Coarsest Topology With Respect to which Functions are Continuous

In summary, the author is trying to solve a homework equation for a function, but is having trouble because it seems that he needs to solve part (b) first.
  • #1
Bashyboy
1,421
5

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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  • #2
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!).
 
  • #3
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)?
 
Last edited:

1. What is the coarsest topology?

The coarsest topology is a type of topology that is defined as the smallest or least restrictive topology that can be placed on a set while still satisfying certain conditions.

2. What does it mean for a function to be continuous with respect to a coarsest topology?

A function is continuous with respect to a coarsest topology if and only if its inverse image preserves open sets in the topology. This means that if an open set in the topology is mapped to another set, the inverse image of that mapped set will also be open in the topology.

3. How is the coarsest topology related to other topologies?

The coarsest topology is often used as a starting point for defining other topologies. It can be seen as a building block for creating more complex topologies by adding additional restrictions or open sets.

4. What are some real-world applications of the coarsest topology?

The coarsest topology is used in a variety of fields, including physics, engineering, and computer science. It is often used in the study of dynamical systems and differential equations, as well as in optimization and control theory.

5. How is the coarsest topology different from the finest topology?

The coarsest topology is defined as the smallest or least restrictive topology, while the finest topology is defined as the largest or most restrictive topology. This means that the coarsest topology has fewer open sets and is less constraining compared to the finest topology.

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