/*! 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 8 Let \(J_{n}:=(0,1 / n)\) for \(n... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 8

Let \(J_{n}:=(0,1 / n)\) for \(n \in \mathbb{N}\). Prove that \(\bigcap_{n=1}^{\infty} J_{n}=\emptyset\).

Short Answer

Expert verified
The intersection of an infinite amount of these sets \(J_{n}\), \(\bigcap_{n=1}^{\infty} J_{n}\), is the empty set.

Step by step solution

01

Understand the definition

The set \(J_{n}\) is defined as the open interval between 0 and \(1/n\). As \(n\) gets bigger, \(1/n\) gets smaller, so the set \(J_{n}\) gets closer to the point 0. Therefore, the question asks whether there is a real number that stays in all intervals \(J_{n}\), i.e., the intersection of all \(J_{n}\).
02

Analyzing the intersection

The idea is to take an arbitrary real number \(x\) and show that one can always find a sufficiently large \(n\) such that \(x\) is not in \(J_{n}\). Let's take any \(x > 0\). Due to the Archimedean property, there exists an \(n \in \mathbb{N}\) such that \(n > 1/x\). This implies \(1/n < x\). Therefore, \(x\) cannot be in the set \(J_{n}\) because \(J_{n}\) includes all points between 0 and \(1/n\) (exclusive). Therefore, \(x\) is not in the intersection \(\bigcap_{n=1}^{\infty} J_{n}\). Since \(x\) was chosen arbitrarily, we conclude that no point is in the intersection. Therefore, \(\bigcap_{n=1}^{\infty} J_{n}=\emptyset\).
03

Final check

A final check can be made by remembering that 0 is not included in any of the \(J_{n}\) sets, and we have shown that no other positive real number can be part of the intersection. This solidifies our claim that the intersection of all such sets \(J_{n}\) is indeed the empty set.

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!

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 \(S \subseteq \mathbb{R}\) be nonempty. Prove that if a number \(u\) in \(\mathbb{R}\) has the properties: (i) for every \(n \in \mathbb{N}\) the number \(u-1 / n\) is not an upper bound of \(S\), and (ii) for every number \(n \in \mathbb{N}\) the number \(u+1 / n\) is an upper bound of \(S\), then \(u=\sup S .\) (This is the converse of Exercise 2.3.8.)

Show that if \(A\) and \(B\) are bounded subsets of \(\mathbb{R}\), then \(A \cup B\) is a bounded set. Show that \(\sup (A \cup B)=\sup \\{\sup A, \sup B\\}\)

Prove the following form of Theorem 2.1.9: If \(a \in \mathbb{R}\) is such that \(0 \leq a \leq \varepsilon\) for every \(\varepsilon>0\), then \(a=0\)

Let \(S\) be a nonempty bounded set in \(\mathbb{R}\). (a) Let \(a>0\), and let \(a S:=\\{\) as: \(s \in S\\}\). Prove that $$ \inf (a S)=a \inf S, \quad \sup (a S)=a \sup S $$ (b) Let \(b<0\) and let \(b S=\\{b s: s \in S\\}\). Prove that $$ \inf (b S)=b \sup S, \quad \sup (b S)=b \inf S $$

If \(a \in \mathbb{R}\) satisfies \(a \cdot a=a\), prove that either \(a=0\) or \(a=1\).
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Calculus

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.