Proving Function Continuity With Compact Sets

Hey guys! Today, we're diving deep into a fascinating problem from real analysis, specifically from Pugh's Real Mathematical Analysis. It's all about proving the continuity of a function $f: ext{ℝ}^m \to ext{ℝ}$ under some pretty interesting conditions. So, buckle up, and let's get started!

The Challenge: Decoding the Problem

Our mission, should we choose to accept it (and we totally do!), is to prove that a function f is continuous given two key properties:

(i) For each compact set K in ℝ^m, its image f(K) is compact in ℝ. (ii) For any nested sequence of compact sets {K_n} in ℝ^m, the image of their intersection is equal to the intersection of their images, i.e., f(∩K_n) = ∩f(K_n).

This problem beautifully marries concepts from real analysis and general topology, particularly focusing on continuity and the properties of compact sets. It’s a classic example of how seemingly abstract mathematical conditions can lead to concrete conclusions about the behavior of functions. We will explore the conditions to show f: ℝ^m → ℝ is continuous and address any concerns about the solution verification.

Laying the Groundwork: Key Concepts and Theorems

Before we jump into the proof, let's refresh some fundamental concepts that will serve as our building blocks:

• Compactness: A set K in ℝ^m is compact if it is both closed and bounded. This is the famous Heine-Borel theorem in action! Compact sets are incredibly well-behaved, and their properties are crucial in analysis.
• Continuity: A function f is continuous at a point x if, for any 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 lead to small changes in the output.
• Nested Sets: A sequence of sets {K_n} is nested if K_(n+1) ⊆ K_n for all n. Think of it like Russian nesting dolls – each set is contained within the previous one.
• Intersection of Compact Sets: The intersection of any collection of closed sets is closed. Furthermore, the intersection of a nested sequence of non-empty compact sets is non-empty.
• Image of a Compact Set under a Continuous Function: If f is continuous and K is compact, then f(K) is compact. This is a cornerstone theorem connecting continuity and compactness.

With these concepts in our toolkit, we're ready to tackle the proof!

The Proof: A Step-by-Step Journey

We'll prove the continuity of f using a proof by contradiction. This means we'll assume f is not continuous at some point and then show that this assumption leads to a logical inconsistency.

1. Assume f is discontinuous at some point x₀ ∈ ℝ^m. This means there exists an ε₀ > 0 such that for every δ > 0, there exists a point x in ℝ^m with |x - x₀| < δ but |f(x) - f(x₀*)| ≥ ε₀. In plain English, no matter how close we get to x₀, there are always points nearby whose function values are not close to f(x₀*)*.
2. **Construct a sequence of points {x_n} that