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Topology generated by a collection of subsets of ##X##

  1. Aug 14, 2014 #1
    The problem statement, all variables and given/known data.

    Let ##X## be a set and ##\mathcal A \subset \mathcal P(X)##. Prove that there is a topology ##σ(A)## on ##X## that satisfies

    (i) every element of ##A## is open for ##σ(A)##

    (ii) if ##\tau## is a topology on ##X## such that every element of ##\mathcal A## is open for ##\tau##, then ##σ(A) \subset \tau##

    The attempt at a solution.

    I suppose the idea is to construct the smallest possible topology which contains ##\mathcal A##

    As the elements of ##\mathcal A## have to be open, then each ##U \in A## has to be in ##σ(A)##. Also, ##X,\emptyset## have to be in ##σ(A)##.

    Since ##σ(A)## has to be closed under finite intersections and arbitrary unions, I'll have to add all sets that come from finite intersections of ##X## with elements of ##\mathcal A##.

    So my idea was to define ##σ(A)=\{S \subset X : S=\bigcap_{i \in I} U_i, U_i \in \mathcal A, I \text{finite}\} \cup \{X\}## but when I've tried to show that ##σ(A)## was a topology I got stuck, so I don't know if that set works.

    It is clear that ##X \in σ(A)##, and that if ##S_j \in σ(A)## for ##j \in J## with ##J## finite, then since ##S_j=\bigcap_{k=1}^{s_j} U_k##, ##\bigcap_{j \in J} S_j=\bigcap_{j \in J} (\bigcap_{k=1}^{s_j} U_k)##, which is clearly a finite intersection of elements in ##σ(A)##. The problem is with arbitrary unions, I couldn't show that arbitrary unions remain in ##σ(A)##. Where is my mistake?
     
  2. jcsd
  3. Aug 14, 2014 #2

    jbunniii

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    There's a much slicker way to do this. Hint: the intersection of an arbitrary number of topologies is a topology.
     
  4. Aug 14, 2014 #3
    ... as long as that arbitrary number isn't 0.
     
  5. Aug 14, 2014 #4

    jbunniii

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    Yes! Good catch. The OP will need to show that there is at least one suitable topology being intersected to form his desired topology.
     
  6. Aug 14, 2014 #5

    Oh, ##\bigcap_{i \in I, \mathcal A \subset \tau_i} T_i##, the intersection of all topologies which contain ##\mathcal A## is a topology (I've verified that intersection of an arbitrary number different from ##0## of topologies is a topology, and if ##\tau## is a topology such that every element in ##\mathcal A## is open in ##\tau##, this implies ##\mathcal A \subset \tau##, but then ##\bigcap_{i \in I, \mathcal A \subset \tau_i} T_i \subset \tau##. Thanks very much for the super hint, is this correct now?
     
  7. Aug 14, 2014 #6

    jbunniii

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    Yes. The intersection is not empty because ##\mathcal{P}(X)## is a topology containing ##A##. Any topology containing ##A## must be one of the ##T_i##, so it contains the intersection.

    This kind of argument is used all the time. "The intersection of all _____ with property P (assuming there is at least one) is a _____ with property P, and therefore it is the smallest such _____." It's an elegant and standard argument, but completely nonconstructive and so it may not give much insight into what the elements of the intersection actually look like.
     
    Last edited: Aug 14, 2014
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