Proof: Convergence Of Midpoint Sequence | Real Analysis

Hey guys! Let's dive into a fun little problem in real analysis. We're going to explore the convergence of a sequence formed by taking the middle value of three other convergent sequences. Sounds intriguing, right? This is a classic problem that tests our understanding of sequence convergence and how to manipulate inequalities. So, let's break it down step by step and make sure we nail the proof. We'll make it super clear and easy to follow, so you can confidently tackle similar problems in the future. Think of this as a journey through the world of sequences, where we'll uncover some neat tricks and techniques. Ready to get started? Awesome, let's jump right in!

Okay, so here's the problem we're tackling today: Suppose we have three sequences, X, Y, and Z, and these sequences are all convergent. This means that each sequence approaches a specific limit as 'n' goes to infinity. Now, we define a new sequence, let's call it 'W', where each term $w_n$ is the middle value among the corresponding terms $x_n$, $y_n$, and $z_n$ from sequences X, Y, and Z, respectively. Our mission, should we choose to accept it (and we do!), is to prove that this new sequence 'W' is also convergent. In simpler terms, we need to show that the sequence formed by taking the middle value of three convergent sequences also converges. This problem is not just a theoretical exercise; it highlights the behavior of sequences and how different sequences interact with each other. It's a fundamental concept in real analysis that has applications in various fields, including numerical analysis and optimization. So, let's put on our thinking caps and get to it!

Alright, to get started, let's formalize our assumptions and set the stage for our proof. We're given that the sequences X = ($x_n$), Y = ($y_n$), and Z = ($z_n$) are all convergent. This means that there exist real numbers x, y, and z such that:

• $x_n$ converges to x
• $y_n$ converges to y
• $z_n$ converges to z

In mathematical notation, this can be written as:

• lim (n→∞) $x_n$ = x
• lim (n→∞) $y_n$ = y
• lim (n→∞) $z_n$ = z

Now, we define our sequence W = ($w_n$) such that each term $w_n$ is the middle value of $x_n$, $y_n$, and $z_n$. To make things a bit clearer, we can write this as:

$w_n$ = mid{$x_n$, $y_n$, $z_n$}

This means that for each n, $w_n$ is the value that falls in the middle when you sort $x_n$, $y_n$, and $z_n$. For example, if $x_n$ = 1, $y_n$ = 3, and $z_n$ = 2, then $w_n$ would be 2.

The goal here is to show that the sequence W also converges. To do this, we'll need to show that there exists a real number w such that:

lim (n→∞) $w_n$ = w

Before we dive into the nitty-gritty details of the proof, let's make a simplifying assumption. Without loss of generality (WLOG), we can assume that x ≤ y ≤ z. This assumption helps us organize our thoughts and doesn't affect the generality of the proof because we can always relabel the sequences if needed. The core idea is that convergence is about what happens as n approaches infinity, and relabeling doesn't change the ultimate behavior of the sequences. So, with this setup in place, let's move on to the heart of the proof.

The crux of our proof lies in demonstrating that the sequence W converges. To do this, we need to show that $w_n$ approaches a limit, say 'w', as n tends to infinity. A natural candidate for 'w' is the middle value among the limits x, y, and z. Given our assumption that x ≤ y ≤ z, it's quite intuitive to think that w should be y. Remember, we assumed WLOG that x ≤ y ≤ z, which means y is indeed the middle value among the limits.

So, our aim is to prove that lim (n→∞) $w_n$ = y. To accomplish this, we'll use the epsilon-delta definition of convergence. This means we need to show that for any given ε > 0 (a tiny positive number), there exists a positive integer N such that for all n > N, the distance between $w_n$ and y is less than ε. Mathematically, we want to show that:

| $w_n$ - y | < ε for all n > N

Now, here's where the fun begins. Since $x_n$ converges to x, $y_n$ converges to y, and $z_n$ converges to z, we know that for any ε > 0, we can find integers $N_1$, $N_2$, and $N_3$ such that:

• | $x_n$ - x | < ε for all n > $N_1$
• | $y_n$ - y | < ε for all n > $N_2$
• | $z_n$ - z | < ε for all n > $N_3$

These inequalities tell us that the terms of the sequences X, Y, and Z get arbitrarily close to their respective limits as n gets large. Our goal now is to use these inequalities to show that $w_n$ also gets arbitrarily close to y.

Let's define N = max{$N_1$, $N_2$, $N_3$}. This means that for all n > N, all three inequalities above hold true simultaneously. This is a crucial step because it allows us to work with all three sequences at the same time. Now, we need to relate $w_n$ to y using the information we have about $x_n$, $y_n$, and $z_n$. This is where the properties of the 'mid' function come into play. We need to consider how the middle value $w_n$ behaves in relation to the limits x, y, and z. So, let's dive deeper into how we can use these inequalities to bound | $w_n$ - y | and show that it is indeed less than ε for all n > N. This is the heart of our proof, and we'll tackle it with precision and clarity. Stay tuned!

Okay, guys, let's get into the nitty-gritty details of the proof. We've set up the foundation, and now it's time to build the main structure. Remember, our goal is to show that | $w_n$ - y | < ε for all n > N, where $w_n$ is the middle value of $x_n$, $y_n$, and $z_n$, and y is the middle value of the limits x, y, and z (with x ≤ y ≤ z).

We know that for n > N = max{$N_1$, $N_2$, $N_3$}, the following inequalities hold:

• | $x_n$ - x | < ε which implies x - ε < $x_n$ < x + ε
• | $y_n$ - y | < ε which implies y - ε < $y_n$ < y + ε
• | $z_n$ - z | < ε which implies z - ε < $z_n$ < z + ε

These inequalities tell us that for large enough n, $x_n$ is close to x, $y_n$ is close to y, and $z_n$ is close to z. Now, we need to figure out how these proximities affect the middle value $w_n$.

Since $w_n$ = mid{$x_n$, $y_n$, $z_n$}, we know that $w_n$ is somewhere between the smallest and the largest of $x_n$, $y_n$, and $z_n$. To bound $w_n$, let's consider the possible scenarios. No matter the actual order of $x_n$, $y_n$, and $z_n$, we can say that:

min{$x_n$, $y_n$, $z_n$} ≤ $w_n$ ≤ max{$x_n$, $y_n$, $z_n$}

This is a fundamental property of the middle value – it's always between the minimum and maximum values. Now, let's use our inequalities to bound the minimum and maximum.

From x - ε < $x_n$, y - ε < $y_n$, and z - ε < $z_n$, we can say:

min{x - ε, y - ε, z - ε} < min{$x_n$, $y_n$, $z_n$}

Since we assumed x ≤ y ≤ z, we have min{x - ε, y - ε, z - ε} = x - ε. So,

x - ε < min{$x_n$, $y_n$, $z_n$} ≤ $w_n$

Similarly, from $x_n$ < x + ε, $y_n$ < y + ε, and $z_n$ < z + ε, we can say:

$w_n$ ≤ max{$x_n$, $y_n$, $z_n$} < max{x + ε, y + ε, z + ε}

Again, since x ≤ y ≤ z, we have max{x + ε, y + ε, z + ε} = z + ε. So,

$w_n$ < z + ε

Combining these, we have:

x - ε < $w_n$ < z + ε

However, this bound isn't directly helping us show | $w_n$ - y | < ε. We need to relate $w_n$ more closely to y. Let's think about how we can refine this. The key insight here is that since $w_n$ is the middle value, it must be