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Math Help - Closure in the usual topology

  1. #1
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    Closure in the usual topology

    p is an element of the real numbers. e>0. Prove that [p-e, p+e] is the smallest closed set containing (p-e, p+e) in the usual topology (aka: the closure of (p-e, p+e) = [p-e, p+e])

    My attempt:
    Proof: Consider p an element os the real numbers and e>0.
    Claim 1: The closure of (p-e, p+e) in the usual topology = [p-e, p+e]
    Claim 1.1: The closure of (p-e, p+e) in the usual topology is a subset of [p-e, p+e]
    Proof: We know (p-e, p+e) is a subset of [p-e, p+e]
    And [p-e, p+e] is a closed neighborhood of p in the usual topology (from proof in class)
    Hence the closure of (p-e, p+e) in the usual topology is a subset (from theorem: if A is a subset of F and F is closed in (X,T), then the closure of A in topology T is a subset of F)
    Claim 1.2: [p-e, p+e] is a subset of the closure of (p-e, p+e) in the usual topology
    Proof: Suppose x is an element of [p-e, p+e]
    Claim 1.2.1: If x is an element of G and G is an element of the usual topology, then G intersects (p-e, p+e)
    [I'm struggling to find the right G. Proving this will show that x is an element of the closure of (p-e, p+e) in the usual topology by a theorem stating that for every G which is an element of a topology, if p is an element of G, then G intersects A <=> p is an element of the closure of A in the topology]
    Last edited by Kiefer; September 18th 2012 at 03:00 PM.
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  2. #2
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    Re: Closure in the usual topology

    Quote Originally Posted by Kiefer View Post
    p is an element of the real numbers. e>0. Prove that [p-e, p+e] is the smallest closed set containing (p-e, p+e) in the usual topology
    I don't follow your proof. But here is a comment.
    Several authors use that statement as the definition of closure.
    Then prove that the closure of a set is the union of the set with its limit points.
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    Re: Closure in the usual topology

    Quote Originally Posted by Plato View Post
    I don't follow your proof. But here is a comment.
    Several authors use that statement as the definition of closure.
    Then prove that the closure of a set is the union of the set with its limit points.
    I don't doubt that you've seen that, but I've never seen it done that way. I've always seen closure(A) defined to be the intersection of all closed sets containing A.
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    Re: Closure in the usual topology

    Quote Originally Posted by johnsomeone View Post
    I don't doubt that you've seen that, but I've never seen it done that way. I've always seen closure(A) defined to be the intersection of all closed sets containing A.
    I do not know to what definition you are referring.
    However the closure of a set is the union of the set with its limit points is the usual definition.
    R.L. Moore who is the "grandfather" of American topology used that definition on page 1 of his 1932 classic Foundations of Point Set Theory.

    So what is your point?
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    Re: Closure in the usual topology

    For A any subset of X, define the closure of A (denoted as cl(A)) by cl(A) = interesection of all closed sets containing A.
    Note that cl(A) is always closed. Also, if A a subset of D, with D closed, then necessarily cl(A) is a subset of D. (Proofs are very easy)

    Consider R with its usual topology. Let U = (p-e, p+e).
    Note that [p-e, p+e] = the complement of { (-infinity, p-e) union (p+e, infinity) }.
    Thus [p-e, p+e] is the complement of an open set, and hence is closed.
    Since [p-e, p+e] is a closed set that obviously contains U (= (p-e, p+e)), it follows that U a subset of cl(U) a subset of [p-e, p+e].

    But then there are only 4 possibilities for cl(U):
    Possibility #1: cl(U) = A1 = (p-e, p+e) (= U).
    Possibility #2: cl(U) = A2 = (p-e, p+e].
    Possibility #3: cl(U) = A3 = [p-e, p+e).
    Possibility #4: cl(U) = A4 = [p-e, p+e].
    Show that possibilities #1-#3 cannot work, since A1, A2, A3 are not closed, and thus cl(U) = A4 = [p-e, p+e].
    There are many ways to show that, but at this elementary description, lets just stick to the basic definitions.

