Question

Let \(X\) be a Banach space. Show that if \(A \subset X\) is totally bounded, then there is a sequence \(\left\\{x_{n}\right\\} \in X\) such that \(x_{n} \rightarrow 0\) in \(X\) and \(A \subset\) sconv \(\left\\{x_{n}\right\\}\) (see Exercise \(1.7)\) In particular, for every compact subset \(A\) of \(X\) there exists a sequence \(\left\\{x_{n}\right\\}\) such that \(x_{n} \rightarrow 0\) and \(A \subset \overline{\operatorname{conv} v}\left\\{x_{n}\right\\}\) (Grothendieck). Hint: We set \(A_{1}=A\), let \(B_{1}\) be a finite \(2^{-2}\) -net in \(A_{1}\). If \(A_{i}\) and \(B_{i}\) were defined for \(i \leq n\), let \(A_{n+1}=\left(A_{n}-B_{n}\right) \cap 2^{-2 n} B_{X} ;\) note that every \(a_{n} \in A_{n}\) is of the form \(a_{n}=a_{n+1}+b_{n}\), where \(a_{n+1} \in A_{n+1}, b_{n} \in B_{n}\). Let \(B_{n+1}\) be a finite \(2^{-2 n}\) -net in \(A_{n+1}\). Therefore, every element \(a \in A=A_{1}\) is of the form \(a=b_{1}+a_{2}=b_{1}+b_{2}+a_{3}=\ldots=\sum_{1}^{n} b_{n}+a_{n+1} .\) Since \(a_{n+1} \in 2^{-2 n} B_{X_{1}}\)

Short answer

Construct a sequence {bₙ} from finite nets covering A. The decomposition ensures convergence to 0 and containment in the closed convex hull.

Step-by-step solution

Understanding Total Boundedness

A set is total bounded in a Banach space if for every ε > 0, there exists a finite number of points such that the set can be covered by balls of radius ε centered at these points.

Defining Subsequences

Set the initial subset as A₁ = A. Let B₁ be a finite 2⁻²-net in A₁. A 2⁻²-net is a set of points such that every point in A₁ is within a distance of 2⁻² from at least one point in B₁.

Inductive Definition

Assume Aᵢ and Bᵢ for i ≤ n have been defined. Let Aₙ₊₁ = (Aₙ - Bₙ) ∩ 2⁻²ⁿ B_X, where B_X is the unit ball in the Banach space. This means each element in Aₙ₊₁ is within 2⁻²ⁿ distance from some finite set of points in B_X.

Finite Net

Let Bₙ₊₁ be a finite 2⁻²ⁿ-net in Aₙ₊₁. This construction ensures that every element in Aₙ₊₁ is of the form aₙ = aₙ₊₁ + bₙ, where aₙ₊₁ ∈ Aₙ₊₁ and bₙ ∈ Bₙ.

Decomposition of Elements

Every element a ∈ A can be written as a sum of elements from the sequence Bᵢ and residual sequences from Aᵢ. Specifically, a = b₁ + a₂ = b₁ + b₂ + a₃ = ... = Σ₁ⁿ bₙ + aₙ₊₁. Here, aₙ₊₁ belongs to 2⁻²ⁿ B_X.

Convergence to Zero

As n → ∞, the residual term aₙ₊₁ in the formulation a = Σ₁ⁿ bₙ + aₙ₊₁ becomes arbitrarily small since they are in 2⁻²ⁿ B_X. Thus, the sequence {xₙ} converges to 0 in X.

Convex Combinations

From the construction, since each element a is expressed in terms of the finite nets Bᵢ, A is contained in the closed convex hull sconv {xₙ}.

Key concepts

Banach space

A Banach space is a type of complete normed vector space. The term 'complete' means that any Cauchy sequence (a sequence where the elements get arbitrarily close to each other) in the space has a limit that is also within the space. This property is critical because it ensures that every sequence that appears to converge actually does.
To visualize, imagine you have a collection of vectors in a multi-dimensional space where every vector has a length (distance from the origin). The concept of a 'norm' provides a way to measure this length. Formally, a norm on a vector space X is a function \(orm{.} \) from X to the non-negative real numbers \(orm{.} \).
Here are a few essential properties of a norm:

• \(orm{a} \) is always non-negative, and \(orm{a} = 0 \) only when the vector a is the zero vector
• For any scalar \(\beta \) and any vector a, \(orm{\beta a} = |\beta| orm{a} \)
• The norm satisfies the triangle inequality: \(orm{a+b} \) ≤ \(orm{a} + orm{b} \)

Additionally, a Banach space allows mathematicians to handle infinite-dimensional spaces while still preserving the structure and properties found in finite-dimensional vector spaces. This is especially useful in functional analysis, where Banach spaces often serve as the foundational framework.

Total Boundedness

Total boundedness is a stronger condition than just boundedness. A set A in a metric space is totally bounded if, for any given distance ε > 0, you can cover the set with a finite number of balls of radius ε. This means that no matter how small ε gets, you can still cover the set with a limited number of tiny balls.
This property is important because it implies that the set can be approximated very closely with a finite subset. In essence, a totally bounded set doesn't spread out indefinitely; it stays 'compact' even when you zoom in. For instance, consider a set of points on a line that keeps clustering closer and closer together. If it is totally bounded, you can always find a finite number of small intervals that cover all points.
In the context of Banach spaces, total boundedness helps in dealing with limits and convergence. When trying to show that each element of a totally bounded set can be represented as sums of sequences converging to zero, total boundedness ensures that these sequences can be carefully controlled and manipulated. Thus, proving results like the one in the exercise becomes feasible by ensuring that our sequences and nets stay within manageable bounds, no matter how finely we need to cover the set.

Sequence Convergence

Sequence convergence is a foundational concept in analysis. A sequence \(orm{a_n} \) in a Banach space converges to an element \(orm{a} \) if the distance between the sequence elements and \(orm{a} \) gets arbitrarily small as n grows larger. Mathematically, \(orm{a_n \rightarrow a} \) means that for any ε > 0, there's an N such that for all n ≥ N, \(orm{a_n - a} < ε \).
Convergence in Banach spaces has some useful implications:

• It ensures that sequences can be approximated as closely as needed by other sequences within the space.
• It makes it possible to isolate and work with limit points and helps to approximate more complex functions with simpler ones.

In the provided exercise, sequence convergence is employed to show that every element of a totally bounded set can be written as sums of sequences converging to zero. This shows that elements of these sets don't 'escape' the space but instead can be tightly controlled in their behavior. So, the sequences in a Banach space not only converge but they can be manipulated to help represent elements of totally bounded or compact subsets effectively.