.
    if x'<x0, then x0-e<x'<x0<=b.
    either way, x' is bounded.
     
  9. Jul 5, 2006 #8

    matt grime

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    You still have the quantifiers wrong, really. E does not have either a max or min so you cannot appeal to that concept.

    E is bounded means that there exist a and be such that all E is contained in [a,b]. That is the order of the quantifiers. And you still have the order of the quantifiers wrong in the epsilon part. As it stands it is trivially true that

    "for any x' in E' there exists x0 in E such that x'!=x0 and there exists e such that x0 in (x'-e,x'+e)"

    and would be true if E' were any set of points and nothing to do with accumulation points because in that order of quntifiers I choose e after I choose x0, so I can choose e to be fantastically large.

    1. Pick a sequence of pairwise distinct elements from A (it is infinite) and use B-W.

    2. Since E is bounded E lies in some interval [a,b], show E' is in that interval (if it didn't some point of E' must lie outsided of [a,b], right....)
     
  10. Jul 5, 2006 #9
    okay, for any x' in E' there exists e such that there exists x0 in E and x'!=x0 x0 is between a and b, and also between x'-e and x'+e, therefore
    a<x'-e<x0<x'<x'+e<b from this we can conclude that:
    a+e<x'<b+e a-e<x'<b-e if we combine these two, we get a<x'<b.

    is this correct?

    btw, you havent answered this question of mine:
    but the theorem i know of bolzano and weierstrass, is that in the end you prove that a subsequence of an infinite and bounded sequnce converges. and here i need to prove that for an infinite and bounded set there exist at least one accumulation point, so i thought to use the same method of proving the former, but with one different thing, that for every domain we choose there will be at least one accumulation point, is this assumption not right?
     
  11. Jul 5, 2006 #10

    matt grime

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    It is still not correct, the inequalities immediately after 'therefore' need not be true at all. In particular E might be just a one point set and equal to E' and you've still not actually carefully stated what a and b are (if they are the sup and inf of E for instance, or just any upper and lower bound.


    I have said that your second mentioned argument is incorrect, or more accurately incomplete. You have not explained what the fact that the domains you choose having acc. points has to do with the set A having an accumulation point, however it is perfectly possible to use the method of repeated bisection if you so wish as long as you state it clearly. However, there is no need to reprove B-W: just use it - pick a sequence of pair-wise distinct elements (it is infinite) and apply B-W, don't go to the length of reproving it.
     
    Last edited: Jul 5, 2006
  12. Jul 5, 2006 #11
    the set A has an accumulation point (at least one) because we have an infinite set, every term in a set is different from one another, this is why we can find an accumulation point for in [a,b] cause there's at least one point in A which is different than another point in A, so for example if x,y are in A and x!=y then a<=x-e<y<x+e<=b.

    about your first assertion, E' is the accumulation points set, so if x' is in E' then there exists e for which x0 is in E and x'!=x by definition x'-e<x0<x'+e, so i dont see how E can be equal to E', if E is a singleton then E' should have at least one point which is different than the term in E.
    the definition of an accumulation point from my textbook is:
    "t is an accumulation point of E if for every neighbourhood e of t there's at least one point in E which is different than t".
    if E={1} then the accumulation point should be different than 1, if we let t=1/2 then for e=1 -1/2<1<3/2, so how can the set of accumulation points of E be equals E, if that were true than, how could the defintion be applied?
    a and b are the upper and lower bounds of the set.
     
  13. Jul 5, 2006 #12

    matt grime

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    er, no. if A is the set {1,2} every element in A is different from every other element in A, and it has exactly two elements.

    accumulation points of [a,b] are related to those of A how? and how are you not assuming A has an accumulation point?


    see previous comment.

    Sorry, should have been if E is a singleton set (or any finite set) then E' is empty.

    This is the whole point. You never acutally use any of the properties or definitions of the things given to you in the first place. You must use the definitions carefully and explain everything. E does not have unique upper and lower bounds, by the way: that is the kind of thing I mean by not using the definitions properly.
     
  14. Jul 5, 2006 #13

    matt grime

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    Can we just forget for one second trying to get your to write proofs out, anbd just focus on the ideas.

