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An infinite union of closed sets?

  1. Apr 11, 2006 #1
    Is this infinite union closed, open or neither? What is your reason?
     
  2. jcsd
  3. Apr 11, 2006 #2

    HallsofIvy

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    Well, if C0= [0, 1] and Cn= a closed subset of [0,1] for all positive integers n, then each Cn is closed and, obviously, the union is C0= [0, 1], a closed set.

    On the other hand, take a look at Cn= [1/n, 1-1/n] (intervals of real numbers) for n any positive integer. Cn is closed for all n, but the union of that collection of closed sets is (0, 1) which is open.

    Or, let Cn= [1/n, 1]. Again, each Cn is closed but the union now is (0, 1] which is neither open nor closed.

    But you left out a possibility: How about "both"? Let Cn= [-n, n] for all positive integers n. Each set is closed and the union is the set of all real numbers which is both open and closed.

    In other words, just like the infinite intersection of open sets, nothing can be said about an infinite union of closed sets!
     
  4. Apr 12, 2006 #3


    Do you mean the union is the set of all postive integers since you allowed n to be a positive integer?

    You say that the union of this set is both closed and open. I can see how it is open because given any set in this infinite union, there will always be a neighbourhood of this union (in the form of a larger set) that is part of this union. But how is it closed? To be closed, you have to show that the complement of this infinite union is open.

    I follow all your other examples. You seemed to be have stated different conclusions so it depends on the situation how an infinite union of closed sets will turn out.
     
    Last edited: Apr 12, 2006
  5. Apr 12, 2006 #4

    matt grime

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    No each of thos sets is an interval, the union is R, exactly like he said.


    The union is R, the complement is empty.

    Several new problems now come to light though:

    1. Neighbourhood of this union. What does that mean?

    2. The set of integers Z in R which you claim is the union is certainly not open like you claim and it is in fact rather obviously closed which you claim not to see. Those are disturbing things. So what did you mean to say, precisely?
     
  6. Apr 12, 2006 #5
    I get this point now. I will continue to use this set as our example.




    I used the word neighbourhood too loosely here. I meant whenever you have a set such as [-n,n] than there will always be another set whether larger such as [-n-1, n+1] or smaller such as [-n+1, n-1] which will lie in this infinite union. Hence the infinite union is open.

    The complement of this set is the empty set as you point out. It is both open and closed according to http://en.wikipedia.org/wiki/Empty_set hence our infinite union could also be closed when taking the complement, the empty set as open.
     
    Last edited: Apr 12, 2006
  7. Apr 12, 2006 #6

    matt grime

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    No, I suggest you need to read the definition of open more closely.

    Openness is a statement involving points and open sets containing points, nothing to do with closed sets containing other closed sets.


    The empty set regarded as a subset of a topological space is open and closed by the definition of topology.
     
  8. Apr 12, 2006 #7

    So the reason why 'Cn= [-n, n] for all positive integers n. Each set is closed and the union is the set of all real numbers which is both open and closed.' is because the complement of this set which is the empty set is both open and closed hence Cn is both closed and open.
     
  9. Apr 13, 2006 #8

    matt grime

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    No, the reason R as a subset of R is open and closed is *because it is*: it satisfies the definition of being an open subset and the definition of being a closed subset.

    X is Open <=> every point is contained in some open neighbourhood N contained in X
    X is closed <=> Insert your favourite definition here.
     
  10. Apr 13, 2006 #9

    So in this case every point in R is contained in some open neighbourhood N ranging from -infinity to infinity. Therefore R as a subset of R is open.

    R as a subset of R is closed because the complement of it is open.
     
  11. Apr 13, 2006 #10

    HallsofIvy

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    Not every one uses "its complement is open" as the definition of closed.
    It is also possible to define "closed" as "contains all of its limit points".
    Of course, if {an} is a convergent sequence of real numbers then its limit is a real number so R is closed under that definition.

    Yet another definition of closed is "contains all of its boundary points" where a "boundary point" is a point such that every neighborhood contains some points in the set and some not in the set. (One can also then say that an open set "contains none of its boundary points). R itself has NO boundary points so it is correct to say that it contains none of its boundary points and that it contains all of its boundary points so it is both open and closed.
     
