/*! 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 10 Prove the version of the Bolzano... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 10

Prove the version of the Bolzano-Weierstrass theorem that applies to sets: Every infinite bounded subset of \(\mathbb{R}^{n}\) has a point of accumulation in \(\mathbb{R}^{n}\).

Short Answer

Expert verified
Every infinite bounded subset of \(\mathbb{R}^{n}\) has an accumulation point in \(\mathbb{R}^{n}\).

Step by step solution

01

Define Boundedness and Accumulation Point

A set is said to be bounded if it is contained within some large ball in \(\mathbb{R}^n\). A point \(x\) in \(\mathbb{R}^n\) is an accumulation point of a set \(A\) if every neighborhood of \(x\) contains infinitely many points of \(A\).
02

Assume Contradiction

Assume there exists an infinite bounded subset \(A \subseteq \mathbb{R}^n\) with no accumulation point. This assumption will lead to a contradiction, helping us prove that such sets do have accumulation points.
03

Apply Heine-Borel Theorem

According to the Heine-Borel theorem, since \(A\) is bounded, it must lie within a closed and bounded set in \(\mathbb{R}^n\), i.e., a compact set. By the properties of compact sets, every sequence in this set has a convergent subsequence.
04

Consider a Sequence Within Set

Let \(\{x_m\}\) be a sequence of distinct points in \(A\). Since \(A\) is compact, \(\{x_m\}\) must have a subsequence \(\{x_{m_k}\}\) that converges to a point \(x\) in \(\mathbb{R}^n\).
05

Contradiction Arising from Sequence

The convergence of the subsequence to point \(x\) implies \(x\) is an accumulation point of \(A\), which is a contradiction to our assumption that \(A\) has no accumulation points.
06

Conclude the Proof

Since assuming no accumulation points leads to a contradiction, it must be true that every infinite bounded subset of \(\mathbb{R}^n\) has a point of accumulation.

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.

Accumulation Point
An accumulation point is like a special guest that keeps drawing in members of a set. Imagine you have a set of points in a particular space, like stars in the sky. An accumulation point is a point where, no matter how small you make your circle around it (think of it like a spotlight), you'll always find an infinite number of stars (points of your set) shining close by. This means these points keep hanging around endlessly, clustering near the accumulation point.
  • It is important because it suggests a form of continuity or clustering within the set.
  • In mathematical terms, if you take any small open neighborhood around the accumulation point, it will contain other points of the set.
Bounded Subset
Picture a playground, and you have a box that can hold all the playground toys, no matter where they are in the park. When a set is bounded, it鈥檚 like saying all these toys can fit into one such box. More formally, a set is bounded if it can be enclosed within a ball of finite radius. This means that there's a limit to how far apart any two points in the set can be from each other.
  • In \( \mathbb{R}^n \), boundedness implies that there exists a real number, within which all points of the set can fit.
  • This concept is crucial for applying the Heine-Borel theorem, which speaks about compactness in terms of both closed and bounded subsets.
Heine-Borel Theorem
The Heine-Borel theorem is a powerful tool that guarantees when a set behaves nicely. In simple terms, it tells us that in the realm of real numbers \( \mathbb{R}^n \), a set is compact if it is closed and bounded鈥攎eaning it鈥檚 not chaotic or infinitely sprawling. Compact sets have special properties, like sequences within them always having convergent subsequences.
- **Why it's important**: This theorem can be thought of as the "order and neatness" rule in mathematics, ensuring that sets don't have loose ends hanging out aimlessly.- **Application in Bolzano-Weierstrass**: The theorem helps pinpoint that an infinite bounded set in \( \mathbb{R}^n \) has all the right ingredients (closed and bounded) to have sequences that converge, setting the stage to prove the Bolzano-Weierstrass theorem.
Convergent Subsequence
Think of convergent subsequences like a group of people moving toward a single destination. In any given set of movements (or a sequence in mathematical terms), a convergent subsequence is a path where everyone decides to come to the same meeting point.
This concept is about finding order and predictability: even if the overall path seems random, there's always a systematic part that congregates to the same place.
  • In \( \mathbb{R}^n \), this means a subsequence where the points get closer and closer to a single point, the limit.
  • This is crucial in proving the Bolzano-Weierstrass theorem, as it tells us that amidst chaos, you can always find an element of convergence.

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 \(E\) be a compact subset of \(\mathbb{R}^{n}\) and let \(\mathbf{x} \in \mathbb{R}^{n}\). Let $$ \operatorname{dist} \mathbf{x}=\inf \\{\|\mathbf{x}-\mathbf{y}\|: y \in E\\} $$ (a) Prove that "inf" can be replaced by "min" in the definition of "dist." (b) Prove that "dist" is a continuous function on \(\mathbb{R}^{n}\). (c) Are parts (a) and (b) valid if \(E\) is assumed closed but not bounded? What if \(E\) is assumed bounded but not closed?

Let \(E \subset \mathbb{R}^{n}\) and let \(\mathbf{x}_{0}\) be an accumulation point of \(E .\) Prove that if \(\lim _{\mathbf{x} \rightarrow \mathrm{x}_{0}} \mathbf{f}(\mathbf{x})\) exists, then there exist \(\varepsilon>0\) and \(M>0\) such that \(\|\mathbf{f}(\mathbf{x})\|

Let \(\mathbf{H}: \mathbb{R}^{2} \Rightarrow \mathbb{R}^{2}\) be defined by \(\mathbf{H}(x, y)=\left(x^{2}-y^{2}, 2 x y\right) .\) Write \(u=x^{2}-y^{2}\) and \(v=2 x y\), so that \(\mathbf{H}(x, y)=(u, v)\). (a) Describe the sets in the \(x y\) -plane that \(\mathbf{H}\) maps onto horizontal lines in the \(u v\) -plane. Do the same for vertical lines. (b) Determine a set \(S \subset \mathbb{R}^{2}\) such that \(\mathbf{H}(S)=[1,2] \times[4,5]\).

Let $$ f(x, y)=\frac{2 x^{2} y}{x^{4}+y^{2}} $$ with \(f(0,0)=0\). Show (a) The limit of \(f(x, y)\) as \((x, y) \rightarrow(0,0)\) along any straight line is \(f(0,0)\). (b) \(f\) is discontinuous at \((0,0)\).

Prove that if \(\mathbf{f}: E \rightarrow \mathbb{R}^{m}\) is uniformly continuous on \(E\), a subset of \(\mathbb{R}^{n}\), and \(E\) is bounded, then \(\mathbf{f}\) is bounded on \(E\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

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.