/*! 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 11 Give an example of a convex set ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 11

Give an example of a convex set whose complement is bounded.

Short Answer

Expert verified
An example is the entire space \( \mathbb{R}^n \), whose complement is the empty set, which is bounded.

Step by step solution

01

Understand Definitions

To solve the problem, we must first understand what a convex set is. A set is convex if, for any two points within the set, the line segment connecting them lies within the set as well. Additionally, a set is bounded if it fits within a sufficiently large ball centered at origin.
02

Identify Complement and Boundedness

The problem asks for a convex set whose complement is bounded. This means we are looking for a set where everything outside this set (the complement) is limited in extent, i.e., does not extend to infinity.
03

Example - The Empty Set

Consider the empty set, denoted by \( \emptyset \). The empty set is technically a convex set because there are no points inside it, so no counterexamples to convexity exist. Its complement is the entire space (say, \( \mathbb{R}^n \)), which is not bounded.
04

Correct Answer - Full Space

Based on the problem constraints, consider using the full space \( \mathbb{R}^n \). Its complement would be the empty set, which *is* bounded because it contains no points, satisfying the condition that its 'complement is bounded'.

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.

Bounded Set
A set is known as a "bounded set" if it can be completely contained within a finite region of space. In simpler terms, if you can imagine drawing a large circle (or using any geometric shape) around the entire set, without ever needing to make the circle infinitely large, then the set is bounded. To illustrate this:
  • Think of a small group of dots drawn on a piece of paper. You can easily draw a circle around them, which means the set of these dots is bounded.
  • In contrast, if the dots continue infinitely in all directions, you would never be able to fully encompass them with a single circle, so the set would not be bounded.
In mathematical terms, a set is bounded if there is a real number that represents the maximum distance between any point in the set and the origin, or origin can be replaced with any other fixed point from which distances are measured. This definition ensures that the set does not "stretch" to infinity in any direction.
Complement of a Set
The "complement of a set" consists of all the elements that are not in the original set, within a universal context. To better understand this, imagine a big universe of elements where a smaller subset is removed. The remaining elements constitute the complement of the original subset.For example:
  • If you have a set containing all vowels from the alphabet, then the complement would be all the consonants, assuming the universal set is the entire English alphabet.
  • More formally, if set A denotes a subset of the universal set U, the complement of A (\( A^c \)) is defined as \( A^c = \{ x \in U : x otin A \} \).
In the context of the given exercise, when considering the entire space \( \mathbb{R}^n \) as our set, its complement is the empty set since there is nothing outside this space.
Empty Set
The "empty set" is a fundamental concept in mathematics. It is the set that contains no elements at all and is often represented by the symbol \(\emptyset\). Here are a few key points:
  • It is unique, meaning there is only one empty set, as it doesn't contain anything that can differentiate it from another empty set.
  • The empty set is considered a subset of every set, which means if you take any set, the empty set is always one of its subsets.
  • Mathematically, it has special properties such as being both an open and closed set, as well as being convex, since there are no points to challenge this property.
Relating to the exercise, the empty set's complement is the entire space \( \mathbb{R}^n \), and conversely, the complement of the whole space is the empty set, which is bounded due to the fact it contains no elements and thereby satisfies the condition of boundedness effortlessly.

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

Assuming that the following continued fraction converges, find its value: $$ \frac{1}{6+} \frac{1}{6+} \frac{1}{6+} \frac{1}{6+} \cdots $$

Find the Maclaurin series for the function $$ f(x)=\int_{0}^{x} \frac{1-\cos t}{t} d t $$ Cultural note: This function is sometimes called the cosine integral and is denoted by \(\operatorname{Cin}(x)\). The nomenclature has not been standardized.

(Ratio test, continuation) If \(\lim _{n \rightarrow \infty}\left|A_{n+1} / A_{n}\right|>1\), then \(\sum_{k=0}^{\infty} A_{k}\) diverges. Use this fact together with the preceding problem to find the radius of convergence of $$ \sum_{k=0}^{\infty}(-1)^{k}(x-1)^{k} $$

Find the Lagrange and Newton forms of the interpolating polynomial for these data: \begin{tabular}{c||r|r|r} \(x\) & \(-2\) & 0 & 1 \\ \hline\(f(x)\) & 0 & 1 & \(-1\) \end{tabular} Write both polynomials in the form \(a+b x+c x^{2}\) to verify that they are identical as functions.

Let \(A\) be an \(n \times(n+2)\) matrix. Show that if the system $$ \begin{array}{llll} A x=0 & x \geq 0 & x \neq 0 & x_{n+2}=0 & x \in \mathbb{R}^{n+2} \end{array} $$ is consistent, then for some \(k
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Geometry

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.