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Some clarifications on my proving

  1. Aug 14, 2005 #1
    Hello. I didn't know if this is the right forum to post this (since this is the Calculus section and my question pertains to real analysis), but I would like some clarifications on a proof I am preparing for the propositon "If a set S is dense, then every point in S is an accumulation point".

    What I did is assume the contrary, that is, "If a set S is dense then none of its points are accumulation points". Then for an arbitrary element x in S it is possible to create a deleted ε-neighborhood for any ε>0 such that (x-ε, x+ε) does not contain any point in S. Now, i let a and b be elements in S such that a < x < b such that x-a = b-x. If I assign ε=b-x (in turn, ε=x-a), then the deleted neighborhood for this choice of ε must be empty to guarantee that x is not an accumulation point, that is, (a,x)U(x,b) should be empty, which means (a,x) is empty and (x,b) is also empty. This means that there are no numbers between a and x and between x and b, which means that S is not dense (since a dense set is defined to be that for any elements c and e and c<e there exists an element d such that c < d < e), which leads to a contradiction since we assume S to be dense. Hence, If a set S is dense, then every real point in S is an accumulation point. ///

    My questions are:

    (1) can I simply assume that x is between a and b since S is dense?
    (2) If I choose my ε to be b-x, are a and b in the neighborhood of x? this also leads to the question that are deleted neighborhoods open sets...

    Thanks for any help.
     
    Last edited: Aug 14, 2005
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  3. Aug 14, 2005 #2

    HallsofIvy

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    What do you mean by "a set S is dense"? I know what it means to say that a set is dense in some larger space. Do you mean dense in the real numbers? You appear to be assuming that S is a set of real numbers. And do you mean to say, then an accumulation point of the set of real numbers?
     
    Last edited: Aug 14, 2005
  4. Aug 14, 2005 #3
    From what I understand, he/she is saying that S is dense in the set of real numbers... continuing further, if a and are real nos. and a < b, there is an s belonging to S such that a < s < b.
     
  5. Aug 14, 2005 #4

    lurflurf

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    This section is called "Calculus & Analysis" so it looks like a good place.
    As has been mentioned Set density is with respect to another set
    as in S is dense it S'
    next the contrary to
    "If a set S is dense, then every point in S is an accumulation point."
    is
    "If S there exist a one point in S that is not an accumulation point, then S is not dense."
    so (assuming you mean S is dense in R)
    Let x be a point in S that is not a point of accumulation and show that S is not dense.
    You are on the right track

    (1) can I simply assume that x is between a and b since S is dense?
    I do not understant this
    (2) If I choose my ε to be b-x, are a and b in the neighborhood of x? this also leads to the question that are deleted neighborhoods open sets...
    neighborhoods are open sets
    deleted neighbor hoods are open
    unions of open sets are open and a deleted neighborhood is a union of open sets in R1.
     
  6. Aug 14, 2005 #5
    HallsOfIvy

    Sorry for not making the definition of Denseness clear. Irony of Truth just spelled it out for me: S is dense in the set of real numbers... continuing further, if a and are real nos. and a < b, there is an s belonging to S such that a < s < b.

    Also, the propostition should be: "If a set S is dense, then every point in S is an accumulation point of set S".

    Sorry again for the incompleteness of my statements and proposition.

    Irony_Of_Truth

    I'm a he ^_^;; Thanks by the way for clearing things up.

    lurflurf

    When I meant "can I assume that a < x < b" I was asking if I could assume that x will be between two numbers, sa a and b, that is in set S such that a < x < b because of the Denseness of S

    ---

    Generally, the thing I'm bothered with is whether or not a and b should be in the deleted ε neighborhood of my arbitrary point x under the assumption that a < x < b, that x-a = b-x and if my chosen ε > 0 would assume a value of ε=x-a (in turn, ε=b-x).

    I would also appreciate it if anyone can find any potholes in my proof :)

    Thanks in advance :)
     
    Last edited: Aug 14, 2005
  7. Aug 14, 2005 #6

    lurflurf

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    well you have found that
    x is not an accumulation point-> there exit (a,x)U(x,b)
    thus S is not dense
    clearly a and be are not is (a,x)U(x,b)
    (you could also say intersection((a,b),S)={x})
    you need not be concerned about denseness other than showing since x is not a point of accumulation S is not dense.
    I do not see how assuming S dense helps
    sketch of proof
    1) let x in S not be a point of accumulation
    2) then there exist a deleted neighbor of x with nothing in common with S
    3) S is not dense in R because (a,x) does not cointain a point in S
    you should not say (a,x) is empty (though perhaps you could if your universe were S) it is better to say (a,x) contains not points from S.
     
  8. Aug 14, 2005 #7
    I see... a and b are not in the deleted ε neighborhood of x when ε=x-a. Thanks for the clarifications. :) I really appreciate the help :)
     
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