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Coarsest Topology With Respect to which Functions are Continuous

  1. Feb 13, 2017 #1
    1. The problem statement, all variables and given/known data
    See attached picture.


    2. Relevant equations


    3. The attempt at a solution
    At the moment, I am dealing with part (a). What I am perplexed by is the ordering of the parts. If the subbasis in part (b) does indeed generate this coarsest topology, why wouldn't showing this be included in part (a)? What does the author intend I do for part (a), if not showing that the subbasis in part (a) generates this coarsest topology. To me it seems that, in order to properly solve part (a), I need to explicitly construct a topology relative to which are functions are continuous . But how do I this if not by constructing a collection of subbasic elements like that in part (b)?
     

    Attached Files:

  2. jcsd
  3. Feb 15, 2017 #2

    fresh_42

    Staff: Mentor

    "What does the author intend?" can only be guessed. I'd start as with almost every exercise: write down (and tell us) the definition of a coarset topology, maybe the one of "spaces" as well, since I missed the word "topologiocal" and the definition of a continuous function.
    This helps to clarify what has to be done, might reveal what the author intended, but most of all, give as a common base to talk (section 2!).
     
  4. Feb 20, 2017 #3
    If ##\tau## and ##\tau'## are two topologies on a given topological space, ##\tau## is said to be coarser than ##\tau'## if and only if ##\tau \subseteq \tau'## (alternatively, one could say that ##\tau'## is finer).

    The ##X_i## are indeed topological spaces (note, I replaced ##\alpha## with ##i##, as ##\alpha## will soon prove too cumbersome)

    Let ##X## and ##Y## be topological spaces. A function ##f : X \rightarrow Y## is said to be continuous, provided ##f^{-1}(V)## is open for every open set in ##Y##.

    Also, one last definition: A collection ##\mathcal{S}## of subsets of ##X## is said to be a subbasis if and only if the union of over all of ##\mathcal{S}## equals ##X##. The topology generated by the subbasis ##\mathcal{S}## is defined to be the collection of all unions of finite intersections of elements in ##\mathcal{S}##?

    Here is a stab at solving part (a). Let ##S_i := \{f^{-1}_i(U_i) ~|~ U_i \mbox{ open in } X_i \}##. Clearly the collection ##\mathcal{S} = \bigcup S_i## serves as a subbasis for a topology on ##A##, since ##f_i^{-1}(X_i) = A## for every ##i##. Let ##\tau_{\mathcal{S}}## denote this topology. Moreover, each ##f_i## is clearly continuous with respect to the topology on ##A## generated by ##\mathcal{S}##, since if ##U_i## is open in ##X_i##, then by definition ##f^{-1}_i(U_i) \in \mathcal{S} \subseteq \tau_{\mathcal{S}}##, implying that by definition ##f^{-1}_i(U_i)## is open in. Whence it follows that each ##f_i## is continuous. Now, suppose that ##\tau \subseteq \tau_{\mathcal{S}}## is a topology on ##A## with respect to which each ##f_i## is continuous. Then for every ##U_i## open in ##X_i##, ##f_i^{-1}(U_i) \in \tau##. But ##f_i^{-1}(U_i) \in \mathcal{S}##, and so this says precisely that ##\mathcal{S} \subseteq \tau## and therefore ##\tau_{\mathcal{S}} \subseteq \tau##, since ##\tau## is closed under arbitrary unions of finite intersections. From this is follows ##\tau_\mathcal{S} = \tau##, that ##\tau_\mathcal{S}## is the coarsest topology with respect to which each ##f_i## is continuous.

    However, as you may notice, in solving part (a) I need first to solve part (b) . Thus the reason for my question in my first post: how are we expected to solve part (a) before part (b)?
     
    Last edited: Feb 20, 2017
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