    Last edited: Apr 13, 2006
  12. Apr 14, 2006 #11
    It worries me somewhat that you could make statements about something that does not exist. For example I am not a bird but I claim that I have a long beak and a short beak. It would be nonsensical for me to speak about beaks in the first place. But this is more philosophy than maths. It might be better to just define the set R as both open and closed, just like the empty set without digging for logical foundations.
     
  13. Apr 14, 2006 #12

    matt grime

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    What does or doesn't exist? The empty set exists just as much as the set of real numbers.

    Or perhaps you need to understand that logical constructs work in certain ways.

    For instance for you to demonstrate that R does not contain all its boundary points you would have to produce one not in R. But as there is none, the statement

    there exists a boundary point of R not in R

    is false, hence

    all boundary points of R are in R

    is true.
     
    Last edited: Apr 14, 2006
  14. Apr 15, 2006 #13
    I was trying to say that the fact that R has no boundary points would mean making agruments about them nonsensical.

    In other words, the boundary points of R does not exist so we shouldn't mention it. We are only allowed to say that they don't exist and nothing else (i.e. can't have a statement starting with "supose R contains all of its boundary points...")


    I understand what you are getting at here.

    Prove: R contains all of its boundary points
    1) Try to show a boundary point of R not in R.
    2) 1 can't happen since R has no boundary points
    Hence R is closed

    Prove: R contains none of its boundary points
    1) Try to show a boundary point of R in R
    2) 1 can't happen since R has no boundary points
    Hence is R is open

    The possible flaw I see in these proofs is that the statement trying to be proved is nonsensical since 'R contains ... its boundary points.' is impossible. So not only step 1 (in each statement) can't happen but what we are trying to prove cannot happen either.
     
  15. Apr 15, 2006 #14

    I have just realised something 'illogical' about the definition of open sets according to 'R contains none of its boundary points'. Surely 'its boundary points' refer to R's boundary points so R has boundary points. But than to state R contains none of its boundary points is confusing.
     
  16. Apr 15, 2006 #15

    Hurkyl

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    Yep. And its set of boundary points is the empty set.
     
  17. Apr 15, 2006 #16

    matt grime

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    Since the definition of open is not 'does not contain any of its boundary points' I'im not going to worry unduly about this.

    However the statement 'if x is in the boundary of R then X' is always going to be true since the hypothesis is false; 'x in the boundary of R is always' false hence it implies anything you want. it's just elementary (in the non-perjorative use of the word logic).
     
    Last edited: Apr 15, 2006
  18. Apr 15, 2006 #17

    HallsofIvy

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    The definition of open set can vary from text book to text book. I have certainly seen texts on metric topology where they introduce the notion of a neighborhood [itex]N_\delta(x)= \{y|d(x,y)< \delta\}[/itex], then define "p is an interior point of set A" as "there exist some [itex]\delta[/itex] such that [itex]N_\delta(x)[/itex] is a subset of A", define "p is an exterior point of set A" as "p is an interior point of the complement of A", and "p is a boundary point of A" as "p is neither an interior point nor an exterior point of A". Then the definition of "open set" is "contains none of its boundary points" and "closed set" is "contains all of its boundary points".

    Since all points in the universal set fall into one and only one of "interior point", "exterior point", "boundary point", and obviously exterior points of A can't be in A, saying that A contains none of its boundary points is exactly the same as saying all of its points are interior points. Since A and its complement have exactly the same boundary points, saying A contains all of its boundary points is exactly the same as saying its complement is open.

    My experience is that students who have been dealing with open and closed intervals since before calculus typically accept those definitions more easily than "all of its points are interior points" for open and "its complement is open" for closed.
     
  19. Apr 16, 2006 #18

    matt grime

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    That was badly worded on my part. If I were to say anything then I should have said: that is not what the definition of open should be.
     
  20. Apr 16, 2006 #19

    I am much more comfortable with "all of its points are interior points" for open and "its complement is open" for closed but than again I have done some calculus and also because they are the exact definitions I have been taught to learn.

    With your prior definitions. Do you have to stick points P to the boundary of a set S and see whether there are points in S which exactly match the points P? If they do than S is closed. If they don't than S is open.
     
    Last edited: Apr 16, 2006
  21. Apr 16, 2006 #20
    So you are implying the material conditional in loigc? For 'if a than b' to be true and when a is false, b can be either true or false. That makes a bit more sense.
     
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