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Chapter 6: Problem 5

Which of the following sets are of type \(\mathcal{F}_{\sigma}\) ? (a) \(\mathrm{N}\) (b) \(\left\\{\frac{1}{n}: n \in \mathbb{N}\right\\}\) (c) The set \(\left\\{C_{n}: n \in \mathbb{N}\right\\}\) of midpoints of intervals complementary to the Cantor set (d) A finite union of intervals (that need not be open or closed)

Short Answer

Expert verified
Sets (b), (c), and (d) are \(\mathcal{F}_{\sigma}\) sets, but (a) is not.

Step by step solution

01

Understand the definition of \(\mathcal{F}_{\sigma}\) sets

A set is of type \(\mathcal{F}_{\sigma}\) if it can be expressed as a countable union of closed sets. This means we need to determine whether each set given in the question can be represented this way.
02

Analyze option (a): \(\mathbb{N}\)

The set of natural numbers, \(\mathbb{N}\), is not closed in the real numbers, because its complement (the real numbers excluding \(\mathbb{N}\)) is not open. Hence, \(\mathbb{N}\) is not a countable union of closed sets, so it is not an \(\mathcal{F}_{\sigma}\) set.
03

Analyze option (b): \(\left\{\frac{1}{n}: n \in \mathbb{N}\right\}\)

The set \(\left\{\frac{1}{n}: n \in \mathbb{N}\right\}\) is finite, as it includes a finite number of points. A finite set of points can be expressed as a countable union of singletons, which are closed. Therefore, this set is \(\mathcal{F}_{\sigma}\).
04

Analyze option (c): Set of midpoints of intervals complementary to the Cantor set

This set consists of countably many points. Any countable set of points can be represented as a countable union of singleton closed sets, hence making it a \(\mathcal{F}_{\sigma}\) set. Therefore, the set of midpoints of intervals complementary to the Cantor set is \(\mathcal{F}_{\sigma}\).
05

Analyze option (d): A finite union of intervals

Any finite union of closed intervals is closed and thus \(\mathcal{F}_{\sigma}\) because a finite set can be considered as a special case of a countable set.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

\(\mathcal{F}_{\sigma}\) sets
In real analysis, an \(\mathcal{F}_{\sigma}\) set is a powerful concept used to understand and classify subsets of the real numbers. Basically, a set is said to be \(\mathcal{F}_{\sigma}\) if it can be expressed as a countable union of closed sets. This classification helps in determining the complexity and properties of the set.
  • "\(\mathcal{F}\)" denotes closed sets ("ferm茅" in French).
  • "\(\sigma\)" indicates "sigma," referring to the countable union.
So, if you can break a set down into smaller closed pieces and unite them together countably, it's \(\mathcal{F}_{\sigma}\). Sets of this type are significant in topology and analysis as they possess measurable properties and often appear in **Lebesgue measurable** sets and function studies.
Cantor set
The Cantor set is a fascinating example in topology and measure theory due to its unusual properties. It is constructed by repeatedly removing the middle third of a line segment, starting with the interval [0, 1].
  • Initially, remove the open middle third of the interval, \( ( rac{1}{3}, rac{2}{3}) \).
  • Continue this process infinitely on the remaining segments.
Despite its intricate form, the Cantor set is uncountable, closed, and it has a Lebesgue measure of zero. This means although it contains no "length" in the traditional sense, it still holds infinitely many points. The Cantor set is a perfect set; it is closed and contains no isolated points, making it a cornerstone example in real analysis for discussing properties of sets and measures.
countable union
A countable union refers to a union of sets that can be indexed by natural numbers. When we say a set is a countable union, we imply that each component of the union is ordered and can be referenced by some countable index set, making the entire union countable. Here's what you need to know:
  • The concept is crucial for defining \(\mathcal{F}_{\sigma}\) sets.
  • It allows infinite processes to still be "manageable" by considering elements in a sequence.
For example, an \(\mathcal{F}_{\sigma}\) set is formed from a countable union of closed sets. Each closed set in the union can be taken singularly and indexed. This method maintains certain properties across the union, such as compactness and continuity being preserved in the larger structure of the set.
closed sets
Closed sets are foundational structures in topology. In the Euclidean space, a set is closed if it contains all its limit points, meaning it's a set where sequences within it converge to points that also lie within it. Important points about closed sets include:
  • They include their boundary points.
  • Complements of open sets are closed.
  • Both finite and arbitrary intersections of closed sets remain closed.
In real analysis, closed sets are used to define continuity, limits, and convergence. They're important building blocks for defining more complex concepts like \(\mathcal{F}_{\sigma}\) sets, where closed sets are collected into union forms to produce more intricate set types, retaining the original closed characteristics to some extent. Understanding closed sets is crucial in the study of topology and measure theory.

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Most popular questions from this chapter

Suppose that \(\bigcup_{n=1}^{\infty} A_{n}\) contains some interval \((c, d)\). Show that there is a set, say \(A_{n_{0}}\), and a subinterval \(\left(c^{\prime} d^{\prime}\right) \subset(c, d)\) so that \(A_{n_{0}}\) is dense in \(\left(c^{\prime} d^{\prime}\right)\).

If \(E\) has measure zero, show that the expanded set $$ c E=\\{c x: x \in E\\} $$ also has measure zero for any \(c>0\).

By altering the construction of the Cantor set, construct a nowhere dense closed subset of \([0,1]\) so that the sum of the lengths of the intervals removed is not equal to 1 . Will this set have measure zero?

Let \(K\) be the Cantor set and let \(\left\\{\left(a_{k}, b_{k}\right)\right\\}\) be the sequence of intervals complementary to \(K\) in \([0,1]\). For each \(k \in \mathbb{N}\), let \(c_{k}=\left(a_{k}+b_{k}\right) / 2\) (the midpoint of the interval \(\left.\left(a_{k}, b_{k}\right)\right)\) and let \(N=\left\\{c_{k}: k \in \mathbb{N}\right\\}\). Prove each of the following: (a) Every point of \(N\) is isolated. (b) If \(c_{i} \neq c_{j}\), there exists \(k \in \mathbb{N}\) such that \(c_{k}\) is between \(c_{i}\) and \(c_{j}\) (i.e., no point in \(N\) has an immediate "neighbor" in \(N\) ). (c) Show that there is an order-preserving mapping \(\phi: \mathbb{Q} \cap(0,1) \rightarrow N\) [i.e.. if \(x

Show that every subset of a set of measure zero also has measure zero.
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