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Closed and open at the same time?

  1. Sep 16, 2012 #1
    Closed and open at the same time!?!?!

    I am doing a reading course with a professor using Rudin's Real and Complex Analysis. I had never seen any topology before so the professor told me to work through the first two chapters of Hocking's Topology.

    I am taking my sweet time with it because I want to be able to absorb everything. I was doing fine until recently when I got to basis and subbassis of a topology.

    "Can we define a topology S for which each [itex]X_{a}[/itex] is an open set? The answer is yes because we may always assign to S the discrete topology in which there are no limit points. In the discrete topology every set is closed, hence every set is open."

    My head exploded from reading that. Could someone please explain to me what the heck happened?
     
  2. jcsd
  3. Sep 16, 2012 #2
    Re: Closed and open at the same time!?!?!

    Closedness is the openness of the complement. There is nothing preventing both a set and its complement from being open at the same time. As a simple example, let X be the disjoint union of two segments on the real line. Then each of the line segments is an open set in X and hence also closed.

    The nonexistence of simultaneously open and closed proper nontrivial subsets is the property characterising connectedness of a topological space.
     
  4. Sep 16, 2012 #3

    Borek

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    Re: Closed and open at the same time!?!?!

    Don't worry, you are not the first one to be confused by topology.
     
  5. Sep 16, 2012 #4
    Re: Closed and open at the same time!?!?!

    WOW, excellent ! Thank's a lot for that link, my somach hurts from laughing so much :rofl: (now I'm going to see the related ones (Rieman hypothesis, ...))

    Now, DeadOriginal, the thing is, in Topology, there is a very precise meaning of what is an open set or a closed set, but it is not the same thing as an open or closed door, that is, they are not mutually exclusive.
    The term open or closed comes from the fact that, for most cases, an open set (or a closed set) is very much what you would expect it to be if thinking about an interval on the real line.
    But you defined your topology in terms of "this is what I will call the open sets in my topology, here they are, they satisfy the rules, so this is what I will call the open sets"
    From there, the closed sets are all the sets whose complement is open.
    For instance, in whatever topology you define, there are always two sets which are both open and closed, that is the null set and the whole set (the set of all the elements)
    In your case, all the sets are open, since they are all open, then they must also all be closed, since any complement of any set must be open
    I am trying to make it clear that "they are all open therefore they are all closed" is not the result of open = closed, but the result that since all the elements of your topology are open (vs, just some of them for other more intuitive topologies (think about the intervals on the real line again)), then as a consequence, their complement must also be open, so they must also be closed
     
  6. Sep 16, 2012 #5

    HallsofIvy

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    Re: Closed and open at the same time!?!?!

    One way of defining open and closed sets is to first define "boundary points".
    p is a boundary point of set A if and only if it is in the closure of A but not in the interior of A. Open sets are sets that contain NONE of their boundary points and closed sets are sets that include ALL of their boundary points.

    Of course, a lot of sets will contain some of their boundary points but not all- such sets are neither open nor closed.

    In order to be both open and closed, a set must contain all of its boundary points and none of them! How is that possible? Very simple- the set must have no boundary points to begin with! One such set is the empty set. Its closure is still the empty set so it has no boundary points. The empty set is both open and closed. Another is the entire base set for the topology- its closure is also itself and its interior is itself- the entire base set for the topology is both open and closed.

    To get more than that you need to deal with spaces that are NOT connected. It can be shown that if a topological space is connected the only sets that are both open and closed are the empty set and the entire set.

    But if a set is NOT connected then its "connectedness components" (the largest connected subsets) are both open and closed. For example, let [itex]X= (0, 2)\cup (3, 4)[/itex] with the metric topology inherited from the real numbers. It is easy to see that the interior of (0, 2) is just (0, 2). And, in fact, that closure of (0, 2) is (0, 2) (It is NOT [0, 2] because 0 and 2 are not members of the topology) and similarly for (3, 4). Both (0, 2) and (3, 4), as well as the empty set and X itself, are both open and closed.
     
  7. Sep 16, 2012 #6
    Re: Closed and open at the same time!?!?!

    Haha. I saw that while I was looking for an explanation on youtube.

    Hmm. Very interesting. Intuitively it makes a lot of sense. I need to ponder more about the definition of both open and closed sets before it becomes completely clear.
     
  8. Sep 16, 2012 #7
    Re: Closed and open at the same time!?!?!

    Do you understand the discrete topology?

    If you like, you can put a metric on it and then define open balls to convince yourself that everything is consistent with the definition of open and closed sets of reals.

    Given any set, define

    [itex]d(x,y) =

    \left\{
    \begin{array}{ll}
    0 & \mbox{if } x = y \\
    1 & \mbox{if } x \neq y
    \end{array}
    \right.
    [/itex]

    Given that metric, what are the open sets? What are the closed sets? Try it out with the underlying set being the integers. Or the reals.

