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Chapter 11: Problem 7

Show that the set \(\mathbb{Q}\) of rational numbers is neither open nor closed.

Short Answer

Expert verified
The set of rational numbers \(\mathbb{Q}\) is not open because for any given rational number, we can always find an irrational number within any interval around it. Similarly, \(\mathbb{Q}\) is not closed because the set of all irrational numbers (which is the complement of \(\mathbb{Q}\)) is not open, as we can find a rational number within any interval around any given irrational number.

Step by step solution

01

Explain the Open Set Definition

Before working on the problem, let's recall what an open set means: a set \(S\) is open if for each \(x\) in \(S\), there exists \(\epsilon > 0\) such that the interval \((x-\epsilon, x+\epsilon)\) is contained entirely within \(S\).
02

Prove that \(\mathbb{Q}\) is not an Open Set

Choosing any rational number \(q\), we can always find an irrational number within any interval \((q-\epsilon, q+\epsilon)\) around \(q\). This is due to the density of the irrational numbers in the real number line. Hence, it is not possible to find an interval around \(q\) that is entirely contained within the rational numbers \(\mathbb{Q}\). Therefore, \(\mathbb{Q}\) is not open.
03

Explain the Closed Set Definition

We recall that a set is closed if its complement is open. The complement of a set \(S\) is the set of all elements not in \(S\). Thus, the complement of \(\mathbb{Q}\) is the set of all irrational numbers.
04

Prove that \(\mathbb{Q}\) is not a Closed Set

As previously noted, the set of irrational numbers is dense in the real numbers. This means for any irrational number \(r\), we can always find a rational number in any interval \((r-\epsilon, r+\epsilon)\) around \(r\). Therefore, we can't find an interval around any irrational number that is entirely contained within the set of irrational numbers. This means that the complement of \(\mathbb{Q}\) is not open, and hence, \(\mathbb{Q}\) is not closed.

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Most popular questions from this chapter

Use the Heine-Borel Theorem to prove the following version of the Bolzano- Weierstrass Theorem: Every bounded infinite subset of \(\mathbb{R}\) has a cluster point in \(\mathbb{R}\). (Note that if a sct has no cluster points, then it is closed by Theorem \(11.1 .8 .\) )

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 \(\\{x \in \mathbb{R}: f(x)<\alpha\\}\) is open in \(\mathbb{R}\) for cach \(\alpha \in \mathbb{R}\).

Let \(\left(K_{n}: n \in \mathbb{N}\right)\) be a sequence of nonempty compact sets in \(\mathbb{R}\) such that \(K_{1} \supseteq K_{2} \supseteq \cdots \supseteq\) \(K_{n} \supseteq \cdots\) Prove that there exists at least one point \(x \in \mathbb{R}\) such that \(x \in K_{n}\) for all \(n \in \mathbb{N} ;\) that is, the intersection \(\bigcap_{n=1}^{\infty} K_{n}\) is not empty.

If \(A \subseteq \mathbb{R}\), let \(A^{\circ}\) be the union of all open sets that are contained in \(A ;\) the set \(A^{\circ}\) is called the interior of \(A\). Show that \(A^{\circ}\) is an open set, that it is the largest open set contained in \(A\), and that a point \(z\) belongs to \(A^{\circ}\) if and only if \(z\) is an interior point of \(A\).
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