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Chapter 11: Problem 2

Exhibit an open cover of \(\mathbb{N}\) that has no finite subcover.

Short Answer

Expert verified
The open cover of the set of natural numbers \(\mathbb{N}\) is given by the collection of sets \(U_i = \{n \in \mathbb{N} : n > i\}\), for \(i \in \mathbb{N}\). This open cover doesn't have a finite subcover since any finite subcollection of the open cover will not include all natural numbers.

Step by step solution

01

Understanding the Concepts

An open cover of a set in a space is a collection of open sets whose union contains the set. A finite subcover is a finite subcollection of the open cover that still covers the entire set. One main property of the set of natural numbers, \(\mathbb{N}\), is that it is infinite.
02

Construct the Open Cover

An example of an open cover of the set \(\mathbb{N}\) of natural numbers would be the collection of sets \(U_i = \{n \in \mathbb{N} : n > i\}\), for \(i \in \mathbb{N}\). In other words, for each natural number \(i\), we can associate an open set \(U_i\) that includes all natural numbers greater than \(i\). The union of all these sets \(U_i\) would then be \(\mathbb{N}\). We can formally express this as \(\mathbb{N} = \bigcup_{i \in \mathbb{N}} U_i\). In other words, the collection \(U_i\) forms an open cover of \(\mathbb{N}\).
03

Demonstrating no Finite Subcover

Let us assume that we have some finite subcollection of the open cover that still covers all of \(\mathbb{N}\): \(U_{i_1}, U_{i_2}, ..., U_{i_k}\). Since this is a finite collection, we have a maximum \(i\), say \(i_m\). But then \(U_{i_m}\) does not cover any number less than or equal to \(i_m\), so not all of \(\mathbb{N}\) is covered. Therefore, this collection \(U_{i_1}, U_{i_2}, ..., U_{i_k}\) is not a subcover of \(\mathbb{N}\). Since our choice of a finite subcollection is arbitrary, we conclude that no finite subcover can cover all of \(\mathbb{N}\).

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Most popular questions from this chapter

Show that if \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous, then the set \(\\{x \in \mathbb{R}: f(x)=k\\}\) is closed in \(\mathbb{R}\) for each \(k \in \mathbb{R}\).

Prove that the intersection of an arbitrary collection of compact sets in \(\mathbb{R}\) is compact.

Show that if \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous. then the set \(\\{x \in \mathbb{R}: f(x) \leq \alpha\\}\) is closed in \(\mathbb{R}\) for each \(\alpha \in \mathbb{R}\)

If \(A \subseteq \mathbb{R}\), let \(A^{-}\) be the intersection of all closed sets containing \(A\); the set \(A^{-}\) is called the closure of \(A\). Show that \(A^{-}\) is a closed set, that it is the smallest closed set containing \(A\), and that a point \(w\) belongs to \(A^{-}\) if and ondy if \(w\) is either an interior point or a boundary point of \(A\).

Exhibit an open cover of the set \(\\{1 / n: n \in \mathbb{N}\\}\) that has no finite subcover.
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