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Let \(\left\\{K_{i}\right\\}\) be a finite family of convex compact sets in \(\mathbf{R}^{n}\) such that every subfamily consisting of \(n+1\) members has a nonempty intersection. Helly's theorem asserts that then \(\bigcap K_{i} \neq \emptyset\). Prove it for \(n=1\). Show an example for \(n=2\) that \(n+1\) is necessary. Hint: Consider the interval between the maximum of the left endpoints of \(K_{i}\) and the minimum of the right endpoints of \(K_{i}\). For the general case, see for example [DGK]. Example: three lines forming a triangle. 1.60 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 \(A \subset \overline{\operatorname{conv}}\left\\{x_{n}\right\\}\) (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 \(a_{n+1} \in 2^{-2 n} B_{X}\)

Short Answer

Expert verified
\(\bigcap K_i eq \emptyset \) when every subfamily of n+1 sets intersects for n=1. An example for n=2 shows \(n+1\) members is necessary.

Step by step solution

01

Prove Helly's theorem for n=1

Consider each set in \{K_i\} to be an interval on the real line, \[K_i = [a_i, b_i]\]. Every subfamily consisting of 2 members (n+1 for n=1) has a nonempty intersection, meaning \[\forall i, j, [a_i, b_i] \cap [a_j, b_j] eq \emptyset.\] Therefore, the intersection must include an interval between the maximum of all left endpoints and the minimum of all right endpoints, \[\bigcap K_{i} = [\max(a_i), \min(b_i)] eq \emptyset.\]
02

Show an example for n=2 where n+1 is necessary

Consider three lines forming a triangle in \mathbf{R}^2\, such as: \[K_1 = L_1, K_2 = L_2, K_3 = L_3\] where \L_1, L_2,\ and \L_3\ are line segments that intersect pairwise but not all three together. In this example, every pair of \K_i\ and \K_j\ has an intersection, but the intersection of all three is empty. This demonstrates that for n=2, n+1 is necessary.
03

Properties of totally bounded sets in Banach spaces

Given a totally bounded set \(A \subset X\), construct a sequence that approximates elements in A. Start with \A_1 = A\ and pick a finite \2^{-2}\ Net \B_1\ in \A_1\. For subsequent sets \A_n\, choose a finite \2^{-2n}\-Net \B_n\. Then, \A_{n+1} = (A_n - B_n) \cap 2^{-2n} B_X\, ensuring every \a_n \in A_n\ is of the form \a_n = a_{n+1} + b_n\. Continue iterating to achieve \A \subset \text{sconv}\{x_n\},'s closure, showing that there exists a sequence \{x_n\} where \ x_n \rightarrow 0\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Helly's Theorem
Helly's Theorem is an important result in convex geometry. It states that for a finite family of convex compact sets in \(\textbf{R}^n\), if every subfamily of at most \(n+1\) sets has a non-empty intersection, then the whole family has a non-empty intersection. This theorem is particularly useful in higher-dimensional geometry and has various applications.
To understand it better, let's consider a simple case when \(n=1\). Here, the sets are intervals on the real line, denoted as \([a_i, b_i]\). According to Helly's Theorem, if every pair of these intervals intersects, there must be a common point that belongs to all intervals.
Banach Spaces
A Banach space is a complete normed vector space. This means that it is a vector space equipped with a norm, and it is complete in the sense that every Cauchy sequence in the space converges to a point within the space.
An example is the space of continuous functions on a closed interval \([a, b]\)\, denoted \(C([a, b])\), with the supremum norm. Banach spaces are fundamental in functional analysis and are used to study various problems in mathematical analysis and its applications.
Totally Bounded Sets
A set \(A\) in a metric space is totally bounded if, for any \(\text{ε}>0\), we can cover \(A\) with a finite number of subsets whose diameter is less than \(ε\).
In simpler terms, a set is totally bounded if it can be covered by a finite number of balls of any chosen small radius. This concept is crucial in analysis as it relates to compactness—a compact set is always totally bounded and closed.
Convex Sets
A set \(K\) in a vector space is convex if, for any two points \(x, y \in K\), the line segment joining \(x\) and \(y\) lies entirely within \(K\). Mathematically, this means that for any \(\text{λ} \in [0,1]\),\[ \text{λ}x+(1- \text{λ})y \in K.\]\
Convex sets are fundamental in various fields such as optimization, economics, and geometry. For example, the feasible region defined by linear inequalities in optimization problems is a convex set.
Infinite Dimensional Geometry
Infinite dimensional geometry involves studying the geometric properties of spaces that have infinitely many dimensions. These spaces are usually Banach or Hilbert spaces and are used in many areas like quantum mechanics, PDEs, and more.
One crucial aspect is understanding how geometric intuition from finite dimensions extends or fails in infinite dimensions. For example, the unit ball in an infinite-dimensional Banach space can be arbitrarily 'spiky', unlike in finite dimensions where it is always 'smooth'.

