How Can Compactness and the Tychonoff Theorem Simplify Functions in Topology?

  • Thread starter cogito²
  • Start date
  • Tags
    Topology
In summary, Topology is a branch of mathematics that studies the properties and relationships of spaces that are preserved under continuous deformations. A topological space is a set of points with a collection of subsets that satisfy certain axioms, and compactness is a property of topological spaces that captures the idea of being "small" or "finite." Compactness is closely related to continuity, and both have various applications in fields such as physics, computer science, and engineering.
  • #1
cogito²
98
0
My question comes from homework from a section on the Tychonoff Theorem. This is the question:

Problem said:
Let [tex]Q = I^A[/tex] be a cube, and let [tex]f[/tex] be a continuous real-valued function on [tex]Q[/tex]. Prove, given [tex]\epsilon > 0[/tex], there is a continuous real-valued function [tex]g[/tex] on [tex]Q[/tex] such that [tex]|f - g| < \epsilon[/tex] and [tex]g[/tex] is a function of only a finite number of coordinates. [Hint: Cover the range of [tex]f[/tex] by a finite number of intervals of length [tex]\epsilon[/tex] and look at the inverse images of these intervals.]
Now I have an idea about how to go about this. I know that [tex]Q[/tex] is compact since [tex]I = [0,1][/tex] is and the Tychonoff Theorem states that the product of compact spaces is compact. I then know that [tex]f(Q) \subset \mathbb{R}[/tex] must be compact, since [tex]f[/tex] is continuous. I then know that it is closed and bounded meaning that there is an interval [tex][-M,M][/tex] such that [tex]f(Q) \subset [-M,M][/tex]. I can then cut that interval into a finite number of (non-disjoint) open sets of length [tex]\epsilon[/tex].

I would then look at their inverse images and see that they are open sets covering [tex]Q[/tex]. I know that those open sets are unions of base sets that are products of sets of which only a finite number are not equal to [tex]I[/tex] (because this is the product topology). The problem for me is here is that the open sets cannot necessarilly be written as products (of sets) so I don't see how I can ignore all but a finite number of coordinates. What I would like is for the open sets to be a finite union of base sets. If that were true than I could find a finite number of coordinates that it belongs to but if not than I don't see how I can define it.

Beyond that I would like to define my function [tex]g[/tex] as a sum of functions [tex]g_i[/tex] each defined on the partitions of [tex][-M,M][/tex]. I would be tempted to just define the functions to have a value of part of the partition but then that would result in a step function that wouldn't be continuous anyway. If I could come up with a set of continuous functions [tex]\{g_i\}[/tex] then my plan is to use the Partition of Unity to combine them into one function [tex]x[/tex] that would be the result of the problem.

So I'm pretty stumped at the moment. I just don't see how to simplify it. Hopefully my explanation is understandable. Any help would be greatly appreciated.
 
Physics news on Phys.org
  • #2


Thank you for your question. The Tychonoff Theorem is a powerful tool in topology and can be used to prove this statement. Here is a possible approach to the problem:

1. As you mentioned, Q is compact since it is the product of compact spaces. Therefore, f(Q) is also compact and thus closed and bounded, as you stated.

2. Let's consider the interval [-M,M] and divide it into a finite number of non-overlapping intervals of length ε. Let's denote these intervals as I_1, I_2, ..., I_n.

3. Now, take the inverse images of these intervals under f, i.e. f^-1(I_1), f^-1(I_2), ..., f^-1(I_n). These are open sets in Q, since f is continuous and the intervals are open.

4. Since Q is a product space, each of these inverse images can be written as a product of open sets in the individual spaces A_i. However, as you noted, we cannot necessarily write these open sets as a product of open sets in A_i with only a finite number of them being non-equal to I. But we can make use of the fact that the inverse images are open in Q and use the product topology to write them as a union of base sets, where only a finite number of them are not equal to I. This is possible because the product topology has a base consisting of all finite intersections of open sets in the individual spaces.

