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Not so open minded open sets

  1. Sep 26, 2005 #1
    This is not a specific homework question so much as it is a general conceptual question.
    My analysis book includes a theorem that states:

    1. The union of any number of open sets is an open set.
    2. The intersection of a finite number of open sets is an open set.

    I follow the proof of property one, however in part of the proof of property two my book directs me to a picture of nested sets and says: "A glance at the above figure will convince you that the intersection is an open interval". The book then goes on to say: "Note that this is where the argument breaks down for infinite intersections: if there were an infinite number of intervals the intersection might not be an open interval. QED".

    I can perhaps buy into the proof that the intersection of a finite number of open sets is an open set, but this in no way convinces me of anything pertaining to the intersection of an infinite number of open sets!

    Perhaps I'm just dense, but I really feel the need for explicit proof that the intersection of an infinite number of open sets may not be open.

    My other thought on this was that maybe there was no way to prove or disprove the infinite case due to the ambiguity of it; that is because the intersection 'might' or 'might not' be open.

    This general concern also reminds me of the notion of raising negative one to an infinite power. There is no way to say if the result is a negative or positive number.

    If anyone could try to help me understand my problem of infinite intersections I would be very appreciative! Also, if anyone has some insight on how to interpret and deal with mathematical ambiguities in general, I would love to hear what you have to say on the topic!

    thanks :smile:
     
  2. jcsd
  3. Sep 26, 2005 #2

    George Jones

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    This might help a little.

    Let A_n be the open interval (of R) (-1/n , 1/n) for n a positve integer. What is the intersection of all the A_n 's?

    Regards,
    George
     
  4. Sep 26, 2005 #3
    isn't that still an open set george? (-1,1)
    is [-1,1) and (-1,1] open sets? never took analysi i tried to learn it on my own.
     
  5. Sep 26, 2005 #4
    Wouldn't the intersection of all the [itex]A_n[/itex]'s be zero?

    So, if that's the case then I take it that zero is not considered an open set? Zero does not seem like it should be an open set.
     
  6. Sep 26, 2005 #5

    Tom Mattson

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    No, look at the first two elements of the sequence:

    [tex]A_1=(-1,1)[/tex]
    [tex]A_2=\left(-\frac{1}{2},\frac{1}{2}\right)[/tex]

    Already the intersection is smaller than [itex](-1,1)[/itex], and it's just going to get whittled down further as n gets larger.
     
  7. Sep 26, 2005 #6

    George Jones

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    Tom has already addressed your first question, so I'll move on to the second.

    [-1,1) isn't open because there no open interval that both contains -1 and is a subset of [-1, 1).

    Regards,
    George
     
  8. Sep 26, 2005 #7

    George Jones

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    Can you show that the intersection is just the set {0}? Here's a start: 0 is an element of A_n for every n. This means that 0 is an element of the intersection, i.e., that {0} is a subset of the intersection. Are there any other elements in the intersection?

    Why is {0} (or {a} for any real number a) not an open set?

    Regards,
    George
     
  9. Sep 26, 2005 #8
    The union/intersection/whatever neurocomp2003 meant to say about [-1, 1) and (-1, 1] is something I can understand.

    Correct me if I'm wrong, but the intersection of the two would be open and the union would be closed, right?

    I also can understand why the set [-1, 1) is neither open or closed, but when George asks: "Why is {0} (or {a} for any real number a) not an open set?" this is a real point of confusion for me.

    My analysis book seems to imply that this intersection of infinite sets might or might not be open. This leads me to think of it as ambiguous because I do not know if the set is or is not open.

    However, George seems to be implying that the set is indeed closed. I do not know why the set {0} is not open...
     
  10. Sep 26, 2005 #9

    George Jones

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    What does your analysis book give for the definition of an open set? For a closed set?

    Regards,
    George
     
  11. Sep 26, 2005 #10

    HallsofIvy

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    How, exactly, does your textbook define "open set".

