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Chapter 5: Problem 10

Prove that if \(f\) is uniformly continuous on a bounded subset \(A\) of \(\mathbb{R}\), then \(f\) is bounded on \(A\).

Short Answer

Expert verified
Because \(f\) is uniformly continuous on a bounded subset \(A\) of \(\mathbb{R}\), there exists a \(\delta > 0\) such that for all \(x, y \in A\) with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < 1\). By selecting a number of points in \(A\), each within \(\delta\) of at least one other, we can show that \(|f(x)| < \max_{1 \leq i \leq n} |f(x_i)| + 1\) for all \(x \in A\), proving that \(f\) is bounded on \(A\).

Step by step solution

01

Definition of Uniform Continuity

First, recall the definition of uniform continuity. A function \(f: A \rightarrow \mathbb{R}\) is said to be uniformly continuous on \(A\) if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y \in A\) with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < \epsilon\). Let's set \(\epsilon = 1\). So, according to the definition, there is a \(\delta > 0\) such that \(|f(x) - f(y)| < 1\) whenever \(|x - y| < \delta\).
02

Boundedness of Set A

Since \(A\) is bounded, there is a real number \(M\) such that \(|x| \leq M\) for all \(x \in A\). Now, take an arbitrary point \(a \in A\). Consider intervals of length \(\delta\) that cover the set \(A\). We can find a finite number of points \(x_1, x_2, \ldots, x_n\) in \(A\) such that every point in \(A\) is within \(\delta\) of at least one \(x_i\).
03

Obtaining an Upper Bound for f

We know from step 1 that \(|f(x) - f(x_i)| < 1\) for any \(x \in A\) where \(|x - x_i| < \delta\). This implies that \(|f(x)| < |f(x_i)| + 1\). Further, by considering all points \(x_1, x_2, \ldots, x_n\), we can say \(|f(x)| < \max_{1 \leq i \leq n} |f(x_i)| + 1\), that holds for all \(x \in A\). This shows that \(f\) is bounded on \(A\).

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