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Topology generated by interior operator

  1. May 3, 2012 #1
    Given an interior operator on the power set of a set X, i.e. a map [itex]\phi[/itex] such that, for all subsets [itex]A,B[/itex] of [itex]X[/itex],

    [tex](IO 1)\enspace \phi X = X;[/tex]

    [tex](IO 2)\enspace \phi A \subseteq A;[/tex]

    [tex](IO 3)\enspace \phi^2A = \phi A;[/tex]

    [tex](IO 4)\enspace \phi(A \cap B) = \phi A \cap \phi B,[/tex]

    I'm trying to show that the set

    [tex]\tau = \left \{ U \in 2^X | \phi U = U \right \}[/tex]

    is a topology for [itex]X[/itex]. I've shown everything except that [itex]\tau[/itex] is closed under arbitrary unions. By (IO 2),

    [tex]\phi\left ( \bigcup_{\lambda \in \Lambda}U_\lambda \right ) \subseteq \bigcup_{\lambda \in \Lambda}U_\lambda = \bigcup_{\lambda \in \Lambda}\phi U_\lambda.[/tex]

    So all that remains is to show that

    [tex]\bigcup_{\lambda \in \Lambda}U_\lambda = \bigcup_{\lambda \in \Lambda}\phi U_\lambda \subseteq \phi\left ( \bigcup_{\lambda \in \Lambda}U_\lambda \right ).[/tex]

    Any hints? I've used all of the axiom so far except (IO 3), so I'm guessing this must be involved somehow ... I've also done the corresponding exercise for a closure operator and got stuck on at the same point.
     
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  3. May 3, 2012 #2

    micromass

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    We know that

    [itex]A_i\subseteq \bigcup A_i[/itex]

    Now take [itex]\phi[/itex] of both sides and take unions.

    You will have to show for this that

    [tex]A\subseteq B~\Rightarrow~\phi A\subseteq \phi B[/tex]

    somehow. For this, notice that

    [tex]A\subseteq B~\Leftrightarrow~A\cap B=A[/tex]
     
  4. May 3, 2012 #3
    Great, thanks micromass! That's just the hint I needed. The only thing that puzzles me now is that I don't think I used (IO 3) anywhere in showing that tau is a topology.

    X is in tau by (IO 1). The empty set is in tau by (IO 2). (IO 4) gives us finite intersections directly. And I used (IO 2) to get arbitrary unions: if each [itex]U_\lambda = \phi U_\lambda[/itex], then

    [tex]\phi\left ( \bigcup_{\lambda \in \Lambda}U_\lambda \right )\subseteq \bigcup_{\lambda\in\Lambda}\phi U_\lambda = \bigcup_{\lambda\in\Lambda}U_\lambda.[/tex]

    Did I use (IO 3) implicitly somewhere, without knowing it? Or is it superfluous, or what is it for?
     
  5. May 3, 2012 #4

    micromass

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    Only showing that you got a topology is not good enough. There are still 2 other things to verify:

    1) [itex]\phi[/itex] is the interior operator of your topology. Indeed, you got a topology, but it might have a completely different interior operator than what we intend.

    2) The topology determined by [itex]\phi[/itex] is unique. Indeed, there might be more topologies with this same interior operator.
     
  6. May 3, 2012 #5
    1) Ah, I see. Here's my proof that [itex]\phi A = A^\circ[/itex], the interior of [itex]A[/itex] with respect to the topology [itex]\tau = \left \{ U \subseteq X \; | \;\phi U = U \right \}[/itex]:

    [tex]\left ( x\in A^\circ \right ) \Leftrightarrow \left ( \left ( \exists U\in\tau \right )\left [ x\in U\subseteq A \right ] \right )[/tex]

    [tex]\Leftrightarrow ( ( \exists U\in\tau ) [ ( x\in U ) \& ( U\subseteq A ) ] )[/tex]

    [tex]\Leftrightarrow \left ( \left ( \exists U\in\tau \right )\left [ (x\in \phi U )\&(\phi U \subseteq \phi A)\right ] \right )[/tex]

    [tex]\Rightarrow (x\in\phi A).[/tex]

    Conversely, by (IO 2),

    [tex](x\in\phi A) \Rightarrow (x\in A).[/tex]

    And by (IO 3), [itex]\phi A = \phi (\phi A)[/itex], so [itex]\phi A \in\tau[/itex], so letting [itex]U = \phi A[/itex], we see that

    [tex](x\in \phi A)\Rightarrow ((\exists U\in\tau)[(x\in U \subseteq A)])[/tex]

    [tex]\Leftrightarrow (x\in A^\circ). \enspace\blacksquare[/tex]

    2) I'm not sure where to start with this one. I suppose it amounts to proving that the definition of the set

    [tex]\tau = \left \{ U \subseteq X \; | \;\phi U = U \right \}[/tex]

    contains a contradiction, so that the set is not well defined. One can prove the existence of a contradiction by finding one, but how to prove the nonexistence of a contradiction?
     
  7. May 4, 2012 #6

    micromass

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    For (2), you must take a topological space [itex](X,\mathcal{T})[/itex] such that [itex]\phi[/itex] is its interior operator. You must prove that a set U is open if and only if [itex]\phi U=U[/itex].
     
  8. May 4, 2012 #7
    Ah, I see. I showed that an arbitrary interior operator gives a topology, defined as the sets which are invariant under this operator, and the definition of interior specified by that topology corresponds to the definition of interior specified by the arbitrary interior operator we started with. Then I'd need to show that the definition of interior, [itex](\cdot)^\circ[/itex], specified by an arbitrary topology, [itex]\tau[/itex], is the specific interior operator, [itex]\phi[/itex], for which the elements of [itex]\tau[/itex] are exactly those sets for which [itex]\phi U = U[/itex]; that is, open sets are exactly those which are equal to their interior.

    For this final part, suppose a set is open. Then it is the union of all of its open subsets, and thus equal to its interior. Conversely, suppose a set is equal to its interior. Then it's the union of all of its open subsets, and a union of open subsets is open, so it must be open.

    (To completely show that [itex](\cdot)^\circ[/itex] is an interior operator, as defined by the axioms above, I'd also have to derive each of these axioms as a property of [itex](\cdot)^\circ[/itex]. I read such a derivation earlier and copied it in my notebook, but it won't hurt to derive it again from scratch, to reinforce the ideas...)
     
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