    Many general topology books define "open set" as a member of the "topology" which is defined in turn as a collection of sets such that
    "any union of sets in the topology is in the topology"
    "any finite intersection of sets in the topology is in the topology"

    If you are dealing with metric topology then you might note that every one of the sets (-1/n, 1/n) is open but the intersection of all such sets is {0} which is not open (it does not contain any interval about 0).
     
  12. Sep 26, 2005 #11
    ah i see so george was saying that the intersection was leading to a limit(n->inf) = [0]?...heh sorry melinda( iwas thinking of union over intersection)
     
  13. Sep 26, 2005 #12
    *Definitions*

    An open set:
    A set is open if every point of the set lies in an open interval entirely contained in the interval.

    A closed set:
    A set is said to be closed if it contains all its limit points.

    I'm ok with the definition of open sets, but I'm still working on really understanding the definition of closed sets.
     
  14. Sep 26, 2005 #13
    Actually, as to the definition of closed sets:
    I could understand if it required the sup and inf to be included in order to be closed, but that's not how the definition reads :confused:
     
  15. Sep 26, 2005 #14

    HallsofIvy

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    Okay- that's specifically talking about real numbers so my "counter-example"
    (-1/n, 1/n) works.

    Another way to define open and closed sets is this:

    x is said to be an "interior" point of a set A if and only is some interval about x,
    (x-ε,x+epsilon;) is a subset of A. (That's the standard definition.)
    x is said to be an "exterior" point of a set A if and only if x is an interior point of the complement of A.
    x is said to be a "boundary" point of a set A if and only if x is neither an interior point nor an exterior point of A.

    It's easy to see that any interior point of A is contained in A, any exterior point of A is not contained in A, and that boundary points may or may not be contained in A. Further, a set and its complement "swap" exterior and interior points but have the same boundary.
    We can define an "open" set as any set that does not contain any of its boundary points- all of its points are "interior" points.
    We can define a "closed" set as any set that contains all of its boundary points.
    Since a set and its complement share boundary points, the complement of any open set is closed and vice-versa.

    Of course it is quite possible for a set to contain some of its boundary points but not all of them- to be neither open nor closed.

    It is even possible for a set to be both open and closed. In that case a set would have to contain all of its boundary points and none of its boundary points- which can happen if and only if it has no boundary points.
     
  16. Sep 26, 2005 #15

    George Jones

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    Take the book's definition and look at what HallsofIvy wrote at the end of his post. This shows that {0} is not an open set.

    One definition of "closed set" is any set whose complement is open.

    What is the book's definition of limit point. That {0} is closed should follow fairly easily from this definition.

    Regards,
    George
     
  17. Sep 26, 2005 #16
    I can see how a set could be neither open nor closed, but I cannot invision something concrete to see how a set could be both open and closed, save defining some set to not contain its boundary points.

    I guess I'm a bit new to this game, because things such as the above still seem way to contrived for my liking.

    I think I need to read a bit more, take another look at my definitions, and try to write a proof or two before I do anything else. Hopefully this will clear up my confusion :blushing:
     
  18. Sep 26, 2005 #17

    George Jones

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    To clear up a possible confusion - in a previous post I referred to HallofIvy's first post.

    There are only 2 "subsets" of R that are both open and closed.

    Regards,
    George
     
  19. Sep 26, 2005 #18
    Since I don't think anyone has properly answered your question (please forgive me, I may be wrong):

    I find that proving number two can come from using DeMorgan's Laws, AFTER proving the analogous theorem to the one you stated for closed sets. Basically, prove that a finite union of closed sets is closed, and use this after applying DeMorgan's Laws to your problem.
     
  20. Sep 26, 2005 #19

    Hurkyl

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    If you take the entire space itself, isn't that open and closed?


    What about the integers? They form a (discrete) topological space. Can you find any sets that are both open and closed within the integers? (Hint: if you simply guess, you'll be right :biggrin:)


    What if the topological space X consists of three disjoint line segments? (I.E.those three line segments are the whole space -- you're not to think of it as being some subset of the line) Can you take a guess at which sets are both open and closed?
     
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