/*! 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 13 Take \(\mathbb{N} \subset \mathb... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 13

Take \(\mathbb{N} \subset \mathbb{R}\) using the standard metric. Find an open cover of \(\mathbb{N}\) such that the conclusion of the Lebesgue covering lemma does not hold.

Short Answer

Expert verified
The open cover \((n-0.5, n+0.5)\) violates the lemma as no \(\delta > 0\) works for subsets like \(\{n, n+1\}\).

Step by step solution

01

Understand the Problem

We need to find an open cover of \(\mathbb{N}\) where the conclusion of the Lebesgue Covering Lemma does not hold. The standard metric on \(\mathbb{R}\) is given by the absolute value of the difference, \(|x-y|\).
02

Recall the Definition of an Open Cover

An open cover of a set \(S\) is a collection of open sets such that every point in \(S\) is contained in at least one of these open sets. For \(\mathbb{N}\), this means every natural number must be in some open interval from the cover.
03

Consider the Lebesgue Covering Lemma

The Lebesgue Covering Lemma states that for a compact metric space \(X\) with an open cover \(\{U_\alpha\}\), there exists a \(\delta > 0\) such that any subset of \(X\) with diameter less than \(\delta\) is entirely contained in some \(U_\alpha\). Since \(\mathbb{N}\) is not compact under the standard metric, the lemma does not automatically apply.
04

Construct an Open Cover that Violates the Conclusion

To violate the conclusion, consider the open cover \(\{(n-0.5, n+0.5) : n \in \mathbb{N}\}\). Each open set \((n-0.5, n+0.5)\) covers exactly one integer \(n\).
05

Verify the Contradiction

In this open cover, for any small \(\delta\), sets with diameter less than \(\delta\) containing more than one element (e.g., \(\{n, n+1\}\)) cannot be contained in a single set of the cover. Thus, no such \(\delta\) can exist that satisfies the lemma for every subset of \(\mathbb{N}\), since \(\mathbb{N}\) is not compact.

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.

Open Cover
An open cover is a fundamental concept in topology. It relates to how we can "cover" a set using open sets. An open set in the context of real numbers (using the standard metric) can be something like an interval
(consider it as a "slice") that does not include its endpoints.
An open cover for a set is a collection of these open sets, ensuring that every point in the original set is encompassed by at least one of the intervals in the open cover.
  • Example: For the set of natural numbers \( \mathbb{N}\), an open cover could be something like \((n-0.5, n+0.5)\). This captures each individual natural number.
  • For this concept to hold, it's important that every point is included in at least one open interval.
Compactness
Compactness is a key property in topology, and it's crucial when working with the Lebesgue Covering Lemma.
In simple terms, a set is compact if it is "small" in a certain topological sense鈥攅ven if it stretches to infinity in actual size.
Mathematically, a metric space is compact if every open cover has a finite subcover. This means you can cover the whole space using only a finite number of open sets.
  • For example, the set \( [0, 1] \) in \( \mathbb{R} \) is compact, as any open cover of this interval must have a finite subcollection that covers it entirely.
  • In contrast, the set of natural numbers \( \mathbb{N} \) is not compact under the standard metric. There are open covers of \( \mathbb{N} \) where no finite subcover can cover all natural numbers comprehensively.
Metric Spaces
When talking about metric spaces, we often think about a system where we can measure distance between points.
A metric space consists of a set together with a metric, which is a function that defines the distance between any two points in the set.
The metric must satisfy specific conditions:
  • It must be positive: the distance between two different points is always positive.
  • Symmetry: the distance from point A to point B is the same as from point B to point A.
  • Triangle inequality: the direct distance between points A and C should not be greater than traveling from A to B and then B to C.
Understanding metric spaces helps in visualizing and comprehending more complex mathematical structures.
Standard Metric
The standard metric on the real numbers \( \mathbb{R} \), is perhaps the most straightforward way to measure distance.
It simply measures the absolute difference between two numbers, \( |x-y| \).
This is the metric we've been using throughout the exercise. It provides a simple idea of proximity or distance, making it intuitive and easy to grasp.
  • For instance, to find the distance between 3 and 7 using the standard metric, we compute \( |3-7| = 4 \).
  • This approach extends to any pair of real numbers and forms the basis for analyzing more general metric spaces.
The standard metric is essential because understanding it helps demystify more complex metrics that we might encounter in advanced contexts.

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 \((X, d)\) be a compact metric space, let \(C(X, \mathbb{R})\) be the set of real-valued continuous functions. Define $$ d(f, g):=\|f-g\|_{u}:=\sup _{x \in X}|f(x)-g(x)| $$ a) Show that d makes \(C(X, \mathbb{R})\) into a metric space. b) Show that for any \(x \in X,\) the evaluation function \(E_{x}: C(X, \mathbb{R}) \rightarrow \mathbb{R}\) defined by \(E_{x}(f):=f(x)\) is \(a\) continuous function.

Let \(C^{1}([a, b], \mathbb{R})\) be the set of once continuously differentiable functions on \([a, b] .\) Define $$d(f, g):=\|f-g\|_{u}+\left\|f^{\prime}-g^{\prime}\right\|_{u}$$ where \(\|\cdot\|_{u}\) is the uniform norm. Prove that \(d\) is a metric.

Let \((X, d)\) be a nonempty metric space and \(S \subset X\) a subset. Prove: a) \(S\) is bounded if and only if for every \(p \in X,\) there exists \(a B>0\) such that \(d(p, x) \leq B\) for all \(x \in S\). b) A nonempty \(S\) is bounded if and only if \(\operatorname{diam}(S):=\sup \\{d(x, y): x, y \in S\\}<\infty\).

Let \(F: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(F(x):=k x+b\) where \(0

Let \(f(x):=x-\frac{x^{2}-2}{2 x}\) (you may recognize Newton's method for \(\sqrt{2}\) ). a) Prove \(f([1, \infty)) \subset[1, \infty) .\) b) Prove that \(f:[1, \infty) \rightarrow[1, \infty)\) is a contraction. c) Apply the fixed point theorem to find an \(x \geq 1\) such that \(f(x)=x,\) and show that \(x=\sqrt{2}\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics 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.