Supremum & Infimum: Using The Archimedean Property

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Hey guys! Let's dive into a Real Analysis problem that many students find tricky: using the Archimedean Property to nail down the supremum and infimum of a set. We'll break it down step by step, making sure you understand not just how but why it works. Trust me, once you get this, you'll feel like a real analysis rockstar!

Understanding the Problem

So, imagine you're given a set, let's call it S. This set is defined as follows:

S = {1/n - 1/m : n, m ∈ ℕ}

Where ℕ represents the set of natural numbers (1, 2, 3,...). Your mission, should you choose to accept it, is to find the supremum (sup S) and the infimum (inf S) of this set. Basically, what's the smallest upper bound and the largest lower bound of S? This is where the Archimedean Property comes into play. It might seem abstract at first, but it's a powerful tool.

To start, let's clarify what supremum and infimum mean. The supremum of a set S is the least upper bound. That means it's the smallest number that is greater than or equal to all the numbers in the set. The infimum, on the other hand, is the greatest lower bound. It's the largest number that is less than or equal to all numbers in the set. Think of it like trying to find the highest and lowest points of a mountain range – but with a mathematical twist!

Now, why do we need the Archimedean Property? Well, this property essentially states that for any real number x, there exists a natural number n such that n > x. In simpler terms, you can always find a natural number bigger than any real number you pick. This might seem obvious, but it's crucial when dealing with limits and bounds, especially when dealing with sets that involve fractions and infinity.

Think of it this way: The Archimedean Property helps us deal with infinitely small quantities. In our set S, we have terms like 1/n and 1/m. As n and m get larger, these fractions get smaller and smaller, approaching zero. The Archimedean Property ensures that we can always find an n or m large enough to make these fractions arbitrarily small, which is essential for determining the supremum and infimum.

So, before we jump into the solution, remember these key ideas:

  • Supremum (sup S): The least upper bound of S.
  • Infimum (inf S): The greatest lower bound of S.
  • Archimedean Property: For any real number x, there exists a natural number n such that n > x.

With these concepts in mind, let's move on to cracking this problem!

Finding the Supremum

Alright, let's find the supremum of the set S = {1/n - 1/m : n, m ∈ ℕ}. We need to figure out the smallest upper bound for all possible values of 1/n - 1/m.

First, let's think about how we can maximize the expression 1/n - 1/m. We want 1/n to be as large as possible and 1/m to be as small as possible. Since n and m are natural numbers, the smallest possible value for n is 1. So, let's set n = 1. Now we have 1 - 1/m.

As m gets larger, 1/m gets smaller, and 1 - 1/m gets closer and closer to 1. So, we can make 1 - 1/m arbitrarily close to 1 by choosing a large enough m. This suggests that the supremum of S is 1.

But how do we prove it rigorously? Here's where the Archimedean Property comes in handy. We need to show two things:

  1. 1 is an upper bound for S.
  2. For any ε > 0, there exists an element in S that is greater than 1 - ε.

Let's tackle the first part. We need to show that 1/n - 1/m ≤ 1 for all n, m in ℕ. Since n ≥ 1, we have 1/n ≤ 1. Also, -1/m ≤ 0 for all m in ℕ. Therefore, 1/n - 1/m ≤ 1 + 0 = 1. So, 1 is indeed an upper bound for S.

Now, let's move on to the second part. We need to show that for any positive ε, there exists an element in S that is greater than 1 - ε. In other words, we need to find n and m such that 1/n - 1/m > 1 - ε.

Let's choose n = 1. Then we need to find m such that 1 - 1/m > 1 - ε. This simplifies to 1/m < ε, or m > 1/ε. Now, here's where the Archimedean Property shines! It guarantees that for any real number 1/ε, there exists a natural number m such that m > 1/ε. So, we can always find such an m.

Therefore, for any ε > 0, we can choose n = 1 and m to be any natural number greater than 1/ε. Then, 1/n - 1/m = 1 - 1/m > 1 - ε. This proves that 1 is the least upper bound of S.

So, we can confidently say that:

sup S = 1

Nice job! Let's move on to finding the infimum.

Finding the Infimum

Okay, now let's find the infimum of the set S = {1/n - 1/m : n, m ∈ ℕ}. This time, we're looking for the greatest lower bound for all possible values of 1/n - 1/m.

To minimize the expression 1/n - 1/m, we want 1/n to be as small as possible and 1/m to be as large as possible. As n gets larger, 1/n gets smaller, approaching zero. The largest possible value for 1/m occurs when m = 1. So, let's consider what happens as n approaches infinity and m = 1. We get 0 - 1 = -1.

This suggests that the infimum of S is -1. But again, we need to prove it rigorously.

We need to show:

  1. -1 is a lower bound for S.
  2. For any ε > 0, there exists an element in S that is less than -1 + ε.

Let's start with the first part. We need to show that 1/n - 1/m ≥ -1 for all n, m in ℕ. Since 1/n ≥ 0 for all n in ℕ, we have 1/n - 1/m ≥ 0 - 1/m. The smallest value of m is 1, so -1/m ≥ -1. Therefore, 1/n - 1/m ≥ -1, which means -1 is indeed a lower bound for S.

Now, let's move on to the second part. We need to show that for any positive ε, there exists an element in S that is less than -1 + ε. In other words, we need to find n and m such that 1/n - 1/m < -1 + ε.

Let's choose m = 1. Then we need to find n such that 1/n - 1 < -1 + ε. This simplifies to 1/n < ε, or n > 1/ε. Once again, the Archimedean Property comes to our rescue! It guarantees that for any real number 1/ε, there exists a natural number n such that n > 1/ε. So, we can always find such an n.

Therefore, for any ε > 0, we can choose m = 1 and n to be any natural number greater than 1/ε. Then, 1/n - 1/m = 1/n - 1 < -1 + ε. This proves that -1 is the greatest lower bound of S.

So, we can confidently say that:

inf S = -1

Fantastic! You've successfully found both the supremum and infimum of the set S using the Archimedean Property. Give yourself a pat on the back!

Wrapping It Up

Alright, let's recap what we've learned. We tackled the problem of finding the supremum and infimum of the set S = {1/n - 1/m : n, m ∈ ℕ}. We used the Archimedean Property to rigorously prove our results.

Here's a quick summary:

  • The supremum of S is 1 (sup S = 1).
  • The infimum of S is -1 (inf S = -1).

We showed that 1 is the least upper bound by demonstrating that for any ε > 0, we can find an element in S greater than 1 - ε. Similarly, we showed that -1 is the greatest lower bound by demonstrating that for any ε > 0, we can find an element in S less than -1 + ε.

The Archimedean Property was crucial in both cases, allowing us to find appropriate values for n and m to satisfy the required inequalities. Remember, the Archimedean Property states that for any real number x, there exists a natural number n such that n > x. This property is fundamental in real analysis and is used extensively when dealing with limits, bounds, and convergence.

So, the next time you're faced with a similar problem, remember these steps:

  1. Understand the definitions of supremum and infimum.
  2. Make an educated guess for the supremum and infimum.
  3. Prove that your guesses are correct by showing they are upper/lower bounds and that you can get arbitrarily close to them using elements from the set.
  4. Don't forget to use the Archimedean Property when you need to find natural numbers that satisfy certain inequalities.

With practice, you'll become more comfortable using the Archimedean Property and other tools from real analysis to solve challenging problems. Keep practicing, and you'll master these concepts in no time!

Great job today! Keep up the awesome work, and I'll see you in the next real analysis adventure!