/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Show that to every open cover \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 31

Show that to every open cover \(\mathcal{C}\) of a compact set \(K\) in a metric space there corresponds a positive number \(L\) (called the Lebesgue number of the cover) such that if \(x, y \in K\) and \(d(x, y)

Short Answer

Expert verified
There exists a positive number \( L \), called the Lebesgue number, such that if \( d(x,y)<L \) for any \( x, y \in K \), both are contained in a member of the cover.

Step by step solution

01

Understanding the Problem

We need to demonstrate that for every open cover of a compact set \( K \) in a metric space, there exists a positive number \( L \), known as the Lebesgue number. This number has the property that any two points in \( K \) that are within a distance \( L \) of each other must both lie in at least one open set of the cover.
02

Property of Compact Sets

Recall that a compact set \( K \) in a metric space has the property that every open cover has a finite subcover. This is crucial for our proof because it ensures the existence of some control over the covering sets.
03

Existence of Finite Subcover

Given that \( \mathcal{C} \) is an open cover of \( K \), we know there exists a finite subcover \( \mathcal{C}' = \{ U_1, U_2, \ldots, U_n \} \) that also covers \( K \). This means that each point in \( K \) must lie inside at least one of the \( U_i \).
04

Cover the Closed Set with Open Balls

For each point \( x \in K \), because the sets \( U_i \) are open, there exists a ball \( B(x, r_x) \) contained in some \( U_i \). Choose \( r_x \) such that \( B(x, r_x) \subseteq U_i \). The collection of these balls for all points in \( K \) serves as an open cover of \( K \).
05

Choose the Minimum Radius

Define \( L = \min \{ r_x : x \in K \} \), where each \( r_x \) is chosen as above. Since \( K \) is compact and the values \( r_x \) are positive (each being the radius of a ball entirely contained in some \( U_i \)), \( L \) is a positive number.
06

Show \( L \) is the Lebesgue Number

For any \( x, y \in K \) where \( d(x, y) < L \), both \( x \) and \( y \) are within distance less than \( L \) of each other. Since the overlap of any such ball \( B(x, L) \) with \( K \) must be entirely contained in some \( U_i \) (by the choice of \( L \) being the minimum radius so that every ball \( B(x, \text{radius}) \) stays within some \( U_i \)), \( x \) and \( y \) must lie in a common member \( U \in \mathcal{C} \).
07