    1. If A is infinte then we can pick out a sequence x_i with x_i=/=x_j for i=/=j, which has a convergent subsequence by B-W, and that is an accumulation point of A

    2. If E is bounded E is a subset of [a,b] for some a,b (which are *not* unique as you keep implying they are). If E' is not a subset of [a,b], then there is a x in E' and not in [a,b]. We can assume wlog that x>b, now, use the fact that x is an accumulation point to derive a contradiction by choosing e(psilon) appropriately.
     
  15. Jul 5, 2006 #14
    the definition of bounded set:
    "A set S of real numbers is called bounded above if there is a real number k such that k ≥ s for all s in S. The number k is called an upper bound of S. The terms bounded below and lower bound are similarly defined.

    A set S is bounded if it is bounded both above and below. Therefore, a set of real numbers is bounded if it is contained in a finite interval."
    from wikipedia, isnt this what i did in my proof?

    for the first does =/= mean "not equal"?
    for 2, i dont know how to choose epsilon, perhaps epsilon should be:
    (b-a)/2?
    but cant get a contradiction, are you sure the only way to prove it is with a contradiction method of proof?
     
  16. Jul 5, 2006 #15

    matt grime

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    What you did in your proof was start with an x in then pick an a and b with x in [a,b] and then do some reasoning. The a and b you can pick for some given x need not be such that A is contained in [a,b] as you seem to want to say. Then you let a and b be the max and min of A. A need not have a max or min. Then you let a and b be *the* upper and lower bounds of A. There is no such thing as *the* upper and lower bound of a bounded set. Any bounded set has infinitely many upper and lower bounds.


    Yes =/= means not equal to. != is computer speak, and not mathematics. In mathematics != reads literally as either factorial equal or uniquely equal.

    I didn't say that contradiction was the only way. I am trying to get you to see what is going on. Just draw a picture:

    here is the number line

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

    here the points of E (obvisouly we want infinitely many, but just think of it like this for now)

    ---x-x---x-----xxxx-----xx-x--x---------

    E is bounded so I can put in an upper and lower bound

    -|--x-x---x-----xxxx-----xx-x--x--|-------

    and E is contained inside those limiting |, one at a, one at b.

    Now, if E', the acc. points is not a subset of [a,b] there must be some point outside of [a,b] in E', so let's assume wlog that y is such a point and that it is strictly larger than b


    -|--x-x---x-----xxxx-----xx-x--x--|---y---

    now, if y is an acc point of E, then I must be able to pick a sequence from E tending to y, but how can I do that? Any limit point of any sequence chosen from A must have limit less than b, so E' must be a subset of [a,b] i.e. it is bounded (and this includes the case when E' is empty).

    All we are saying is that if x_n converged to x and x_n is less than b for all n, then so is x. That is all that is going on.
     
  17. Jul 6, 2006 #16
    a<=x0<=b
    x0=(b-a)/2 which is the middle point of the line,
    |x-x0|<b-(b-a)/2=(b+a)/2
    (b-a)/2-(b+a)/2<x<(b+a)/2+(b-a)/2
    so we get that x from E' is also bounded.
    ( or to be preicse, we get a contradiction, that x<b).
    is this right, now?
     
  18. Jul 6, 2006 #17

    matt grime

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    what are you triying to do? why is x0 the midpoint of an arbitrary interval that contains E? what has that got to do with the accumulation point x? Note I am guessing x is an accumulation point.
     
  19. Jul 6, 2006 #18
    my mistake, im totally confused.
    shouldnt i be picking a point x0 between [a,b] such that |x-x0|<e when x is acc point?
    i know that x0-x<x0-b by the assumption x>b, then if x0=(b-a)/2
    then |x0-x|<|x0-b|=|b-x0|=(b+a)/2
    then x<(b-a)/2+(b+a)/2=b

    i think that could easily also pick any other point (b-a)/n in [a,b], is it wrong to do so?
     
  20. Jul 6, 2006 #19

    matt grime

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    Why are you doing this? Just explain the thinking behind your argument. What are you trying to show? Why are you picking any points of E in [a,b]? What is your intention for doing this?

    The proof really is one line (and is not really a proof by contradiction).

    Let x be greater than b, i.e. a point not in [a,b], where E is contained in [a,b], then as there is no element of E in the interval (b,x] so x is not an acc. point of E, and we can repeat the argument for x <a, hence E' is a subset of [a,b]. Didn't even have to mention epsilon once.
     
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