    Claim: A2 is not closed.
    Proof:
    ASSUME A2 = (p-e, p+e] is closed.
    Then A2's complement, call it V, is open. V = (-infinity, p-e] union (p+e, infinity).
    Since V is open in R, using that the basis for the topology is the open intervals, have that every x in V is contained in some open interval W that's entirely contained in V.
    Let x = p-e in V. Then there exists an interval W containing x such that W is contained in V.
    Let W = (x-a, x+b), where a and b are some positive real numbers such that x in W subset V.
    Note that could have one or both of a & b being infinity, but if that were possible, then so would it be possible to choose a W' contained in W that also "worked" but that didn't have infinity at an interval endpoint.
    Note that a & b both positive is required in order that x be in W.
    But then W = (p-e-a, p-e+b) is a subset of V = (-infinity, p-e] union (p+e, infinity).
    But that's clearly impossible, as will now show, and this contradiction will prove that the initial assumption was false, and thus that A2 is not closed.
    Let D = min{2e, b}. Then D positive since both b and 2e are.
    Let y = p-e+D/2. Will show that y in W, but y not in V, contradicting that W is a subset of V.
    Note that D>0, so D/2>0, so from p-e < p-e+D/2(=y) < p-e+D/2+D/2 we know that p-e < y < p-e+D.
    Then p-e-a < p-e < y < p-e+D <= p-e+b, proving that y in W = (p-e-a, p-e+b).
    Also, p-e < y < p-e+D <= p-e+2e = p+e. Thus y in (p-e, p+e) which is a subset of A2.
    But then, since y is in A2, y cannot be in V, since V is the complement of A2.
    So we've proven that y is in V (since W was initially choosen to be a subset of V), and that y is not in V.
    Thus the assumption that A2 is closed has led to a contradiction.
    Therefore A2 is not closed, and the claim is proven.

    The exact same argument proves that A1 and A3 are not closed.

    Therefore, cl(U) = A4 = [p-e,p+e], since A4 is the *only* closed set between U and the closed set A4.

    Therefore cl(p-e,p+e) = [p-e,p+e].
    Last edited by johnsomeone; September 18th 2012 at 03:54 PM.
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    Re: Closure in the usual topology

    Quote Originally Posted by johnsomeone View Post
    For A any subset of X, define the closure of A (denoted as cl(A)) by cl(A) = interesection of all closed sets containing A.
    Note that cl(A) is always closed. Also, if A a subset of D, with D closed, then necessarily cl(A) is a subset of D. (Proofs are very easy)

    Consider R with its usual topology. Let U = (p-e, p+e).
    Note that [p-e, p+e] = the complement of { (-infinity, p-e) union (p+e, infinity) }.
    Thus [p-e, p+e] is the complement of an open set, and hence is closed.
    Since [p-e, p+e] is a closed set that obviously contains U (= (p-e, p+e)), it follows that U a subset of cl(U) a subset of [p-e, p+e].

    But then there are only 4 possibilities for cl(U):
    Possibility #1: cl(U) = A1 = (p-e, p+e) (= U).
    Possibility #2: cl(U) = A2 = (p-e, p+e].
    Possibility #3: cl(U) = A3 = [p-e, p+e).
    Possibility #4: cl(U) = A4 = [p-e, p+e].
    Show that possibilities #1-#3 cannot work, since A1, A2, A3 are not closed, and thus cl(U) = A4 = [p-e, p+e].
    There are many ways to show that, but at this elementary description, lets just stick to the basic definitions.

    Claim: A2 is not closed.
    Proof:
    ASSUME A2 = (p-e, p+e] is closed.
    Then A2's complement, call it V, is open. V = (-infinity, p-e] union (p+e, infinity).
    Since V is open in R, using that the basis for the topology is the open intervals, have that every x in V is contained in some open interval W that's entirely contained in V.
    Let x = p-e in V. Then there exists an interval W containing x such that W is contained in V.
    Let W = (x-a, x+b), where a and b are some positive real numbers such that x in W subset V.
    Note that could have one or both of a & b being infinity, but if that were possible, then so would it be possible to choose a W' contained in W that also "worked" but that didn't have infinity at an interval endpoint.
    Note that a & b both positive is required in order that x be in W.
    But then W = (p-e-a, p-e+b) is a subset of V = (-infinity, p-e] union (p+e, infinity).
    But that's clearly impossible, as will now show, and this contradiction will prove that the initial assumption was false, and thus that A2 is not closed.
    Let D = min{2e, b}. Then D positive since both b and 2e are.
    Let y = p-e+D/2. Will show that y in W, but y not in V, contradicting that W is a subset of V.
    Note that D>0, so D/2>0, so from p-e < p-e+D/2(=y) < p-e+D/2+D/2 we know that p-e < y < p-e+D.
    Then p-e-a < p-e < y < p-e+D <= p-e+b, proving that y in W = (p-e-a, p-e+b).
    Also, p-e < y < p-e+D <= p-e+2e = p+e. Thus y in (p-e, p+e) which is a subset of A2. Thus y can't be in V, since V is the complement of A2.
    Thus the assumption that A2 is closed has led to a contradiction.
    Therefore A2 is not closed, and the claim is proven.