    Did you learn about continuity yet? What happens if you apply the epsilon-delta definition of continuity to any function whose domain is a discrete space?

    This is a valuable example to spend time with.
     
    Last edited: Sep 16, 2012
  9. Sep 16, 2012 #8
    Re: Closed and open at the same time!?!?!

    I am currently learning continuity in my analysis course. I haven't gotten up to metrics yet in topology but they are basically on the next page and the concept seems simple so I will look into this.
     
  10. Sep 16, 2012 #9
    Re: Closed and open at the same time!?!?!

    If you look at the definition of open and closed, rather than what the words make you think of, it's not even the least bit surprising that sets can be both open and closed at the same time.

    What might be less clear is the motivation for the definitions. For me, it's the fact that pre-images of open sets being open is equivalent to continuity in a metric space (or R^n, if you prefer). In geometry, you could say two things are the same if there is a rigid motion (isometry) that puts one right on top of the other. Topology just replaces the rigid motion by a continuous and continuously invertible map and says two things are the same if there's one of those between them. So, then, you ask what structure such a map should preserve, and the answer, according to what I said above is that it preserves the "open set structure", that is to say, which sets are open. So, if you start with a set and you want it to have some topology, it ought to mean that you should be able to make sense of which things it's homeomorphic to, so that means you need to decide which sets are open and closed. And just open is good enough because we can agree that closed sets are complements of open ones. You could just allow the open sets to be any old sets, but you want it to satisfy enough basic properties to have continuous maps that resemble the ones we are familiar with. So, you take some of the properties that you can prove for open sets in metric spaces (what happens when you intersect them or union them), and you get the definition of a topology on a set.

    Anyway, there's no real reason to expect that the complement of an open set couldn't also be open. Even, if you want to think of an open set as a set that has no boundary points, if you think of a subset of R^n as having a topology, like two disjoint open balls, it's clear that that could easily happen. The complement of one open ball is the other open ball, so each open ball is both open and closed in that subspace of R^n.
     
  11. Sep 17, 2012 #10
  12. Sep 17, 2012 #11

    quasar987

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    Re: Closed and open at the same time!?!?!

    That Hitler clip was iNsAaAaAaNe..! hahaha
     
  13. Sep 17, 2012 #12

    mathwonk

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    Re: Closed and open at the same time!?!?!

    a set S is open if for every p in S, any point near p is also in S.

    But if you spread out al the points of your space so none of them are near any others, then the only way for a point of your space to be near p is for it to equal p. In particular then if q is near p and p is in S, then q (=p) is also in S.

    This shows that every set S containing p is open.
     
  14. Sep 18, 2012 #13

    HallsofIvy

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    Re: Closed and open at the same time!?!?!

    You don't need a metric to define the discrete topology. Given any set, X, we define all subsets of X to be open. Of course, since the complement of any subset of X is a subset of X, it follows that all subsets of X are also closed.
     
  15. Sep 18, 2012 #14
    Re: Closed and open at the same time!?!?!

    Hmm. Would it be correct to think of this as an interval like the way HallsofIvy described it? Like say [itex]p\in (a,b)[/itex], and [itex]S=\{x|x\in(a,b)\}[/itex] then S is open if the points within (a,b) close to p are also in S. Is that how I should see it?

    Maybe my confusion lies within the definition of a discrete topology. The way my book introduces it to me is by first presenting the definition of a topology: The set S has a topology provided that for every p in S and every subset X of S the question Is p a limit point of X? can be answered. It then tells me that when the answer is always no then we have the discrete topology.

    Now what confuses me in what you said is say we have A as the collection of all subsets of X that are open and since we defined all subset of X to be open then the complement X-A=∅. Now I understand thanks to your previous explanation and by definition that ∅ is open and closed but how can ∅ being closed follow with all of the subsets of X also being closed?
     
  16. Sep 18, 2012 #15
    Re: Closed and open at the same time!?!?!

    Hi DeadOriginal, I don't really understand your last question.
    I think you are getting confused by too intricate consequences of definitions, or equivalent definitions or 'sufficiency conditions'.
    A Topology T over a set S is just a subset of P(S), that is, it is a set whose elements are all subsets of your original set, with also those rules:
    it is closed under intersection (if A and B are in T, then so is their intersection)
    it is also closed for arbitrary Union (doesn't have to be countable)
    and, the null set and the whole set S belong to T
    this is it...
    You define your topology like this, there are several subsets in P(S) you can choose (that satisfy the rules) and there you have it
    The elements of T, are called the open sets
    then you define the closed sets as the sets whose complement is open.
    So, by definition, the null set and the whole set S are always open (this is the third rule) and since they are both the complement of each other, well, their complement is also open by definition, so they are closed.
    One topology you could define over some set S is the trivial topology
    it just has the unique required null set and whole set. they are open and closed.
    another trivial one is the so called discrete topology where you say 'every set is open', it works: the null set and the whole set are open, any intersection or union you can think of is still a subset of S so it is indeed open, everything is open.
    Therefore, the complement of any of those sets (say, X), since it still belongs to P(S) is open, and that makes X closed.
    So in the discrete topology, all sets are closed because they are all open.
    but the null set (and the whole set) are always open and closed in any topology because it is the basic definition/rule
     
  17. Sep 18, 2012 #16

    lavinia

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    Re: Closed and open at the same time!?!?!

