Finding Infimum And Supremum Of Set S = { (n-1)/n | N ∈ ℕ }

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Introduction

In the realm of real analysis, understanding the bounds of a set is crucial. Specifically, the concepts of infimum and supremum provide a way to define the greatest lower bound and least upper bound, respectively. This article delves into the set S = { (n-1)/n | n ∈ ℕ }, where ℕ represents the set of natural numbers, and aims to determine its infimum and supremum, if they exist. To do this, we will rigorously analyze the behavior of the elements in the set as n varies over the natural numbers, utilizing fundamental definitions and properties of real numbers. The process will involve examining the set's elements, establishing potential bounds, and demonstrating that these bounds satisfy the infimum and supremum criteria. This exercise not only reinforces the understanding of these concepts but also provides insights into the nature of sequences and their limits. The journey through this exploration will enhance your appreciation for mathematical rigor and analytical thinking.

Defining the Set S

First, let's define the set S clearly. It comprises elements of the form (n-1)/n, where n belongs to the set of natural numbers (ℕ). The natural numbers are the set of positive integers, typically starting from 1 (i.e., ℕ = {1, 2, 3, ...}). Therefore, the set S can be expressed as:

S = { (1-1)/1, (2-1)/2, (3-1)/3, (4-1)/4, ... } = { 0, 1/2, 2/3, 3/4, ... }

Each element in S is generated by substituting a natural number n into the expression (n-1)/n. For instance:

  • When n = 1, the element is (1-1)/1 = 0.
  • When n = 2, the element is (2-1)/2 = 1/2.
  • When n = 3, the element is (3-1)/3 = 2/3.
  • When n = 4, the element is (4-1)/4 = 3/4.

And so on. This representation helps us visualize the elements of S and observe their behavior. As n increases, the elements of S also increase, approaching a certain value. This observation is crucial in identifying potential infimum and supremum. The expression (n-1)/n can also be rewritten as 1 - (1/n), which provides further insight into the set's behavior. As n becomes very large, 1/n approaches 0, and thus (n-1)/n approaches 1. This suggests that 1 might be the supremum of the set S. However, to rigorously prove this, we need to demonstrate that 1 is an upper bound and the least upper bound. Similarly, we need to find the infimum, which is the greatest lower bound of the set. Understanding the set's elements and their trend is the first step in this analytical process.

Identifying Potential Infimum and Supremum

To identify the infimum and supremum of the set S, we first observe the behavior of the elements (n-1)/n as n varies across the natural numbers. We've already seen that the set S can be written as {0, 1/2, 2/3, 3/4, ...}. The smallest element in this set occurs when n = 1, which gives us (1-1)/1 = 0. As n increases, the value of (n-1)/n also increases. This suggests that 0 is the smallest element and a potential infimum. To confirm this, we need to show that 0 is a lower bound for S and that no number greater than 0 can be a lower bound.

Now, let's consider the behavior of (n-1)/n as n becomes very large. We can rewrite (n-1)/n as 1 - (1/n). As n approaches infinity, 1/n approaches 0, and therefore (n-1)/n approaches 1. This suggests that 1 is a potential supremum. However, we must prove that 1 is an upper bound for S and that no number smaller than 1 can be an upper bound. We can intuitively see that (n-1)/n will always be less than 1 because 1/n is always greater than 0 for any natural number n. Thus, 1 - (1/n) is always less than 1. This gives us a strong indication that 1 is indeed the supremum.

In summary, based on our observations, we hypothesize that:

  • The infimum of S is 0.
  • The supremum of S is 1.

To rigorously establish these, we need to prove that these values satisfy the formal definitions of infimum and supremum. This involves demonstrating that 0 is the greatest lower bound and 1 is the least upper bound for the set S. The subsequent sections will provide the detailed proofs for these claims.

Proving the Infimum

To prove that the infimum of S is 0, we need to demonstrate two key properties:

  1. 0 is a lower bound for S. This means that every element in S is greater than or equal to 0.
  2. 0 is the greatest lower bound. This means that any number greater than 0 cannot be a lower bound for S.

