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Chapter 11: Problem 6

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

Short Answer

Expert verified
By showing that A's complement is not an open set, it can be inferred that A is not a closed set. After adding 0 to A, however, the complement of the new set can be seen as an open set, showing that A union 0 is indeed a closed set.

Step by step solution

01

Understanding Open and Closed Sets

An open set is a set where, for each point in the set, there exists a neighborhood fully contained in the set. Conversely, a closed set is a set where its complement is an open set in the metric space.
02

Evaluating Closeness of A

To show that A is not closed, it is necessary to show that its complement is not an open set. Consider the complement of set A, which includes all real numbers which are not of the form 1/n. One such real number is 0. But no matter how small a neighborhood around 0 is chosen, it includes members of A (as 1/n can get arbitrarily small), indicating that it is not fully contained in the complement of A, thus showing that the complement of A is not an open set. Therefore, by definition, A is not a closed set.
03

Evaluating Closeness of A union 0

Now reconsider the set A union 0. The complement of A now excludes 0, debating the point raised earlier about the neighborhoods around 0. For every other real number not of the form 1/n, choosing a sufficiently small neighborhood around it will ensure the neighborhood is fully contained in the complement of the enlarged set, showing that the complement is an open set and thus showing that the enlarged set (A union 0) is a closed set.

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