    A metric space has the discrete topology if for each point there is an epsilon small enough so that the open ball of radius epsilon around the point contains only that point. So by definition of openness in a metric space, each point is open.

    Since no other points can enter this open ball, all Cauchy sequences inside the ball are constant. But in a metric space a set is closed if all Cauchy sequences in the set converge to a point in the set. Therefore each point is also closed.

    Note that a discrete metric space may not be complete, for instance the set, {1/n|n an integer} is not complete in the usual metric.
     
    Last edited: Sep 18, 2012
  18. Sep 18, 2012 #17
    Re: Closed and open at the same time!?!?!

    I am definitely getting confused with all of the definitions and such.

    My question is about this part: "another trivial one is the so called discrete topology where you say 'every set is open', it works: the null set and the whole set are open, any intersection or union you can think of is still a subset of S so it is indeed open, everything is open.
    Therefore, the complement of any of those sets (say, X), since it still belongs to P(S) is open, and that makes X closed."

    We have that the null set is open/closed and the whole set is open. Then the complement of those sets would be the null set right? Now we just said it is open/closed and I understand the whole set S is open but I cant seem to see how it is closed.

    So we are saying S is open. Now X is the collection of subsets of S that are open so its complement S-X=[itex]X^{c}[/itex] is closed. What I am seeing in my head is [itex]X\cup X^{c}=S[/itex]. I can understand that since all sets of S is open, we have that [itex]X^{c}[/itex] is open and is closed because it is the complement of the open sets. I don't see how this says X which is the collection of the open sets is also closed.
     
  19. Sep 18, 2012 #18
    Re: Closed and open at the same time!?!?!

    Hi again DeadOriginal,
    this (your post) is good, because we are probably getting to what actually confuses you (the different ways (more or less general/complex) not really helping getting rid of the confusion)

    No, a closed set is (so called/defined) closed because its complement is open.
    not because its complement is void (in which case the only possible closed set would be S), or anything else. if its complement happens to belong to the set of open sets, then you call it 'closed'

    void and S being open are not logical derivations of the rest, they are requirements. when you say "here is the set of sets I declare to be a topology over S (here is the list of what we will call the open sets in this set)", you just have to include void and S, it is the strict minimum.
    After that you must also make sure that if you included anything else than void and S, then their intersections and infinite unions also must belong to this set.
     
  20. Sep 18, 2012 #19
    Re: Closed and open at the same time!?!?!

    So then say we have X in S as the collection of all open subsets of S. Its complement would then be represented as S-X. Since X is not closed, S-X is not open but how does this make X closed?


    Waiiiiiitttttttttt........ Since the empty set is closed and X is the complement of it then X must be open. Then since the empty set is open and X is the complement of it then it must be closed too. Is that what I'm supposed to see?
     
  21. Sep 18, 2012 #20
    Re: Closed and open at the same time!?!?!

    No, forget about S and void as having anything to do with open and closed
    you start with a Set on which you will define a topology, which means, you will say "here are the subsets of S that I will call OPEN"
    you only have to follow some rules about intersection and union of those sets so that when you intersect any 2 sets you just said 'those are open', then their intersection must also belong to the list you provided as your definition of what are the open sets according to this topology I define on S. (and you must also make sure that uncountable union also works)
    then, you must say 'the void set and the whole set also belong to the set of open sets' (because this is part of the requirements, of the definition, of what is an open set in topology)"

    so let's see that again
    void is open, because the definition of open explicitly says "the whole set and the void set are open"
    S is open because the definition of open says 'the whole set and the void set are open"

    then, the definition of closed set is "the complement of this set must be open"
    so,
    the complement of void is the whole set, which by definition of 'open' is open. therefore, void is closed
    the complement of the whole set is void, which by definition of 'open' is open, therefore, the whole set is closed

    now, for the discrete topology, any subset of S is open
    so, take X, a subset of S. by the definition of the discrete topology, it is an open set, since all of them are open.
    Then, what about its complement ? its complement is a subset of S, but we have just said that all subsets of S are open.
    So the complement of any subset of S is an open subset of S. that makes it a closed subset.
    Hence, all subsets open => all subsets closed
     
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