Let's start with the first property. We need to show that for all n ∈ ℕ, (n-1)/n ≥ 0. Since n is a natural number, n is always positive. Also, for n ≥ 1, we have n - 1 ≥ 0. Therefore, (n-1)/n is the ratio of two non-negative numbers, with a positive denominator, making the entire expression non-negative. Thus, (n-1)/n ≥ 0 for all n ∈ ℕ. This confirms that 0 is a lower bound for S.

Now, let's move on to the second property. We need to show that for any number x > 0, x cannot be a lower bound for S. To do this, we will show that there exists an element in S that is less than x. Consider an arbitrary x > 0. We need to find a natural number n such that:

(n-1)/n < x

Rewriting this inequality, we get:

1 - (1/n) < x

1 - x < 1/n

n < 1/(1-x)

We can solve for n:

n > 1/(1-x)

For this, we first need to clarify that x must be less than 1, because if x >=1 then 1-x would be negative or zero, making 1/(1-x) non-positive or undefined, and thus not giving us a meaningful range for n. However, we can still proceed by considering the inequality 1 - 1/n < x, and rewrite it as: 1 - x < 1/n n > 1/(1-x) If x >= 1, then certainly 0 is the infimum because all elements of the set are non-negative and 0 is an element in the set S when n=1.

If 0 < x < 1, then we consider n > 1/(1-x). By the Archimedean property, there exists a natural number n that satisfies this inequality. Thus, for any x > 0, we can find an n such that (n-1)/n < x. This means that x cannot be a lower bound for S. Since 0 is a lower bound and no number greater than 0 is a lower bound, 0 is the greatest lower bound, i.e., the infimum of S.

Therefore, we have proven that inf(S) = 0.

Proving the Supremum

To prove that the supremum of S is 1, we need to demonstrate the following two properties:

  1. 1 is an upper bound for S. This means that every element in S is less than or equal to 1.
  2. 1 is the least upper bound. This means that any number smaller than 1 cannot be an upper bound for S.

Let's first show that 1 is an upper bound for S. We need to prove that for all n ∈ ℕ, (n-1)/n ≤ 1. We can rewrite (n-1)/n as 1 - (1/n). Since n is a natural number, n ≥ 1, and thus 1/n > 0. Therefore, 1 - (1/n) < 1. This shows that every element in S is less than 1, and thus less than or equal to 1. So, 1 is an upper bound for S.

Next, we need to show that 1 is the least upper bound. This means that for any number y < 1, y cannot be an upper bound for S. To prove this, we will show that there exists an element in S that is greater than y. Let y be any real number such that y < 1. We need to find a natural number n such that:

(n-1)/n > y

Rewriting this inequality, we get:

1 - (1/n) > y

1 - y > 1/n

n > 1/(1-y)

Since y < 1, 1 - y is a positive number. By the Archimedean property, for any positive real number, there exists a natural number n greater than its reciprocal. Therefore, there exists a natural number n such that n > 1/(1-y). This means that we can always find an element (n-1)/n in S that is greater than y. Thus, y cannot be an upper bound for S.

Since 1 is an upper bound and no number smaller than 1 is an upper bound, 1 is the least upper bound, i.e., the supremum of S.

Therefore, we have proven that sup(S) = 1.

Conclusion

In conclusion, we have successfully determined the infimum and supremum of the set S = { (n-1)/n | n ∈ ℕ }. Through rigorous proofs, we have established that:

  • The infimum of S, inf(S), is 0.
  • The supremum of S, sup(S), is 1.

These results highlight the behavior of the set S, which consists of elements that approach 1 as n increases, but never actually reach 1. The infimum, 0, is the smallest element in the set, achieved when n = 1. The supremum, 1, is the least upper bound, which the elements of the set get arbitrarily close to but never exceed. This exploration provides a clear example of how the concepts of infimum and supremum are applied in real analysis to understand the bounds of a set. The techniques used here, such as leveraging the Archimedean property, are fundamental in mathematical analysis and are essential for understanding the properties of real numbers and sets. The ability to determine these bounds is critical in many areas of mathematics, including calculus, topology, and functional analysis. This exercise not only reinforces theoretical knowledge but also enhances problem-solving skills in mathematical contexts. Understanding these concepts is vital for further studies in advanced mathematical topics and their applications in various scientific and engineering fields.