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Subbasis vs. basis

  1. Feb 11, 2009 #1
    This is probably a stupid question.. but

    Can someone tell me the difference between subbasis and basis.. in topology?? I know the definitions...

    So Subbasis is defined to be the collection T of all unions of finite intersections of elements of S (subbasis)

    sooo... S is pretty much a topology on X which is a collection of subsets of X whose union equals X.

    Basis, however... is
    If X is a set, basis on X is a collection B of subsets of X (basis elements) s.t.
    1. for each x [tex]\in[/tex] X, there is at least one basis element B containing x.
    2. If x belongs to the intersection of two basis elements B1 and B2, then there is a basis element B3 containing x such that B3[tex]\subset[/tex] B1[tex]\cap[/tex]B2.

    Right? So pretty much... A subset U of X is said to be open in X if for each x [tex]\in[/tex] U, there is a basis element B [tex]\in[/tex] [tex]B[/tex] such that x [tex]\in[/tex] B and B [tex]\subset[/tex] U.

    But I'm still not understanding this quite... so well..

    Can someone explain this to me??


    Thank You!
     
  2. jcsd
  3. Feb 12, 2009 #2
    The topology generated by a subbasis is the topology generated by the basis of all finite intersections of subbasis elements. It seems like you just need an example, though, so here's one.

    A basis for the standard topology of R is the collection of all open intervals. A subbasis for R is the collection [tex]\{(a,\infty):a\in \mathbb{R}\} \cup \{(-\infty,b):b\in \mathbb{R}\}[/tex]. The reason that this is a subbasis for R is because finite intersections of elements in this set are precisely the basis elements of R. For instance, we have [tex](a,b) = (-\infty,b)\cap (a,\infty) [/tex]. Subbases are often important because they offer an easier way of expressing a topology. In the previous example, there are many fewer sets in the sub-basis than there are in the basis.
     
  4. Feb 12, 2009 #3
    I think there is a distiction that can help you avoid confusion.

    If you have a set [tex]X[/tex] and NO topology on it
    every [tex]S[/tex] subset of the set [tex]P(X)[/tex] of the parts of [tex]X[/tex]
    can be taken and you can generate e topology that has [tex]S[/tex] as subbasis.

    If you have a set [tex]X[/tex] and NO topology on it then
    if you have a set [tex]B \subset P(X)[/tex]
    that satisfies the 2 property 1. ans 2. you mentioned you can generate a topology on
    [tex]X[/tex] that has [tex]B[/tex] has basis.


    Now if you have a set [tex]X[/tex] and HAVE a topology [tex]T[/tex] on it
    a set [tex]S \subset P(X)[/tex] is a subbasis of [tex](X,T)[/tex] if and only if (by definition)
    for every open set [tex]A[/tex] and for every [tex]a \in A[/tex] exists
    [tex]S_1, \ldots, S_n \in S[/tex] such that
    [tex]a \in S_1 \cap \cdots S_n \subset A[/tex].


    Now if you have a set [tex]X[/tex] and HAVE a topology [tex]T[/tex] on it
    a set [tex]B \subset P(X)[/tex] is a basis of [tex](X,T)[/tex] if and only if (by definition)
    for every open set [tex]A[/tex] and for every [tex]a \in A[/tex] exists
    [tex]B_1 \in B[/tex] such that
    [tex]a \in B_1 \subset A[/tex].



    There is a slight difference to understand.
    Please read it carefully and think on it and you'll get the concept.
    Hope this helps.
     
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