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Chapter 2: Problem 9

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

Short Answer

Expert verified
The intersection of the sets \(\bigcap_{n=1}^{\infty} K_{n} = \emptyset\), because for any given real number \(x\), there will always be a natural number \(n\) such that \(n > x\), meaning there cannot be a number common to all sets \(K_{n}\).

Step by step solution

01

Definition of the Given Set

We have the sets \(K_{n}\) where \(K_{n}:=(n, \infty)\). This means each set \(K_{n}\) includes all real numbers that are greater than \(n\). Note that as \(n\) increases, the lower bound of each set \(K_{n}\) increases.
02

Understanding Intersection of the Sets

The intersection of the sets \(K_{n}\) for \(n=1,2,3,...\) such that \(\bigcap_{n=1}^{\infty} K_{n}\) refers to the set of numbers which exist in every set \(K_{n}\).
03

Proof of the Intersection

Suppose there was a common number \(x\) that existed in all sets \(K_{n}\). Then, \(x > n\) for every natural number \(n\). However, for any real number \(x\), we can always find a natural number \(n\) such that \(n > x\). This is a contradiction. Therefore, no such common number \(x\) can exist in all sets \(K_{n}\). This means the intersection of all these sets is an empty set.

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Most popular questions from this chapter

Let \(S_{2}=\\{x \in \mathbb{R}: x>0\\} .\) Does \(S_{2}\) have lower bounds? Does \(S_{2}\) have upper bounds? Does inf \(S_{2}\) exist? Does sup \(S_{2}\) exist? Prove your statements.

Modify the argument in Theorem \(2.4 .7\) to show that there exists a positive real number \(u\) such that \(u^{3}=2\).

What rationals are represented by the periodic decimals \(1.25137 \ldots 137 \ldots\) and \(35.14653 \cdots 653 \cdots ?\)

Let \(S \subseteq \mathbb{R}\) be nonempty. Show that if \(u=\sup S\), then for every number \(n \in \mathbb{N}\) the number \(u-1 / n\) is not an upper bound of \(S\), but the number \(u+1 / n\) is an upper bound of \(S\). (The converse is also true; see Exercise 2.4.3.)

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