Conclusion

Since \( L \) is a positive number and satisfies the condition for each pair \( x, y \) with \( d(x, y) < L \), every such pair is covered by at least one open set in \( \mathcal{C} \). Thus, \( L \) is indeed a Lebesgue number for the cover \( \mathcal{C} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understand Compact Sets
The concept of compact sets is central in topology, especially in metric spaces. A set is known as "compact" if every open cover of the set has a finite subcover.
In simpler terms, if you can cover a compact set with a collection of open sets, you can choose a finite number of those to cover the whole set.
This property is crucial because it implies that compact sets are "manageable" in size and structure within a metric space.
  • Compactness helps in establishing the existence of certain properties, like the Lebesgue number.
  • It's particularly useful when dealing with continuous functions and convergence.
Compact sets can be seen as a generalization of closed and bounded sets in Euclidean spaces, highlighting their role in connecting geometry and analysis.
The Role of Open Covers
An open cover is a collection of open sets whose union includes the entire set that is being covered. If you have a set, say a subset of a metric space, an open cover is a way of wrapping this set with open sets, like a patchwork quilt.
Each piece of the quilt, or open set, contributes to covering the whole set.
  • An open cover of a set determines how the set can be pieced together using open sets.
  • This concept is fundamental in topology and is used to prove various properties about sets, such as compactness.
Because any compact set in a metric space has the property that any open cover has a finite subcover, understanding open covers is essential to utilizing the compactness criterion.
Understanding Metric Spaces
A metric space is a set equipped with a distance function, called a metric. This function gives a measure of the "distance" between any two points within the set. A key characteristic of metric spaces is their structure, enabling a rigorous way to define and analyze concepts like continuity, convergence, and compactness.
  • The metric satisfies specific properties: non-negativity, identity of indiscernibles, symmetry, and the triangle inequality.
  • Metric spaces provide a framework for discussing concepts of closeness and limits, central to calculus and analysis.
Understanding the metric is crucial for grasping the notion of compact sets, as it allows you to talk precisely about how far objects are from each other, enabling concepts like open covers and finite subcovers to be rigorously applied.
Navigating Finite Subcover
When dealing with open covers of a set, especially for compact sets, the idea of a finite subcover becomes vital. A finite subcover is a selection of a finite number of open sets from an open cover that still manages to completely cover the compact set.
This concept is crucial when working with compact sets because it gives a controlled way to manage potentially infinite processes and ensures that any infinite collection of cover sets can be simplified to a finite one.
  • The existence of a finite subcover demonstrates the finite nature of open covers on compact sets.
  • It ensures that discussions about covering compact sets don't get lost in infinite details.
In essence, while the entire open cover might be infinite, the power of compactness is in ensuring a finite subcover, making the concept of a finite subcover indispensable when navigating the landscape of real analysis and topology.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f: X \rightarrow X\) be a continuous mapping from a compact space \(X\) into itself. Define the sequence of sets $$ X_{1}=f(X), X_{2}=f\left(X_{1}\right), \ldots, X_{n}=f\left(X_{n-1}\right) $$ Let \(K=\bigcap_{i=1}^{\infty} X_{i} .\) Show that \(K\) is nonempty, compact, and invariant under \(f\) in the sense that \(f(K)=K\).

Show that a metric space \((X, d)\) is compact if and only if for every family \(\mathcal{F}\) of closed subsets of \(X\) for which $$ \bigcap_{F \in \mathcal{F}} F=\emptyset $$ there must be a finite collection \(F_{1}, F_{2}, \ldots, F_{m}\) of sets in \(\mathcal{F}\) so that $$ \bigcap_{i=1}^{m} F_{i}=\emptyset $$

This problem requires a bit of knowledge about complex numbers and transcendental numbers. We exhibit two sets \(R\) and \(T\) in \(\mathbb{R}^{2}\) such that \(R\) and \(T\) are congruent and each is congruent to their union \(S=R \cup T\). For each complex number \(z\), let \(t(z)=z+1, r(z)=e^{i} z\). Thus \(t\) is just a right translation by 1 unit and \(r\) is a rotation by 1 radian. Let \(S\) consist of those points that can be obtained by a finite number of applications of \(t\) and \(r\) starting from the origin. Each member of \(S\) can be represented as a polynomial in \(e^{i}\) with positive integer coefficients. (For example, if we translate five times, then rotate twice, then translate once more, the resulting point can be represented as \(5 e^{2 i}+1 .\) Since \(e^{i}\) is transcendental, the representation is unique. Let \(R\) consist of those points that have no constant term in their representation, and let \(T=S \backslash R\). Prove that \(t(S)=T\) and \(r(S)=R\) so \(R, T\) and \(S=R \cup T\) are pairwise isometric. Note that the isometries involved are isometries of \(\mathbb{R}^{2}\) onto \(\mathbb{R}^{2}\), not just isometries among the sets \(R, S\), and \(T\).

Use the Heine-Borel property to prove that if \(f: X \rightarrow Y\) is continuous and \(X\) is compact, then the set \(f(X)\) is compact in \(Y\).

Show that if \(E\) and \(F\) are closed subsets of a metric space \((X, d)\), at least one of which is compact, then there are points \(e \in E\) and \(f \in F\) so that $$ d(e, f)=\inf \\{d(x, y): x \in E, y \in F\\} $$ Show that if the sets \(E\) and \(F\) are not compact, but merely closed or complete, then this would not necessarily be true.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.