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Most popular questions from this chapter

Let \(X\) be a Banach space and \(G\) be a subspace of \(X\) that is a \(G_{\delta}\) set in \(X\). Show that \(G\) is closed in \(X\). Hint: Let \(S=\bar{G} \backslash G\), where \(\bar{G}\) denotes the closure of \(G .\) Since \(G\) is a \(G_{\delta}\) set in \(X, G\) is \(G_{\delta}\) in \(\bar{G}\). Hence \(G=\bigcap G_{n}\), where \(G_{n}\) are open subsets in \(\bar{G}\). Therefore, \(S=\bigcup\left(\bar{G} \backslash G_{n}\right)\) and each \(G_{n}\) is dense in \(\bar{G}\) because it contains \(G .\) Thus, each \(\bar{G} \backslash G_{n}\) is nowhere dense in \(\bar{G}\), so \(S\) is of first category in \(\bar{G}\). We will show that \(S\) is an empty set. If \(x_{0} \in S\), consider the set \(G^{*}=\left\\{x_{0}+x ; x \in G\right\\}\). Note that \(G^{*} \subset S\). Indeed, any point \(x_{0}+x, x \in G\), is in \(\bar{G}\) since \(\bar{G}\) is a linear set. If for some \(x \in G\) we have \(x_{0}+x \in G\), then by linearity of \(G\) we have \(x_{0} \in G_{1}\) a contradiction. Therefore, \(G^{*} \subset S\) and \(G^{*}\) is of first category in \(\bar{G} .\) Thus, the shift \(G\) of \(G^{*}\) is also of first category in \(\bar{G}\). Hence, the whole \(\bar{G}\) as a union of \(G\) and \(S\), which are both of first category in \(\bar{G}\), is of first category in itself. This is a contradiction, since \(\bar{G}\) is a complete metric space.

Show that the linear dimension (the cardinality of a Hamel basis) of the space \(\ell_{p}, p \in[1, \infty)\), is the continuum \(c\). Hint: First, note that \(\operatorname{card}\left(\ell_{p}\right)=c .\) Therefore, the linear dimension of \(\ell_{p}\) is less than or equal to \(c .\) On the other hand, note that if \(\lambda<1\), then the vector \(\left(\lambda, \lambda^{2}, \ldots\right) \in \ell_{p}\), and these vectors form a linearly independent set.

Let \(A\) be a totally bounded set in a Banach space \(X .\) Show that \(\operatorname{conv}(A)\) is totally bounded. Therefore, \(\overline{\operatorname{conv}}(A)\) is compact. Hint: Let \(B\) be a finite \(\delta\) -net for \(A .\) It is straightforward to check that \(\operatorname{conv}(B)\) is a \(\delta\) -net for \(\operatorname{conv}(A)\). Since \(\operatorname{conv}(B)\) lies in a finite- dimensional space and is bounded, it is totally bounded, and its \(\delta\) -net would produce a \(2 \delta\) -net for \(\operatorname{conv}(A)\). Note also that the closure of a totally bounded set is totally bounded.

Is the convex hull of a closed set in \(\mathbf{R}^{2}\) closed? Hint: No, check \(A=\left\\{\left(x, \frac{1}{x}\right) ; x>0\right\\} \cup\\{(0,0)\\}\).

Let \(X\) be the normed space obtained by taking \(c_{0}\) with the norm \(\|x\|_{0}=\sum 2^{-i}\left|x_{i}\right| .\) Show that \(X\) is not a Banach space. Note that this shows that \(\|\cdot\|_{0}\) is not an equivalent norm on \(c_{0}\). Hint: The sequence \(\\{(1,1, \ldots, 1,0, \ldots)\\}_{n=1}^{\infty}\) is Cauchy and not convergent since the only candidate for the limit would be \((1,1, \ldots) \notin c_{0}\).

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