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Finer/coarser topologies

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose [itex]\tau_1[/itex] and [itex]\tau_2[/itex] are 2 topologies on a set [itex]X[/itex] and that [itex]\tau_2 \subseteq \tau_1[/itex]. We say that [itex]\tau_1[/itex] is stronger/finer than [itex]\tau_2[/itex] and that [itex]\tau_2[/itex] is weaker/coarser than [itex]\tau_1[/itex].

    Show, directly from the definitions, that if:

    (a) [itex]A \subseteq X[/itex] is closed in [itex](X,\tau_2)[/itex] then [itex]A[/itex] is closed in [itex](X,\tau_1)[/itex] ;

    (b) [itex](Y,\tau_Y)[/itex] is another topological space and [itex]f[/itex] is a continuous map from [itex](Y,\tau_Y)[/itex] to [itex](X,\tau_1)[/itex] then [itex]f[/itex] is continuous from [itex](Y,\tau_Y)[/itex] to [itex](X,\tau_2)[/itex].

    3. The attempt at a solution

    For (a), if [itex]A \subseteq X[/itex] is closed in [itex](X,\tau_2)[/itex] then, by definition, [itex]\partial A \subseteq A[/itex], but this is precisely the definition of [itex]A[/itex] being closed in [itex](X,\tau_1)[/itex] (the definition is dependent on the set, not the topology).

    Is this OK - is there a better way to show it?

    For (b), [itex]f:(Y,\tau_Y) \to (X,\tau_1)[/itex] is continuous if for every open set [itex]A\subseteq X,\; f^*(A)[/itex] is open in [itex]Y[/itex]. Again, isn't this just the definition of [itex]f:(Y,\tau_Y) \to (X,\tau_2)[/itex] being continuous?
     
  2. jcsd
  3. Dec 6, 2011 #2
    This last comment isn't true, at least not the way I understand it. What is your definition of the boundary of A? It must depend on the topology of the space in some way. Where have you used the fact that [itex] \tau_2 \subseteq \tau_1 [/itex]?
    The definition of closed I learned for a closed set is F is closed iff [itex] X \setminus F [/itex] is open. You could try starting there.

    Again, where have you used the fact that [itex] \tau_2 \subseteq \tau_1 [/itex]? Maybe you're just omitting these because they seem clear, but I think it's important to mention where you use each piece of information.
     
  4. Dec 7, 2011 #3
    [itex]\tau_2 \subseteq \tau_1[/itex] means every [itex]\tau_2[/itex]-open set is [itex]\tau_1[/itex]-open so for (a):

    [itex]A[/itex] is closed in [itex](X,\tau_2) \implies A^c[/itex] is [itex]\tau_2[/itex]-open in [itex]X[/itex]

    [itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies A^c[/itex] is [itex]\tau_1[/itex]-open in [itex]X[/itex] (since [itex]\tau_2 \subseteq \tau_1[/itex])

    [itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies A[/itex] is closed in [itex](X,\tau_1)[/itex]

    Does that look better?

    However is every [itex]\tau_1[/itex]-open set [itex]\tau_2[/itex]-open? As for (b):

    [itex]f:(Y,\tau_Y) \to (X,\tau_1)[/itex] is continuous [itex]\implies[/itex] for every [itex]\tau_1[/itex]-open set [itex]A \subseteq X,\; f^*(A)[/itex] is [itex]\tau_Y[/itex]-open in [itex]Y[/itex]

    Does this imply the following?

    [itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies[/itex] for every [itex]\tau_2[/itex]-open set [itex]A \subseteq X,\; f^*(A)[/itex] is [itex]\tau_Y[/itex]-open in [itex]Y[/itex] (since [itex]\tau_2 \subseteq \tau_1[/itex]??)


    [itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \implies f:(Y,\tau_Y) \to (X,\tau_2)[/itex] is continuous
     
  5. Dec 7, 2011 #4
    Looks good.
    This would only be true if [itex] \tau_1 \subseteq \tau_2 [/itex], in which case the 2 topologies are equal.

    This is fine, but you seem unsure, so maybe it would be better to start with an arbitrary open set O in [itex] (X, \tau_2) [/itex] and prove that [itex] f^{-1}(O) [/itex] is open in Y.
     
  6. Dec 7, 2011 #5
    The thing I was unsure about was every [itex]\tau_1[/itex]-open set... implying every [itex]\tau_2[/itex]-open set...

    (as you say, wouldn't this only be true if [itex] \tau_1 \subseteq \tau_2 [/itex], in which case the 2 topologies are equal?)
     
  7. Dec 7, 2011 #6
    You've got it backwards. You don't need every set in [itex] \tau_1[/itex] to be in [itex] \tau_2 [/itex] to prove the statement in part b). As I said before, try starting with an arbitrary open set O in [itex] (X, \tau_2) [/itex] and prove that [itex] f^{-1}(O) [/itex] is open in Y. Writing it out is the best way to make this clear.
     
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