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Minimal basis of a topology

  1. Mar 25, 2008 #1
    I can't seem to find this result in any of my textbooks. Given any basis B for a topology T on X, is there a minimal subset M of B that also is a basis for T (in the sense that any proper subset of M is not a basis for T)? If so, is Zorn's Lemma needed to prove this?
    Is the same true of subbases?

    Attempt at proof using Zorn's Lemma:
    Let B be a basis for a topology T on X. Let A be the collection of all bases for T that is a subcollection of B. A is not empty because B is in A. Partially order A by set containment (i.e. D < E iff D contains E). Let C = {C_i} be a totally ordered subcollection of A. Let K = n(C_i) (intersection). We must show that K is a basis for T. Let U be a T-open set, and let x be in U. Since each C_i is a basis for T, then for each i, there exists C in C_i such that x is in C is a subset of U. Wait, C needs not be the same, and C needs not be in K.

    Is the assertion false? What's a counterexample?
    Last edited: Mar 25, 2008
  2. jcsd
  3. Mar 26, 2008 #2
    How about this weaker assertion: Every topology T contains a minimal basis B for T (in the sense that any proper subset of B is not a basis for T).

    This must be true, right? And the same for subbases? But Zorn's Lemma still doens't work.
  4. Mar 27, 2008 #3
    My question even stumped topologist Henno Bradsma. He said:

    "I found a result that a metric space has a minimal subbase (proved by van Emde Boas).
    So probably not all spaces have them...
    All finite spaces have a unique minimal base. This is all U_x, where U_x = /\{U: U open and x in U},
    for x in X. Note that this argument works in all spaces where all intersections of open sets are open
    (sometimes called Alexandrov spaces), so e.g. it's true in discrete spaces.
    I think no base for R can be minimal, e.g., so the general result for bases seems false to me.

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