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Chapter 11: Problem 4

Prove that \((0,1]=\bigcap_{n=1}^{\infty}(0,1+1 / n)\), as asserted in Example \(11.1 .6(\mathrm{a})\).

Short Answer

Expert verified
The sets \( (0,1] \) and \( \bigcap_{n=1}^{\infty}(0,1 +1/n) \) are identical. By examining the limit of the intersected sets as n approaches infinity, it's clear that they close in towards 1, replicating the exact properties of the left-hand set \( (0,1] \).

Step by step solution

01

Understanding Set Definitions

Firstly, let's understand these two sets are defined in mathematical terms. The set \( (0,1] \) means it's an open interval, including every number that is strictly more than 0 and less than or equal to 1. The other set \( \bigcap_{n=1}^{\infty}(0,1+1 / n) \) is the intersection over infinite intervals, where for all natural numbers n starting from 1, each interval is (0, 1 + 1/n).
02

Analyzing the Right-Hand Set

Now focus on the right side, \( \bigcap_{n=1}^{\infty}(0,1+1 / n) \). As n approaches infinity for \( (0,1+1/n) \), the end of each interval ultimately closes in on 1. Hence, the entirety of these intervals is in fact the set (0,1].
03

Analyzing the Left-Hand Set

Clearly, the left side set is \( (0,1] \), which was already established that is equal to the right side set. These intervals have all the number from 0 (exclusive) and 1 (inclusive).
04

Equality of the Sets

For any x belongs in \( (0,1] \), x will also belong to each of the intervals \( (0,1+1/n) \) for every n. Simultaneously, any x belongs to \( \bigcap_{n=1}^{\infty}(0,1+1 / n) \) must also belong in \( (0,1] \). Therefore, these two sets are equal.

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Most popular questions from this chapter

Show that the set \(\mathbb{Q}\) of rational numbers is neither open nor closed.

Show that \(A=\\{1 / n: n \in \mathbb{N}\\}\) is not a closed set, but that \(A \cup\\{0\\}\) is a closed set.

A point \(x \in \mathbb{R}\) is said to be a boundary point of \(A \subseteq \mathbb{R}\) in case every neighborhood \(V\) of \(x\) contains points in \(A\) and points in \(\mathcal{C}(A)\). Show that a set \(A\) and its complement \(\mathcal{C}(A)\) have exactly the same boundary points.

Show that the set \(\mathbb{N}\) of natural numbers is a closed set.

Show that if \(G\) is an open set and \(F\) is a closed set, then \(G \backslash F\) is an open set and \(F \backslash G\) is a closed set.
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