Coarsest Topology With Respect to which Functions are Continuous

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SUMMARY

The discussion centers on the concept of coarsest topology in relation to continuity of functions within topological spaces. The participants emphasize the necessity of defining a coarsest topology and the role of subbases in generating such topologies. Specifically, the collection of subsets serving as a subbasis must cover the entire space, and the topology generated from this subbasis is crucial for demonstrating the continuity of functions. The confusion arises from the ordering of parts in the homework assignment, where understanding part (b) is essential for solving part (a).

PREREQUISITES
  • Understanding of topological spaces and their properties
  • Familiarity with the definitions of coarsest topology and continuous functions
  • Knowledge of subbases and their role in topology
  • Ability to work with unions and intersections of sets in a topological context
NEXT STEPS
  • Study the definition and properties of coarsest topology in detail
  • Learn how to construct subbases and their generated topologies
  • Explore examples of continuous functions between topological spaces
  • Review the relationship between different topologies on the same space
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Mathematics students, particularly those studying topology, educators teaching advanced mathematics, and anyone interested in the foundational concepts of continuity and topological structures.

Bashyboy
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Homework Statement


See attached picture.

Homework Equations

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)?
 

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Bashyboy said:

Homework Statement


See attached picture.

Homework Equations

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)?
"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!).
 
fresh_42 said:
definition of a coarset topology

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).

fresh_42 said:
maybe the one of "spaces" as well, since I missed the word "topologiocal"

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

fresh_42 said:
definition of a continuous function

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)?
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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