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Chapter 2: Problem 4

Show that \(|x-a|<\varepsilon\) if and only if \(a-\varepsilon

Short Answer

Expert verified
The statement \(|x-a|<\varepsilon\) is equivalent to \(a-\varepsilon<x<a+\varepsilon\). This is because both are saying exactly the same thing: the distance between \(x\) and \(a\) on the number line is less than \(\varepsilon\).

Step by step solution

01

- Understanding the Absolute Value

The first thing to understand is that absolute value of a difference \(|x - a|\) represents the distance between \(x\) and \(a\) on the real number line. This distance is always positive or zero, never negative.
02

- Breaking Down the First Statement

Now let's consider the first statement \(|x - a| < \varepsilon\). This means that the distance between \(x\) and \(a\) is less than \(\varepsilon\). On the number line, this implies that \(x\) is located within the interval \((a - \varepsilon, a + \varepsilon)\), i.e., \(a - \varepsilon < x < a + \varepsilon\). This demonstrates that the first statement implies the second statement.
03

- Breaking Down the Second Statement

Let's consider the second statement \(a - \varepsilon < x < a + \varepsilon\). This statement shows that \(x\) lies between \(a - \varepsilon\) and \(a + \varepsilon\) on the number line, i.e., the distance between \(x\) and \(a\) is less than \(\varepsilon\). Formally said, \(|x - a| < \varepsilon\). This demonstrates that the second statement implies the first statement.
04

- Conclusion

Since we have shown that the first statement implies the second and the second implies the first, we have proven that the two statements are equivalent, i.e., \(|x - a| < \varepsilon\) if and only if \(a - \varepsilon < x < a + \varepsilon\).

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Most popular questions from this chapter

(a) If \(a

Let \(X\) and \(Y\) be nonempty sets and let \(h: X \times Y \rightarrow \mathbb{R}\) have bounded range in \(\mathbb{R}\). Let \(F: X \rightarrow \mathbb{R}\) and \(G: Y \rightarrow \mathbb{R}\) be defined by $$ F(x):=\sup \\{h(x, y): y \in Y\\}, \quad G(y):=\sup \\{h(x, y): x \in X\\} $$ Establish the Principle of the Iterated Suprema: $$ \sup \\{h(x, y): x \in X, y \in Y\\}=\sup \\{F(x): x \in X\\}=\sup \\{G(y): y \in Y\\} $$ We sometimes express this in symbols by $$ \sup _{x, y} h(x, y)=\sup _{x} \sup _{y} h(x, y)=\sup _{y} \sup _{x} h(x, y) . $$

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