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Figure out some properties of density of a set

  1. Jan 22, 2005 #1
    I am trying to figure out some properties of density of a set, but I keep confusing myself.

    I know the definition of dense is: A set E of real numbers is said to be dense if every interval (a, b) contains a point of E. Could I rephrase the definition so that every interval (a, b) contains INFINITELY many points of E?

    I am trying to justify that by saying that if (a, b) is dense, then there must be a point c in E such that a < c < b. But then (a, c) must be dense, since it is contained in (a, b), and so we could find another point d such that a < d < c, and keep going until we find infinitely many points.

    But that leads me to another question: If you have two sets, E1 and E2 that are dense, what do you know about E1 intersect E2? Would E1 intersect E2 be dense if E1 is contained in E2? Is there a way to prove the conclusion about the intersection?

    Also, if a set E is dense, then what do you know about the set A, where E is contained in A?

    Thanks in advance for any insight!
     
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  3. Jan 22, 2005 #2

    dextercioby

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    The part of topology that i know (maybe Matt and/or Hurkyl could prove themselves more rigurous) states that given a topological space (a set+topological structure) A,a subset B(it can be even A,that's a trivial case) is called DENSE EVERYWHERE IN THE SET "A",if for any point "x" in the set A,there is a point "y" from the subset B which is included in a vecinity of "x".
    For metric spaces (topology induced by metric/distance) this definition can take a rigurous form...
    R is a topological space with the structure induced by the metric (the modulus/absolute value).

    So i cannot really follow you on DENSE SETS.The way i know it,u must have a set (topological space) and a subset about whom u can say it is dense everywhere...I cannot remember any "density" without "everywhere density"... :confused:

    Daniel.

    P.S.The advice:read more into maths...Me too,i know... :tongue2:
     
  4. Jan 22, 2005 #3

    Hurkyl

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    The phrase "the rationals are dense in the reals" is a fairly standard, I believe. I'm also mildly unsettled by just calling a set "dense", unless the ambient space was previously made clear.


    Several of your questions are quite straightforward to prove, once you write down your questions mathematically, and apply the definitions.

    I think you already suspect that the intersection of dense sets might not be dense: have you tried much to come up with an example of two dense sets whose intersection isn't?

    Your proof of your first question looks like a good start... but a proof by contradiction would be easiser.
     
  5. Jan 22, 2005 #4
    How would I go about doing a proof by contradiction? Suppose that (a, b) is dense. Also suppose that there are a finite number of elements of E in (a, b). To me this doesn't seem easier. Where would I go from here?

    Also, I know that (1, 2) intersect (3, 4) would not be dense, but that (1, 2) intersect (1.4, 1.7) would be dense. So the intersection of two dense sets will be dense if one set is contained in the other?
     
  6. Jan 22, 2005 #5

    Hurkyl

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    Draw a picture of a specific example -- that might suggest something.


    By the definition you gave, (1, 2) is not dense: for example, the interval (7, 96) does not contain a point of (1, 2).


    In general, if one set contains another set, what is their intersection?
     
  7. Jan 22, 2005 #6

    HallsofIvy

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    It doesn't make sense to say "(a,b) is dense". Dense in what space? It certainly IS dense in itself- every set is. It certainly is NOT dense in the space of all real numbers or any interval containing (a,b).

    To answer your original question: the definition of "A is dense in B" is "every neighborhood of B contains a member of A" could, in fact, be replaced with "every neighborhood of B contains an infinite number of members of A" IF every neighborhood necessarily has another neighborhood as subset. That's true in any metric space- in particular in the real numbers with the usual topology. It is not, however, true in a space with, for example, the "indiscreet topology" in which the only neighborhoods are the empty set and the entire space.
     
  8. Jan 22, 2005 #7
    Actually, the definition of density implies that a dense interval (a,b) must contain an infinitely many points. Take any point one the number axis, now you can locate that number by two other numbers to any arbitrary degree of accuracy, which means that you can find a number which is arbitrarily close to any other number in the set.

    If you have two sets which are dense, then their intersection would also be dense if some part of one set is a subset of the other. If their intersection is a null set then it is not dense because it doesnt contain any elements.
     
  9. Jan 22, 2005 #8

    Hurkyl

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    Given any set not dense in the reals, I can find two sets that are dense in the reals, and have that set as their intersection. :tongue2:
     
  10. Jan 23, 2005 #9
    the definition i learned was "A is dense in B if the [tex]\overline{A} = B[/tex] " (closure of A is B). or any [tex]a \in A[/tex] is within [tex]\epsilon[/tex] of a [tex]b \in B[/tex] so that looks like it might work
     
  11. Jan 23, 2005 #10

    HallsofIvy

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    Assuming you are talking about a metric space (as I said before the theorem "if a set A is dense in set B [every neighborhood of set B contains a point of set A] then every neighborhood of set B contains an infinite number of points of set A" is untrue in the indiscreet topology), then it is simplest to prove this by contradiction.

    Suppose N&epsilon;(p) is a neighborhood of B (centered on P with radius &epsilon;) which contains only a finite number of points of A. Then there exist a point of A, call it x, which is closest to P (the set of distances from from points of A to P is finite and so contains a smallest member). Let &delta;= d(P,x), the distance from P to X. Then the neighborhood N&delta;(P) contains NO points of A contradicting the hypothesis that A is dense in B.
     
  12. Jan 23, 2005 #11
    yeah but i didn't think it was necessary to go into a lot of topology since the original post didn't mention it & i just assumed that she/he is just in an intro analysis course.

    just in case people aren't confused enough, here's the algebraic definition of density, which is equivalent to the one given in analysis courses:
    Let V be a (left) vector space over a division ring D. A subring of R of the endomorphism ring [tex]Hom_D(V,V)[/tex] is called a dense ring of endomorphisms of V (or a dense subring of [tex]Hom_D(V,V)[/tex]) if for every positive integer n, every linearly independent subest [tex]{u_1, u_2, ... , u_n}[/tex] of V and every arbitrary subset [tex]{v_1, v_2, ... , v_n}[/tex] of V, there exists [tex]\theta \in R[/tex] such that [tex]\theta(u_i) = v_i, (i=1, 2, ..., n)[/tex]

    if V has the discrete topology & we take the product of Vs then density in this topology is the same thing. or if V is a normed space & we put the strong operator topology on [tex]Hom_D(V,V)[/tex] & we do [tex]|\theta(u_i) - v_i| < \epsilon[/tex] & let [tex]\epsilon \rightarrow 0 [/tex] we get algebraic density also. there's no point to any of that here but since a lot of people were throwing in stuff that has nothing to do with a little density problem (which is obviously in R with the usual metric, standard topology, etc) i thought i throw it out there for people.
     
  13. Jan 24, 2005 #12

    matt grime

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    You algebraic defintion elides some very important points, not least the fact that you're talking about norms tending to zero. What if it were a division ring over the algebraic closure of F_p?
     
  14. Jan 24, 2005 #13
    it's not "my" definition, it's the standard definition given in a standard algebra text (the one by hungerford). not that it matters though, i still don't nderstand why you people want to go off on tangents pontificating about the "finer points" of topology when it clearly has nothing to do with the original post, and probably does nothing but confuse the original poster.
     
  15. Jan 25, 2005 #14

    matt grime

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    Ahh, i failed to notice the assumption of the existence of the norm, idiot that I am.
     
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