Question

Let \(f\) be a real uniformly continuous function on the bounded set \(E\) in \(R^{1}\). Prove that \(f\) is bounded on \(E\). Show that the conclusion is false if boundedness of \(E\) is omitted from the hypothesis.

Short answer

#Answer# 1. If \(f\) is a uniformly continuous function on the bounded set \(E\), then \(f\) is bounded on \(E\). 2. The conclusion that \(f\) is bounded on \(E\) is false if the boundedness of \(E\) is omitted from the hypothesis. An example of this is the function \(f(x) = x\) on the set of real numbers, \(R\). Although this function is uniformly continuous, it is not bounded.

Step-by-step solution

Proof by contradiction

Let's assume that \(f\) is not bounded on \(E\). This means, there exists a sequence \(\{x_n\} \subseteq E\) such that \(|f(x_n)| \rightarrow \infty\) as \(n \rightarrow \infty\).

Uniform Continuity definition

Since \(f\) is uniformly continuous on \(E\), for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for every \(x, y \in E\), if \(|x - y| < \delta\), then \(|f(x) - f(y)| < \epsilon\).

Bounded set properties

We know that \(E\) is a bounded set, so it has a supremum (least upper bound) and infimum (greatest lower bound). In other words, there exist \(L, U \in R\) such that \(L \leq x, y \leq U\) for all \(x, y \in E\). We define \(\Delta = U - L\).

Create a contradiction

Choose \(\epsilon = 1\). Since \(f\) is uniformly continuous on \(E\), there exists a \(\delta > 0\). Now, create a new sequence \(\{y_n\}\) such that \(|y_n - x_n| = \frac{\delta}{2}\) and \(y_n \in E\). By our previous statement regarding uniform continuity, for every \(n\), we have \(|f(y_n) - f(x_n)| < 1\). However, given that \(|f(x_n)| \rightarrow \infty\), this implies that \(|f(y_n)| \rightarrow \infty\) as well, which contradicts the uniform continuity of \(f\). Therefore, our assumption that \(f\) is not bounded on \(E\) is wrong, and we can conclude that \(f\) is indeed bounded on \(E\).

Counterexample when E is unbounded

Now, let's consider a case where \(E\) is unbounded. Take the function \(f(x) = x\) on the set of real numbers, \(R\). The function is uniformly continuous, as for any \(\epsilon > 0\), we can choose \(\delta = \epsilon\). But clearly, the function is not bounded on \(R\). Therefore, when the boundedness of \(E\) is omitted from the hypothesis, the conclusion is false.

Key concepts

Bounded Set in Real Analysis

In real analysis, a bounded set is fundamental to understanding concepts like continuity and limits. A set of real numbers is said to be bounded if there exists a real number that serves as a boundary beyond which no elements of the set can be found.

For example, consider the set \( E = \{ x \in \mathbb{R} \,|\, a \leq x \leq b \} \) where \( a \) and \( b \) are real numbers. \( E \) is bounded because all elements are contained within the interval from \( a \) to \( b \). We define these boundary points as the infimum (inf) or greatest lower bound, and supremum (sup) or least upper bound, respectively.

In the context of uniform continuity, a function defined on a bounded set \( E \) carries an assurance of both the function's and its outputs' containment within some bounds. Conversely, if a set is unbounded—there's no limit to the values it can contain—proving uniform continuity does not guarantee that its function will be bounded, as demonstrated in the exercise.

Proof by Contradiction

The proof by contradiction is a powerful logical tool in mathematics used to establish the validity of a statement. The technique involves assuming the opposite of what you want to prove and then showing that this assumption inevitably leads to an inconsistency or contradiction. In this way, one can conclude that the original assumption must be false and therefore the statement being proven must be true.

Applying this method to the exercise, it was assumed that the function \( f \) was not bounded on the set \( E \). If this were true, we could find a sequence within \( E \) where the function values increase without bound. However, by also knowing that the function is uniformly continuous on \( E \), a contradiction arises, as uniform continuity implies that the function’s values cannot diverge as they did in the assumption. The emerging contradiction proves that the assumption was false, and thus, the function must indeed be bounded on the set \( E \).

Properties of Supremum and Infimum

The concepts of supremum (sup) and infimum (inf) are essential in real analysis for characterizing the bounds of sets. Every non-empty set of real numbers that is bounded above has a least upper bound known as the supremum. Similarly, every non-empty set of real numbers that is bounded below has a greatest lower bound known as the infimum.

These properties are especially relevant when discussing bounded sets. The supremum and infimum are not necessarily elements of the set but are limits that no member of the set can exceed. In the provided exercise, using the properties of supremum and infimum is crucial in showing that a uniformly continuous function on a bounded set must be bounded. The values between sup \( (E) \) and inf \( (E) \) define a range within which all function values must lie, due to the uniform continuity of the function. This range is crucial in constructing the argument that led to the enactment of the contradiction in Step 4 of the solution.