Bounded Continuous Function On Bounded Set Is Uniformly Continuous

Bounded Continuous Function on a Bounded Set is Uniformly Continuous: A Detailed Explanation

This article will rigorously prove the statement: A continuous function on a bounded and closed set is uniformly continuous. This is a fundamental theorem in real analysis with significant implications in various fields, including calculus and numerical analysis. Understanding this concept is crucial for grasping the nuances of continuity and its variations. We'll explore the proof, discuss its implications, and provide examples to solidify your understanding.

Understanding the Concepts

Before delving into the proof, let's clarify the key terms:

• Continuous Function: A function f is continuous at a point x if for every ε > 0, there exists a δ > 0 such that if |x - y| < δ, then |f(x) - f(y)| < ε. A function is continuous on a set if it's continuous at every point in that set.

• Uniformly Continuous Function: A function f is uniformly continuous on a set S if for every ε > 0, there exists a δ > 0 such that for all x, y in S, if |x - y| < δ, then |f(x) - f(y)| < ε. The crucial difference is that δ depends only on ε, not on the specific point x.

• Bounded Set: A set S is bounded if there exists a real number M such that |x| ≤ M for all x in S.

• Closed Set: A set S is closed if it contains all its limit points. Intuitively, this means that if a sequence of points in S converges to a limit, then that limit is also in S.

The Proof: A Continuous Function on a Closed and Bounded Set is Uniformly Continuous

We'll prove this using proof by contradiction. Assume f is a continuous function on a closed and bounded set S, but f is not uniformly continuous.

1. Negation of Uniform Continuity: If f is not uniformly continuous, then there exists an ε > 0 such that for every δ > 0, there exist points x, y in S with |x - y| < δ, but |f(x) - f(y)| ≥ ε.

2. Constructing Sequences: Let's choose a sequence of δ values: δₙ = 1/n for n = 1, 2, 3… For each δₙ, we can find points xₙ, yₙ in S such that |xₙ - yₙ| < δₙ = 1/n, but |f(xₙ) - f(yₙ)| ≥ ε.

3. Bolzano-Weierstrass Theorem: Since S is bounded, the sequence {xₙ} has a convergent subsequence {xₙₖ}. Let's denote the limit of this subsequence as x. Because S is closed, x is also in S.

4. Convergence of the Corresponding Subsequence: Since |xₙₖ - yₙₖ| < 1/nₖ and nₖ → ∞, we have that yₙₖ also converges to x.

5. Contradiction: Since f is continuous at x, for our chosen ε, there exists a δ such that if |x - z| < δ, then |f(x) - f(z)| < ε/2. However, for sufficiently large nₖ, both xₙₖ and yₙₖ are within δ of x. This means |f(xₙₖ) - f(x)| < ε/2 and |f(yₙₖ) - f(x)| < ε/2. By the triangle inequality: |f(xₙₖ) - f(yₙₖ)| < ε, which contradicts our initial assumption that |f(xₙ) - f(yₙ)| ≥ ε.

6. Conclusion: Therefore, our initial assumption must be false. A continuous function on a closed and bounded set must be uniformly continuous.

Implications and Examples

This theorem is incredibly useful in analysis. It assures us that under specific conditions, the approximation of function values can be controlled uniformly across the entire domain. This is particularly relevant in numerical methods where uniform continuity is often a prerequisite for convergence and accuracy.

For example, consider the function f(x) = x² on the interval [0, 1]. This function is continuous on a closed and bounded set, and hence uniformly continuous. However, the same function on the unbounded interval [0, ∞) is continuous but not uniformly continuous.

This theorem provides a powerful tool for understanding and working with continuous functions, significantly impacting various areas within mathematics and its applications. Understanding the proof and its implications is crucial for anyone pursuing advanced studies in analysis or related fields.