5. Now, we have a finite union of base sets that covers each of the inverse images. We can choose one base set from each inverse image and define g_i to be the function that takes the value of the corresponding base set, and 0 everywhere else. Since each g_i is defined on only a finite number of coordinates, their sum will also be a function of only a finite number of coordinates.

6. Finally, we can use the partition of unity to combine these g_i's into one function g, as you suggested. This function will be continuous, since each g_i is continuous on its respective base set and the partition of unity ensures that the sum is continuous everywhere.

I hope this explanation helps and clarifies any doubts you may have. Please feel free to ask any further questions if needed.
 
  • #3



Topology and compactness are important concepts in mathematics, particularly in the field of analysis. Topology deals with the study of the properties of spaces that are preserved under continuous transformations, while compactness is a property of a space that ensures that certain functions defined on that space have nice properties. In this question, we are dealing with the Tychonoff Theorem, which states that the product of compact spaces is compact. This theorem is a powerful tool in topology and is often used to prove various results in analysis.

To solve this problem, we need to use the fact that Q = I^A is compact, as I is compact and the product of compact spaces is compact. We also know that f is continuous on Q, which means that f(Q) \subset \mathbb{R} is compact, closed, and bounded. This allows us to cover f(Q) with a finite number of intervals of length \epsilon. Now, we need to look at the inverse images of these intervals and use the fact that they are open sets covering Q. This means that they can be written as unions of base sets, where only a finite number of these sets are not equal to I. This is because we are dealing with the product topology, where the open sets are defined as unions of products of sets.

However, the problem arises when we try to simplify these open sets into finite unions of base sets. This is because the open sets may not necessarily be written as products of sets, making it difficult to ignore all but a finite number of coordinates. To overcome this, we can use the Partition of Unity, which allows us to combine a finite number of continuous functions to create a new continuous function. By defining our function g as a sum of functions g_i, each defined on the partitions of [-M,M], we can use the Partition of Unity to combine them into one function that satisfies the conditions of the problem.

In summary, to solve this problem, we need to use the fact that Q is compact and f is continuous, and then use the properties of compactness and topology to cover f(Q) with a finite number of intervals. From there, we can use the Partition of Unity to combine a finite number of continuous functions to create a new function g that satisfies the conditions of the problem. This may seem like a complex process, but it is a common approach in topology and analysis problems. With practice, you will become more familiar with these concepts and be able to
 

FAQ: How Can Compactness and the Tychonoff Theorem Simplify Functions in Topology?

What is topology?

Topology is a branch of mathematics that studies the properties and relationships of spaces that are preserved under continuous deformations such as stretching, bending, and twisting. It is concerned with the study of shapes, spaces, and their transformations.

What is a topological space?

A topological space is a set of points with a collection of subsets, called open sets, that satisfy certain axioms. These axioms determine which subsets are considered "open" and allow for the study of continuity and convergence of functions on the space.

What is compactness?

Compactness is a property of topological spaces that captures the idea of being "small" or "finite." A space is compact if every open cover of the space has a finite subcover. Intuitively, this means that the space cannot be "spread out" infinitely.

How is compactness related to continuity?

Compactness is closely related to continuity in topology. In fact, a continuous function on a compact space will always have a maximum and minimum value. This property is known as the Extreme Value Theorem and is a consequence of compactness.

What are some applications of topology and compactness?

Topology and compactness have many applications in various fields of science, including physics, computer science, and engineering. In physics, topology is used to study the properties of materials, while compactness is used in the study of dynamical systems. In computer science, topology is applied in data analysis and image processing. In engineering, topology optimization is used to design efficient and robust structures.

Similar threads

Replies
12
Views
2K
Replies
9
Views
1K
Replies
2
Views
2K
Replies
2
Views
793
Replies
8
Views
2K
Replies
8
Views
687
Replies
11
Views
780
Back
Top