Answered: Prove that Cl(Q) = R in the standard…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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$**Problem Statement:** Prove that $ Cl(\mathbb{Q}) = \mathbb{R} $ in the standard topology on $ \mathbb{R} $. **Explanation:** This problem asks us to show that the closure of the rational numbers $\mathbb{Q}$ in the real numbers $\mathbb{R}$ is the entire set of real numbers $\mathbb{R}$, within the context of the standard topology. **Approach:** 1. **Understanding the Closure:** The closure of a set $A$ in a topological space is the smallest closed set containing $A$ or equivalently, the set of all limit points of $A$ along with the points in $A$. 2. **Rational Numbers:** The set of rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$, meaning that between any two real numbers, no matter how close, there exists a rational number. 3. **Density Argument:** To show that $ Cl(\mathbb{Q}) = \mathbb{R} $, consider any real number $x \in \mathbb{R}$. For every $x$, and for any $\epsilon > 0$, there exists a rational number $q$ such that $|x - q| < \epsilon$. Thus, every real number is a limit point of $\mathbb{Q}$. 4. **Conclusion:** Since every real number can be approximated arbitrarily closely by rational numbers, the closure of $\mathbb{Q}$ in $\mathbb{R}$ is indeed $\mathbb{R}$. This illustrates a fundamental property of the rational numbers within the real number system pertaining to their distribution.$

Transcribed Image Text:**Problem Statement:** Prove that $ Cl(\mathbb{Q}) = \mathbb{R} $ in the standard topology on $ \mathbb{R} $. **Explanation:** This problem asks us to show that the closure of the rational numbers $\mathbb{Q}$ in the real numbers $\mathbb{R}$ is the entire set of real numbers $\mathbb{R}$, within the context of the standard topology. **Approach:** 1. **Understanding the Closure:** The closure of a set $A$ in a topological space is the smallest closed set containing $A$ or equivalently, the set of all limit points of $A$ along with the points in $A$. 2. **Rational Numbers:** The set of rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$, meaning that between any two real numbers, no matter how close, there exists a rational number. 3. **Density Argument:** To show that $ Cl(\mathbb{Q}) = \mathbb{R} $, consider any real number $x \in \mathbb{R}$. For every $x$, and for any $\epsilon > 0$, there exists a rational number $q$ such that $|x - q| < \epsilon$. Thus, every real number is a limit point of $\mathbb{Q}$. 4. **Conclusion:** Since every real number can be approximated arbitrarily closely by rational numbers, the closure of $\mathbb{Q}$ in $\mathbb{R}$ is indeed $\mathbb{R}$. This illustrates a fundamental property of the rational numbers within the real number system pertaining to their distribution.

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