Continuous On A Compact Set Is Uniformly Continuous

Continuous on a Compact Set is Uniformly Continuous: A Comprehensive Guide

This article will rigorously prove the theorem: a function continuous on a compact set is uniformly continuous. This is a fundamental result in real analysis with significant implications in calculus and beyond. We'll explore the theorem's meaning, the proof, and its practical significance. Understanding this theorem is crucial for anyone studying advanced calculus or analysis.

What Does it Mean?

Before diving into the proof, let's clarify the terms.

• Continuous: 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. In simpler terms, small changes in the input (x) lead to small changes in the output (f(x)). A function is continuous on a set if it's continuous at every point in that set.

• Uniformly Continuous: 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 key difference here is that δ depends only on ε, not on the specific point x. This means the same δ works for the entire set.

• Compact Set: A set S is compact if every open cover of S has a finite subcover. Intuitively, a compact set is "bounded" and "closed." For subsets of the real numbers, compactness is equivalent to being closed and bounded.

The theorem states that if a function is continuous on a compact set, the stronger condition of uniform continuity holds. This means that the "smoothness" of the function is consistent across the entire compact set.

Proof:

We will prove this by contradiction.

1. Assume the opposite: Suppose f is continuous on a compact set S, but f is not uniformly continuous.

2. Formalize the negation of uniform continuity: This means there exists an ε > 0 such that for every δ > 0, there exist points x and y in S with |x - y| < δ, but |f(x) - f(y)| ≥ ε.

3. Construct sequences: For each n ∈ ℕ, let δ = 1/n. Then there exist points xₙ and yₙ in S such that |xₙ - yₙ| < 1/n and |f(xₙ) - f(yₙ)| ≥ ε.

4. Use compactness: Since S is compact, the sequence {xₙ} has a subsequence {xₙₖ} that converges to some point x ∈ S. Because |xₙ - yₙ| < 1/n, the subsequence {yₙₖ} also converges to x.

5. Apply continuity: Since f is continuous at x, we have lim (k→∞) f(xₙₖ) = f(x) and lim (k→∞) f(yₙₖ) = f(x).

6. Reach a contradiction: This implies that lim (k→∞) |f(xₙₖ) - f(yₙₖ)| = 0, which contradicts our earlier condition that |f(xₙ) - f(yₙ)| ≥ ε for all n.

7. Conclusion: Our initial assumption must be false. Therefore, if f is continuous on a compact set S, then f is uniformly continuous on S.

Implications and Applications:

This theorem is a cornerstone of real analysis. Its implications are far-reaching:

• Extending results: Many theorems that rely on uniform continuity can now be applied to continuous functions on compact sets.
• Simplifying proofs: The theorem often simplifies proofs by allowing us to replace the stronger condition of uniform continuity with the weaker condition of continuity on a compact set.
• Numerical analysis: Uniform continuity guarantees that numerical approximations of continuous functions on compact sets will converge reliably.

Understanding this theorem is essential for a deep grasp of real analysis and its applications in various fields, including calculus, differential equations, and numerical methods. It highlights the importance of compactness in guaranteeing uniformity and stability in mathematical functions.