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Chapter 11: Problem 3

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

Short Answer

Expert verified
An open cover of the set \(A = \{1 / n: n \in \mathbb{N}\}\) could be \(G = \{(1 / (n+1), 1 / (n-1)): n \in \mathbb{N}, n > 1\}\) and this open cover does not contain any finite subcover.

Step by step solution

01

- Define the Set

Rewrite the given set as \(A = \{1 / n: n \in \mathbb{N}\}\), where \(\mathbb{N}\) denotes the natural numbers.
02

- Define an Open Cover

Craft an open cover \(G\) for the set \(A\). One potential open cover could be defined as \(G = \{(1 / (n+1), 1 / (n-1)): n \in \mathbb{N}, n > 1\}\). This defines \(G\) as a collection of open intervals.
03

- Prove No Finite Subcover

To prove no finite subcover exists, assume the contrary i.e., there exists a finite subcover of \(G\). So it must contain a maximum indexed interval (1 / m, 1 / (m-1)) for some \(m\). But then, the number \(1/m\) in set \(A\) will not be covered by this subcover, leading to a contradiction. Hence, no finite subcover exists for the given open cover of the set \(A\).

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Most popular questions from this chapter

Give an example of a function \(f: \mathbb{R} \rightarrow \mathbb{R}\) such that the set \(\\{x \in \mathbb{R}: f(x)=1\\}\) is neither open nor closed in \(\mathbb{R}\).

Let \(I:=[a, b]\) and let \(f: I \rightarrow \mathbb{R}\) and \(g: I \rightarrow \mathbb{R}\) be continuous functions on \(I\). Show that the set \(\\{x \in I: f(x)=g(x)\\}\) is closed in \(\mathbb{R}\).

Show that each point of the Cantor set \(\mathbb{F}\) is a cluster point of \(\mathbb{F}\).

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

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