Question

(a) Show that every totally bounded set is bounded. (b) Let $$ \delta_{i r}=\left\\{\begin{array}{ll} 1 & \text { if } i=r \\ 0 & \text { if } i \neq r_{+} \end{array}\right. $$ and let \(T\) be the subset of \(\ell_{\infty}\) consisting of the sequences \(\mathbf{X}_{r}=\left\\{\delta_{i r}\right\\}_{i=1}^{\infty}\) \(r \geq 1 .\) Show that \(T\) is bounded, but not totally bounded.

Short answer

Use the definitions of bounded and totally bounded sets, and create a short answer based on the provided solution. Answer: No, it is not true that if a set is bounded, then it must be totally bounded. While it is true that every totally bounded set is also bounded, the converse is not always true. The given solution demonstrates a subset T of the l_∞ space that is bounded but not totally bounded, which shows that the conditions are not equivalent.

Step-by-step solution

Definition of totally bounded sets

A set S is totally bounded if, given any positive number ε, there exists a finite number of open balls with their radius being ε that can cover the entire set S. Formally, for any ε > 0, there exist n ∈ ℕ and x_1, x_2, ..., x_n ∈ S such that S ⊆ ⋃_{i=1}^n B(x_i, ε), where B(x_i, ε) is the open ball with center x_i and radius ε.

Definition of bounded sets

A set S is bounded if there exists a positive number M that bounds the distance between any two points in the set. In other words, S is bounded if there exists M > 0 such that ∀x, y ∈ S, d(x, y) ≤ M, where d(x, y) is the distance between x and y.

Prove that every totally bounded set is bounded

Let ε > 0 be arbitrary, and let S be a totally bounded set. By definition, there exist n ∈ ℕ and x_1, x_2, ..., x_n ∈ S such that S ⊆ ⋃_{i=1}^n B(x_i, ε). Now, consider any two points x, y ∈ S. Then, there must exist i, j ∈ {1, 2, ..., n} such that x ∈ B(x_i, ε) and y ∈ B(x_j, ε). Since x and y are in these open balls, we have d(x, x_i) < ε and d(y, x_j) < ε. Applying the triangle inequality, we have d(x, y) ≤ d(x, x_i) + d(x_i, x_j) + d(x_j, y) < ε + d(x_i, x_j) + ε Now, let D = max{d(x_i, x_j) : 1 ≤ i, j ≤ n} and M = 2ε + D. Then, we have d(x, y) < M for all x, y ∈ S. This shows that S is bounded. Hence, every totally bounded set is bounded. #Part (b):#

Prove that T is bounded

Firstly, let's prove that T is bounded. Consider any two sequences in T, say, x_r and x_s where r,s ∈ ℕ. Then, for any i in ℕ, | x_{r_i} - x_{s_i} | = | δ_{ir} - δ_{is} | There are three cases: 1. i ≠ r and i ≠ s: | δ_{ir} - δ_{is} | = 0 2. i = r and i ≠ s: | δ_{ir} - δ_{is} | = 1 3. i = s and i ≠ r: | δ_{ir} - δ_{is} | = 1 Hence, we have that for any x_r and x_s in T, the supremum norm || x_r - x_s ||_∞ = max { | x_{r_i} - x_{s_i} | : i ∈ ℕ } = 1. Thus, the set T is bounded.

Prove that T is not totally bounded

To prove that T is not totally bounded, let's consider the positive number ε = 1/2. Let's assume that there exists a finite number of open balls that cover T, say: T ⊆ ⋃_{k=1}^n B(a_k, ε), where each a_k is a sequence with components a_{k_i}. As each open ball has radius ε = 1/2, this means that the maximum distance between any two sequences x and y in each open ball is 1. However, we know from Step 1 that || x_r - x_s ||_∞ = 1 for r ≠ s. So, for any two sequences x_r and x_s with r ≠ s, only one of them can belong to each open ball B(a_k, ε). Therefore, there must be at least one open ball for each sequence x_r for r ∈ ℕ, which requires an infinite number of open balls to cover T. This contradicts our assumption that a finite number of open balls cover T. Thus, T is not totally bounded. Consequently, we have shown that the given subset T of l_∞ is bounded and not totally bounded.

Key concepts

Bounded Sets in Real Analysis

Understanding bounded sets is fundamental in real analysis, as it deals with the way sets behave under the lens of distances and limits. A set in a metric space is considered bounded if there is a real number that serves as an upper bound for the distance between any two points within the set.

Formally, a set S in a metric space is bounded if there exists a positive number M such that for every pair of points x, y in S, the distance d(x, y) between them is less than or equal to M. This concept is not only intuitive but crucial for various proofs and theorems in real analysis.

For instance, in the context of the given exercise, a totally bounded set is shown to be bounded as part of the proof. This is significant because it ties total boundedness, which involves covering the set with finitely many balls of a given radius, to the more general concept of boundedness which simply requires a universal upper limit on all distances within the set.

Students often confuse these two concepts but recognizing the distinction is important. Every totally bounded set is necessarily bounded; however, the converse isn't always true. This asymmetry often appears in problems and identifies the subtleties within metric space analysis.

Supremum Norm

In the realm of real analysis, norms are functions that assign lengths or sizes to objects in a vector space. The supremum norm, also known as the maximum norm or infinity norm, is a particular type of norm that is very relevant when dealing with sequences of real numbers or functions.

For any bounded sequence x, the supremum norm is defined as ||x||_∞ = sup{|x_i| : i ∈ ℕ}, where |x_i| represents the absolute value of the i-th term of the sequence. Essentially, it measures the 'largest' element in the sequence in terms of absolute value.

In the exercise provided, the supremum norm is used to demonstrate the boundedness of a set of sequences. Each element in the sequence is compared, and the largest absolute difference is found to be 1. It is critical to note how this norm reflects the concept of boundedness within the context of sequences and thereby firmly establishes whether a set is bounded or not in a specific normed space.

Triangle Inequality in Metric Spaces

The triangle inequality is a core principle in metric spaces that describes a fundamental relationship between the distances of points. It asserts that for any three points x, y, and z in a metric space, the distance from x to z is at most the sum of the distances from x to y and from y to z. Symbolically, it can be written as d(x, z) ≤ d(x, y) + d(y, z).

This relation is reminiscent of the inequality in a geometric triangle where the length of one side must be less than or equal to the sum of the lengths of the other two sides. The triangle inequality is used to great effect in the solution of the exercise to show that a totally bounded set is also bounded; by applying this inequality, one can place an upper limit on the distance between any two points within the set.

The triangle inequality often works hand in hand with the concept of the supremum norm to examine the structure of metric spaces within real analysis, providing a robust framework for understanding the relationships between points in these spaces.