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Chapter 11: Problem 8

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 \(\mathbb{R}\).

Short Answer

Expert verified
The sine function \(f(x) = \sin(x)\) is a function that generates a set of x-values where the function is equal to 1 that is neither open nor closed.

Step by step solution

01

Understand the function

The first step is to understand what kind of function could yield a set of x-values for which \(f(x) = 1\) is neither open nor closed. To get such a set, it would mean that the set contains some of its limit points, but doesn't contain others and that there are points that are not an interior point.
02

Propose a function

A simple function that meets this requirement is the sine function \(f(x) = \sin(x)\). The sine function oscillates between -1 and 1.
03

Analyze the function

The set of points where \(f(x) = 1\) is \(\{x \in \mathbb{R}: x = n\pi + \frac{\pi}{2}, n \in \mathbb{Z}\}\). Considering the definition of open and closed sets, this set is neither open nor closed. For example, there doesn't exist a small open interval around the point \(x = \frac{\pi}{2}\) that is entirely within this set so it's not an open set. Also the limit point \(x=\pi\) is not in the set, thus it's not a closed set.

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Most popular questions from this chapter

Show that a set \(G \subseteq \mathbb{R}\) is open if and only if it does not contain any of its boundary points.

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 \(k \in \mathbb{R}\).

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

If \((S, d)\) is a metric space, a subset \(A \subseteq S\) is said to be bounded if there exists \(x_{0} \in S\) and a number \(B>0\) such that \(A \subseteq\left\\{x \in S: d\left(x, x_{0}\right) \leq B\right\\}\). Show that if \(A\) is a compact subset of \(S\), then \(A\) is closed and bounded.

Find an infinite collection \(\left\\{K_{n}: n \in \mathbb{N}\right\\}\) of compact sets in \(\mathbb{R}\) such that the union \(\bigcup_{n=1}^{\infty} K_{n}\) is not compact.
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