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Chapter 4: Problem 3

Suppose \(n \in \mathbb{Z}^{+}\) and \(K_{1}, K_{2}, \ldots, K_{n}\) are compact sets. Show that \(\bigcup_{i=1}^{n} K_{i}\) is compact.

Short Answer

Expert verified
The union \( \bigcup_{i=1}^{n} K_i \) is compact because it is both closed and bounded.

Step by step solution

01

Understand the Definition of Compactness

A set is compact if it is closed and bounded in Euclidean space. For a union of compact sets to be compact, we need to ensure the union is closed and bounded too.
02

Check if the Union is Bounded

Since each set \( K_i \) is compact, each one is bounded. This means there exist real numbers \( M_i \) such that for all \( x \in K_i \), \( ||x|| < M_i \). Therefore, for the union \( \bigcup_{i=1}^{n} K_i \), it is bounded by \( \max(M_1, M_2, \ldots, M_n) \) since each element of the union must be in one of the \( K_i \) sets.
03

Check if the Union is Closed

Each \( K_i \) is a closed set since it is compact. A finite union of closed sets is also closed. Thus, \( \bigcup_{i=1}^{n} K_i \) is closed.
04

Combine Boundedness and Closure for Compactness

Since \( \bigcup_{i=1}^{n} K_i \) is both closed and bounded in a Euclidean space, it satisfies all the criteria for compactness. Therefore, the union is compact.

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Key Concepts

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

Closed Sets
In topology, a set is defined as closed if it contains all its limit points. This means that if a sequence of points in the set has a limit, that limit is also included in the set.
Closed sets can also be thought of as the complement of open sets within a given space.
In a Euclidean space, a closed set can be equally described by having the property that it contains its boundary.
Here are some properties of closed sets:
  • The union of a finite number of closed sets is closed.
  • The complement of a closed set is open, and vice versa.
  • Every closed set is the intersection of arbitrary collections of closed sets.
Understanding these properties is crucial because they help us prove that certain operations on closed sets, such as their finite union, will yield another closed set, as required for examining compactness.
Bounded Sets
A set in a Euclidean space is bounded if there is a real number (let's call it \( M \)) such that all points in the set are within a distance \( M \) from the origin.
This means simply that the set doesn't stretch out infinitely in any direction.
Bounded sets can be considered 'trapped' within some large ball around a chosen center.
  • It's possible to find a ball of finite radius that contains the entire set.
  • A set is unbounded if no such ball exists.
  • All points of a bounded set are within a fixed distance from each other.
In proving that a union of compact sets is compact, the concept of boundedness ensures that the union does not "go on forever" and maintains some finite space.
Union of Sets
The union of sets is a fundamental operation in set theory, involving the combination of all elements from two or more sets, without repetition.
If you take sets \( A \) and \( B \), their union is denoted as \( A \cup B \).
It contains every element that is in \( A \), in \( B \), or in both.
  • For any finite collection of closed sets, their union resulted is also a closed set.
  • The union of sets doesn't require the sets to be related or overlap.
  • For compactness, ensuring the union remains within the required boundaries and limits is key.
When we talk about compact sets, noting the union remains closed and bounded increases our confidence in their collective compactness.
Euclidean Space
Euclidean space is a mathematical concept that extends the familiar two-dimensional plane to higher dimensions, allowing generalization for any positive integer \( n \).
It's the setting where notions like distance, angles, and geometric figures are well-defined using coordinates.
In \( n \)-dimensional Euclidean space:
  • The notion of a "straight line" extends as a concept of connecting points in the shortest path possible.
  • Distances are calculated using the formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + \ldots} \) for any two points \((x_1, y_1, \ldots)\) and \((x_2, y_2, \ldots)\).
  • Compactness in Euclidean space simply means closed and bounded.
Working within a Euclidean framework, certain attributes like being closed and bounded become powerful tools to determine the compactness of sets, especially in mathematical proofs and exercises connecting to real-world phenomena.

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Most popular questions from this chapter

For \(n=1,2,3, \ldots,\) let \(I_{n}=\left(-\frac{1}{n}, \frac{n+1}{n}\right) .\) Is $$ \bigcap_{n=1}^{\infty} I_{n} $$ open or closed?

Let \(U \subset \mathbb{R}\) be a nonempty open set. Show that \(\sup U \notin U\) and \(\inf U \notin U\).

We say a collection of sets \(\left\\{D_{\alpha}: \alpha \in A\right\\}\) has the finite intersection property if for every finite set \(B \subset A\), $$ \bigcap_{\alpha \in B} D_{\alpha} \neq \emptyset $$ Show that a set \(K \subset \mathbb{R}\) is compact if and only for any collection \(\left\\{E_{\alpha}: \alpha \in A, E_{\alpha}=C_{\alpha} \cap K\right.\) where \(C_{\alpha} \subset \mathbb{R}\) is closed \(\\}\) which has the finite intersection property we have $$ \bigcap_{\alpha \in A} E_{\alpha} \neq \emptyset $$

Suppose \(U \subset \mathbb{R}\) is a nonempty open set. For each \(x \in U,\) let $$ J_{x}=\bigcup(x-\epsilon, x+\delta) $$ where the union is taken over all \(\epsilon>0\) and \(\delta>0\) such that \((x-\epsilon, x+\delta) \subset U\). a. Show that for every \(x, y \in U,\) either \(J_{x} \cap J_{y}=\emptyset\) or \(J_{x}=J_{y}\). b. Show that $$ U=\bigcup_{x \in B} J_{x} $$ where \(B \subset U\) is either finite or countable.

Suppose \(A_{i} \subset \mathbb{R}, i \in \mathbb{Z}^{+},\) and let $$ B=\bigcup_{i=1}^{\infty} A_{i} $$ Show that $$ \bigcup_{i=1}^{\infty} \overline{A_{i}} \subset \bar{B} $$ Find an example for which $$ \bar{B} \neq \bigcup_{i=1}^{\infty} \overline{A_{i}} $$
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