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Math Help - Open Sets on the Real Line

  1. #1
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    Talking Open Sets on the Real Line

    A theorem in Taylor (General Theory of Functions and Integration) states:

    R is the real line

    Theorem: If E is an open set in R, E can be expressed as a countable (possibly finite) union of disjoint sets E1, E2,... where each En is one of the following:
    1) {x: a<x<b}
    2) {x: x<b}, b in R
    3) {x: a<x}, a in R

    Forget the proof. I can't even understand the theorem. I try to come up with an example but the furthest I get is:

    (0,2) is the sum of (0,1) and (1,2), which is wrong because 1 is missing. The only way I can see is (0,1) and [1,2) but this isn't one of the conditions above.

    Any ideas?
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  2. #2
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    I'm not an expert in analysis, but isn't (0,2) already of type 1? You don't need to decompose it further, do you?
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    Quote Originally Posted by Ackbeet View Post
    I'm not an expert in analysis, but isn't (0,2) already of type 1? You don't need to decompose it further, do you?
    Intersting observation. But then the theorem is meaningless. It could be expressed as the intersection of (0,2) and 0, but the theorem is still vacuous. No?: If E is an open set it can be expressed as intersection of E and 0, and E and 0 are disjoint (if E doesn't contain 0, as in example).
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  4. #4
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    Not all open sets are open intervals. How about this set:

    (0,2)\cup(3,5)?

    It's open (that is, every point in the set contains a neighborhood that's also in the set), but it's not an open interval.

    What this theorem is saying is that all open sets of any kind can be written as combinations of only certain kinds of open sets.
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  5. #5
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    If you will give a page number in Taylor, I will look up what you are allowed to use.

    Here is the basic idea. A component is a maximally connect subset. Any component of an open set is open. Two distinct components must be disjoint otherwise they would not be maximum.
    Example: (-\infty,0)\cup(1,2)\cup (5,10) is an open set with three components.

    Given any open, set each of its points must belong to one of its components. Remember they are pair-wise disjoint. Because the set of rationals is countable that collection of components must be countable.
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  6. #6
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    Quote Originally Posted by Plato View Post
    If you will give a page number in Taylor, I will look up what you are allowed to use.

    Here is the basic idea. A component is a maximally connect subset. Any component of an open set is open. Two distinct components must be disjoint otherwise they would not be maximum.
    Example: (-\infty,0)\cup(1,2)\cup (5,10) is an open set with three components.

    Given any open, set each of its points must belong to one of its components. Remember they are pair-wise disjoint. Because the set of rationals is countable that collection of components must be countable.
    Thanks for response. I was just about ready to type the whole damn thing (printed out LaTex code from tutorial):

    Taylor, Fist Ed (1965) Theorem 2-3 IV, pg 55

    As for the examples, it seems to me that if you start off by defining E as a union of disjoint open sets, the theorem is trivial: A union of disjoint open sets can be expressed as a union of the same disjoint open sets.

    Since others are interested, maybe I'll struggle through copying the theorem from Taylor
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  7. #7
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    Maybe I just don't get what you are asking for?
    The proof is clearly in Taylor. As I said above to key is maximally connect subsets.
    So what is your point?
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  8. #8
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    The Theorem from Taylor

    "THE STRUCTURE OF OPEN SETS IN R

    The nature of open sets in R is made explicit by the following theorem, which has no counterpart for open sets in Rk,  k\geq{2}. (yeahhhhh!)

    Theorem 2-3 IV. If E is an open set in R such that E\neq 0, and E\neq R, E can be expressed as a countable (possibly finite) union of disjoint sets E1, E2,..., where each En is a non-empty open set of one of the following types:
    (i) a finite open interval {x : a<x<b},
    (ii) a left-semiinfinite open interval {x : x<b}, where b\in R,
    (iii) a right-semiinfinite open interval {x : a<x}, where a\in R."
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  9. #9
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    Quote Originally Posted by Plato View Post
    Maybe I just don't get what you are asking for?
    The proof is clearly in Taylor. As I said above to key is maximally connect subsets.
    So what is your point?
    I'm just looking for a simple example of the theorem. If E is (0,2) give an example of E1, E2, ... Note E satisfies the premise.

    Ackbeet says E1=(0,2) and E2=0, then E is the union of two disjoint open sets. Is that it? Then I'm totally at a loss as to what the theorem is all about.

    Sorry, but I don't understand maximally connected subsets, and you don't haveto explain, just write me off as dumb.
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  10. #10
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    If E=(0,2) then as Ackbeet mentioned, E is already of type 1, so your decomposition would just be E itself. If you take Ackbeet's example, E=(0,2) \cup (3,5) then your decomposition will be E=E_1 \cup E_2 with E_1 = (0,2) and E_2 = (3,5). The point of this theorem is to give us information about open sets which we do not know beforehand to have this structure. For example, if f:\mathbb{R} \to \mathbb{R} is a continuous function, then \{x \in \mathbb{R} : f(x) > a\} is an open set, and this theorem gives us a lot of information about how it 'looks'.
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  11. #11
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    Structure of Open Sets on the Real Line

    Was just about to give up on this when I found the following:

    Elements of the Theory of Functions ... - Google Books

    According to Kolmogorov:

    Theorem: Every open set on the real line is the sum of a countable number of disjoint intervals.

    Now that makes perfectly good sense. There is no mention of disjoint open intervals. So (0,2) is (0,1) and [1,2).

    It may be that Taylor's version is a more encompassing one familiar only to the cognizanti. If not, it makes you wonder about his proof.
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    Is a sum allowed to have only one summand in it? Like a union with only one "unionand"?
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  13. #13
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    Quote Originally Posted by Ackbeet View Post
    Is a sum allowed to have only one summand in it? Like a union with only one "unionand"?
    According to Kolmogorov, "a countable number:" Theorem: Every open set on the real line is the sum of a countable number of disjoint intervals.
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  14. #14
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    I was thinking more along the lines of the Taylor theorem you're asked to prove. (0,2) can be written as the union of set types 1, 2, 3, even if it only takes one set to be joined in the union.
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  15. #15
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    There are only three basic forms of connected open proper subsets of real numbers: (-\infty,a),~(a,b),~\&~(a,\infty).
    The theorem means: Any open set of real numbers can be written as a countable union of pairwise disjoint basic open connected sets.

    The set [1,\infty )\backslash \mathbb{Z}^ +   = \bigcup\limits_{n = 1}^\infty  {\left( {n,n + 1} \right)} therefore  [1,\infty )\backslash \mathbb{Z}^ + is an open set.

    The set (2,3) is a basic open set by definition.
    Last edited by Plato; January 31st 2011 at 02:28 PM.
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