Proof Of Uniqueness Of Sequence Limits

Hey guys! So, I've been diving deep into the fascinating world of Real Analysis lately, and one of the concepts that really got me thinking is the uniqueness of the limit of a sequence. It sounds simple enough, right? A sequence can only converge to one specific number. But how do we actually prove that? I decided to put my understanding to the test and came up with a logical proof using a truth table approach to convince myself. It's always the best way to really nail these concepts, don't you think?

I wanted to share my proof with you all here in Plastik Magazine to get your thoughts. Is it solid? Does it need any tweaks? Let's break it down together. So, the fundamental idea we're trying to prove is that if a sequence $(a_n)$ converges to a limit $L$, then it cannot converge to any other limit, say $M$, where $M eq L$. This is a cornerstone property in calculus and real analysis, and understanding the proof helps solidify our grasp on limits themselves.

The Setup: What We Assume and What We Want to Show

Before we jump into the proof, let's lay down the groundwork. We're working within the realm of sequences and their limits. The definition of a sequence converging to a limit $L$ is crucial here. It states that for every $\epsilon > 0$, there exists a natural number $N$ such that for all $n > N$, the absolute difference between $a_n$ and $L$ is less than $\epsilon$, i.e., $|a_n - L| < \epsilon$. This definition is the bedrock upon which our proof will be built. We're essentially saying that the terms of the sequence get arbitrarily close to the limit $L$ as $n$ gets large enough.

Now, to prove uniqueness, we'll employ a common proof technique: proof by contradiction. We'll assume the opposite of what we want to prove and show that this assumption leads to a logical inconsistency. So, let's assume that our sequence $(a_n)$ has two distinct limits, $L$ and $M$, with $L \neq M$. Our goal is to show that this assumption cannot hold true, thereby proving that the limit, if it exists, must be unique.

Constructing the Proof: A Logical Approach

Let's dive into the heart of the proof. We are given a sequence $(a_n)$ that converges to $L$. This means, by definition, for any $\epsilon_1 > 0$, there exists an $N_1 \in \mathbb{N}$ such that for all $n > N_1$, we have $|a_n - L| < \epsilon_1$.

Simultaneously, let's assume $(a_n)$ also converges to $M$, where $M \neq L$. This means, by definition, for any $\epsilon_2 > 0$, there exists an $N_2 \in \mathbb{N}$ such that for all $n > N_2$, we have $|a_n - M| < \epsilon_2$.

Our goal is to show that these two conditions cannot coexist if $L \neq M$. To do this, we need to choose our $\epsilon$ values strategically. Since $L$ and $M$ are distinct real numbers, the distance between them, $|L - M|$, is a positive value. Let's denote this positive distance by $d = |L - M| > 0$.

Now, consider choosing $\epsilon_1 = \frac{d}{2}$ and $\epsilon_2 = \frac{d}{2}$. These are positive values, so they are valid choices for our epsilon in the limit definitions.

With these choices, we have:

1. For $\epsilon_1 = \frac{d}{2}$: There exists $N_1$ such that for all $n > N_1$, $|a_n - L| < \frac{d}{2}$.
2. For $\epsilon_2 = \frac{d}{2}$: There exists $N_2$ such that for all $n > N_2$, $|a_n - M| < \frac{d}{2}$.

Now, let's consider an index $n$ that is greater than both $N_1$ and $N_2$. Let $N = \max(N_1, N_2)$. For any $n > N$, both of the above inequalities hold true.

So, for $n > N$, we have $|a_n - L| < \frac{d}{2}$ and $|a_n - M| < \frac{d}{2}$.

Let's think about the distance between $L$ and $M$. We know $|L - M| = d$. We can use the triangle inequality to relate this distance to the terms $a_n$. Consider the expression $|L - M|$. We can rewrite this as:

$|L - M| = |L - a_n + a_n - M|$

Using the triangle inequality, $|x + y| \le |x| + |y|$, we get:

$|L - M| \le |L - a_n| + |a_n - M|$

Since $|L - a_n| = |a_n - L|$, we have:

$|L - M| \le |a_n - L| + |a_n - M|$

Now, for $n > N$, we know that $|a_n - L| < \frac{d}{2}$ and $|a_n - M| < \frac{d}{2}$. Substituting these into the inequality:

$|L - M| < \frac{d}{2} + \frac{d}{2}$

$|L - M| < d$

But we defined $d = |L - M|$ at the beginning. So, the inequality becomes:

$d < d$

This is a clear contradiction! The statement $d < d$ is logically impossible.

The Truth Table Logic Behind It

While I didn't literally construct a truth table with 'True' and 'False' columns for each propositional variable, the underlying logic mirrors a truth table's function. Let's think of the statements involved:

• P: Sequence $(a_n)$ converges to $L$. (Assumed True)
• Q: Sequence $(a_n)$ converges to $M$. (Assumed True for contradiction)
• R: $L \neq M$. (Assumed True for contradiction)

Our proof shows that the conjunction of P, Q, and R leads to a logical falsehood (e.g., $d < d$). In propositional logic, if a premise $A$ implies a contradiction (False), then the premise $A$ must be false. Here, our premise is "$(a_n)$ converges to $L$ AND $(a_n)$ converges to $M$ AND $L \neq M$". Since this leads to a contradiction, at least one of these statements must be false.

We start by assuming P is true (the sequence converges to $L$). We then assume Q and R are also true (it also converges to a different limit $M$). Our derivation $d < d$ is the