    The exact same argument proves that A1 and A3 are not closed.
    Therefore, cl(U) = A4 = [p-e,p+e], since A4 is the *only* closed set between U and the closed set A4.
    Therefore cl(p-e,p+e) = [p-e,p+e].
    Again what is your point?
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    Re: Closure in the usual topology

    Quote Originally Posted by Plato View Post
    Again what is your point?
    Really? I wasn't clear?

    Thread's Initial Post: "Prove that [p-e, p+e] is the smallest closed set containing (p-e, p+e) in the usual topology (aka: the closure of (p-e, p+e) = [p-e, p+e])"

    My post:
    "For A any subset of X, define the closure of A (denoted as cl(A)) by cl(A) = interesection of all closed sets containing A.
    ...
    Therefore cl(p-e,p+e) = [p-e,p+e]."

    See the point?
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    Re: Closure in the usual topology

    Quote Originally Posted by johnsomeone View Post
    Really? I wasn't clear?
    Thread's Initial Post: "Prove that [p-e, p+e] is the smallest closed set containing (p-e, p+e) in the usual topology (aka: the closure of (p-e, p+e) = [p-e, p+e])"
    My
    "For A any subset of X, define the closure of A (denoted as cl(A)) by cl(A) = interesection of all closed sets containing A.
    ...
    Therefore cl(p-e,p+e) = [p-e,p+e]."
    See the point?
    No frankly I do not no see your point.
    You have no idea how the original poster defines closure .
    You are are assuming that you limited experiences with this definition is the only one.
    Now I have taught topology for well over forty years.
    I know of no widely used textbook that uses the definition of closure you quoted.
    As I said before R L Moore is the 'grandfather' of American topology.
    Look at his genealogy . Can you find a major American topologist not on that list?
    They all use the closure of a set is the union of the set with its limit points.
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  9. #9
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    Re: Closure in the usual topology

    Quote Originally Posted by Plato View Post
    No frankly I do not no see your point.
    You have no idea how the original poster defines closure .
    You are are assuming that you limited experiences with this definition is the only one.
    Now I have taught topology for well over forty years.
    I know of no widely used textbook that uses the definition of closure you quoted.
    As I said before R L Moore is the 'grandfather' of American topology.
    Look at his genealogy . Can you find a major American topologist not on that list?
    They all use the closure of a set is the union of the set with its limit points.
    No need to get so touchy. I haven't taught the subject for 40 years, but neither am I just pulling my claims out of thin air. Why would you suspect I would be? I said what I said for a reason. I didn't doubt your claim, but you seem compelled to try to refute mine. Well, you're mistaken - which is much worse for you than it is for me, seeing as you've been teaching the subject for 40 years and all...

    I went through some of my books, and found several examples of both definitions (and a few others that defined it using interior/exterior/boundary - and one that used the Kuratowski Closure Axioms to define a topology in the first place). I was mistaken - I had seen it before, decades ago as an undergraduate the textbook "Basic Topology" by Armstrong.

    Of course, within a page or two (usually more like a paragraph or two), all these definitions are proven to be equivalent anyhow, so it's a purely a matter of taste by the author.

    Among the books using my definition were 3 Dover books: "Tensor Analysis on Manifolds" by Bishop & Goldberg, "Algebraic Topology" by Maunder, and "Topology" by McCathy. I doubt those are frequently used in classes, but Munkres also uses my definition, and that is a commonly read text. Lastly, my stated defintion was used by one of my graduate texts, which is also one of the absolute classic texts used by many (heck maybe most) first year graduate students, and that was authored by a famous point set topologist/analyst: Big Rudin.

    Finally, look again at the original problem: "Prove that [p-e, p+e] is the smallest closed set containing (p-e, p+e) in the usual topology". Solving that problem by looking at the intersection of all closed sets containing that interval is the correct, proper, optimal, and in every sense best way to answer that problem as it was phrased - completely REGARDLESS of how anyone defines the closure of a set, or even if no one had ever even concocted such a definition.

    Thus your "frank" comments "No frankly I do not no see your point" about my approach (where I laid out precisely how my approach precisely answers the problem) tells me that either you don't have a good handle on topology (which I doubt) - or that you're just being childish. I assume the later, but neither befits an experienced teacher.
    Last edited by johnsomeone; September 18th 2012 at 06:56 PM.
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