Constant Function Proof: Nested Interval Theorem

Hey everyone! Today, we're diving deep into a fascinating concept in calculus: proving that a continuous function, which hits local extrema at every single point, is actually a constant function. This might sound a bit mind-bending at first, but we'll break it down step-by-step using the powerful Nested Interval Theorem. We're tackling exercise 70 from chapter 11 of Spivak's Calculus (4th edition), and I know some of you might be scratching your heads at the solution manual's approach. Don't worry, we'll demystify it together and explore how one might even stumble upon such a proof.

The Core Idea: Local Extrema and Constant Functions

Let's start with the basics. What does it mean for a function to attain a local extremum at a point? Simply put, it means that at that point, the function reaches either a local maximum (a peak) or a local minimum (a valley) within a small neighborhood around that point. Now, imagine a function that does this everywhere. It's like the function is constantly trying to climb up or down, but it's doing so at every single location. Intuitively, if a continuous function is always turning around (going up then down, or down then up), but it has to do it at every point, it suggests the function can't actually go anywhere! It's stuck being flat – a constant function. Our mission is to rigorously prove this intuition using the Nested Interval Theorem. The proof that a continuous function which attains local extrema at every point is constant isn't immediately obvious, but by dissecting the logic and building upon fundamental theorems, we'll see how beautifully it unfolds.

Breaking Down the Problem

Before we jump into the formal proof, let's outline the general strategy. We need to show that if a continuous function ${f}$ has a local extremum at every point in its domain, then ${f(x)}$ is the same for all ${x}$ in the domain. The key idea is to assume the opposite – that there exist two points where the function values are different – and then use the Nested Interval Theorem to derive a contradiction. This contradiction will then force us to accept that our initial assumption was false, and therefore the function must be constant. The Nested Interval Theorem will be our main tool here, allowing us to zoom in on intervals where the function's behavior can't possibly be consistent with the existence of local extrema everywhere if the function isn't constant. We'll also need to leverage the continuity of the function, which guarantees that small changes in the input result in small changes in the output. This interplay between continuity and local extrema is what ultimately drives the proof.

The Nested Interval Theorem: Our Guiding Star

So, what exactly is this Nested Interval Theorem we keep mentioning? In essence, it states that if you have a sequence of closed, bounded intervals on the real number line, where each interval is contained within the previous one, and the lengths of the intervals shrink towards zero, then there exists exactly one point that belongs to all of the intervals. Think of it like zooming in on a map. Each zoom level gives you a smaller area, but there's always one specific location that remains in focus. This theorem is a cornerstone of real analysis, and it's going to be crucial in our proof. It allows us to construct a sequence of intervals where the function's behavior becomes increasingly constrained, ultimately leading to our desired contradiction. Understanding how the Nested Interval Theorem applies is key to grasping the entire proof strategy. We'll be building nested intervals based on the assumption that the function isn't constant and then showing that this assumption leads to a situation where the function can't have local extrema at every point within those intervals.

The Proof Unveiled: Step-by-Step

Alright, let's get our hands dirty and dive into the proof itself. Remember, our goal is to show that if a continuous function ${f}$ has a local extremum at every point, then it must be constant. We'll proceed by contradiction.

1. Assume the Opposite: Suppose, for the sake of contradiction, that ${f}$ is not constant. This means there exist two points, let's call them ${a}$ and ${b}$, in the domain of ${f}$ such that ${f(a) \neq f(b)}$. Without loss of generality, let's assume that ${f(a) < f(b)}$. This sets the stage for our contradiction argument. If the function isn't constant, we have a starting point to build our nested intervals.

2. Construct the First Interval: Consider the interval ${[a, b]}$. Since ${f}$ has a local extremum at every point, it has a local extremum at some point, say ${x_1}$, within the open interval ${(a, b)}$. This means there exists a small neighborhood around ${x_1}$ where ${f(x_1)}$ is either the maximum or the minimum value of the function within that neighborhood. This is the first step in leveraging the local extrema property. We're identifying a point where the function must