Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Question
**Title: Proving the Openness of the Interval \(E = (a, +\infty)\) in \(\mathbb{R}\)** **Objective:** To prove that the interval \(E = (a, +\infty)\) is an open set in the real number system \(\mathbb{R}\). We will show all the details of the proof. **Introduction to Open Sets:** In the context of real numbers, a set \(S\) is open if, for every point \(x \in S\), there exists an \(\epsilon > 0\) such that the interval \((x - \epsilon, x + \epsilon) \subseteq S\). **Proof:** We need to show that for any \(x \in (a, +\infty)\), there exists an \(\epsilon > 0\) for which \((x - \epsilon, x + \epsilon) \subseteq (a, +\infty)\). 1. **Choose any \(x \in (a, +\infty)\).** - By definition, \(x > a\). 2. **Define \(\epsilon = \frac{x - a}{2}\).** - Since \(x > a\), \(\epsilon\) is positive. 3. **Consider the interval \((x - \epsilon, x + \epsilon)\).** - Calculate \(x - \epsilon = x - \frac{x - a}{2} = \frac{2x - x + a}{2} = \frac{x + a}{2}\). - Note that \(\frac{x + a}{2} > a\) because \(x > a\). 4. **Verify that \((x - \epsilon, x + \epsilon) \subseteq (a, +\infty)\):** - From above, \(x - \epsilon = \frac{x + a}{2} > a\), ensuring \(x - \epsilon > a\). - Therefore, the entire interval \((x - \epsilon, x + \epsilon)\) is contained within \((a, +\infty)\). **Conclusion:** Since for every \(x \in (a, +\infty)\), there exists an \(\epsilon > 0\) such that \((x - \epsilon, x + \epsilon
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Transcribed Image Text:**Title: Proving the Openness of the Interval \(E = (a, +\infty)\) in \(\mathbb{R}\)** **Objective:** To prove that the interval \(E = (a, +\infty)\) is an open set in the real number system \(\mathbb{R}\). We will show all the details of the proof. **Introduction to Open Sets:** In the context of real numbers, a set \(S\) is open if, for every point \(x \in S\), there exists an \(\epsilon > 0\) such that the interval \((x - \epsilon, x + \epsilon) \subseteq S\). **Proof:** We need to show that for any \(x \in (a, +\infty)\), there exists an \(\epsilon > 0\) for which \((x - \epsilon, x + \epsilon) \subseteq (a, +\infty)\). 1. **Choose any \(x \in (a, +\infty)\).** - By definition, \(x > a\). 2. **Define \(\epsilon = \frac{x - a}{2}\).** - Since \(x > a\), \(\epsilon\) is positive. 3. **Consider the interval \((x - \epsilon, x + \epsilon)\).** - Calculate \(x - \epsilon = x - \frac{x - a}{2} = \frac{2x - x + a}{2} = \frac{x + a}{2}\). - Note that \(\frac{x + a}{2} > a\) because \(x > a\). 4. **Verify that \((x - \epsilon, x + \epsilon) \subseteq (a, +\infty)\):** - From above, \(x - \epsilon = \frac{x + a}{2} > a\), ensuring \(x - \epsilon > a\). - Therefore, the entire interval \((x - \epsilon, x + \epsilon)\) is contained within \((a, +\infty)\). **Conclusion:** Since for every \(x \in (a, +\infty)\), there exists an \(\epsilon > 0\) such that \((x - \epsilon, x + \epsilon
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