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Chapter 1: Problem 53

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

Short Answer

Expert verified
The convex hull of the set A is not closed.

Step by step solution

01

Understand the Problem

We need to determine if the convex hull of a closed set in \(\textbf{R}^2\) is always closed. To show the contrary, we should find an example where the convex hull of a closed set is not closed.
02

Analyze the Given Set

Consider the set A provided in the hint: \(\text{A} = \{ (x, \frac{1}{x}) ; x>0 \} \cup \{ (0,0) \},\) which is a closed set in \( \textbf{R}^2 \).
03

Determine the Convex Hull

The convex hull of a set is the smallest convex set that contains it. For set A, consider the points along the curve and the origin (0,0). The convex hull will also include all the line segments connecting any two points in the set.
04

Verify Closedness

Given \(\text{A}\), the convex hull includes points along the line \(\text{y} = 0\) for all \(\text{x} \geq 0\). We need to check if this new set includes all its limit points. Consider a sequence of points inside the convex hull approaching some limit point. If this sequence converges to a point not in the convex hull, the convex hull is not closed.
05

Conclusion Using Example

Notice that the points \( \text{(x, 0)} \) for \( x > 0 \) are included in the convex hull, but as \(x \to 0 \), \((x, 0)\) approaches \((0,0)\), which is a limit point. However, there are other points (such as \((x, \frac{1}{x})\)) that do not converge to zero. Hence, the convex hull fails to contain such limit points and is not closed.

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Key Concepts

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

Convex Sets
A set is convex if, for any two points within the set, the line segment connecting them also lies within the set.
This means that if you pick any two points, say \(A\) and \(B\), and draw a straight line between them, every single point on that line must also be inside the set.
This property is crucial in understanding concepts like the convex hull, which is the smallest convex set containing a given set. Imagine a rubber band stretched around the outside of the set; the shape that it takes defines the convex hull.
Limit Points
Limit points, also known as accumulation points, are points where every neighborhood around them contains at least one point from the set different from itself.
In simpler terms, if you zoom in very closely around a limit point, you'll still find points from the original set.
This concept is important when determining whether a set is closed. If a set contains all its limit points, it is closed.
Take, for example, the sequence defined by \( \text{A} = \{ (x, \frac{1}{x}) ; x > 0 \} \cup \{ (0,0) \}. \) Here, \( (0,0) \) is a limit point because as \(x \to 0\), the points \( (x, \frac{1}{x}) \) get arbitrarily close to \( (0,0) \).
Closed Sets
A set is closed if it contains all its limit points.
This means that for any point approaching the set (from the inside or the outside), if such point is included in the set, then the set is considered closed.
Closed sets are essential in analysis because they provide stable boundaries and limits.
For instance, the set \( \text{A} = \{ (x, \frac{1}{x}) ; x > 0 \} \cup \{ (0,0) \} \) is closed because it includes the limit point \( (0,0) \).
Counterexamples in Analysis
Counterexamples are invaluable in mathematics because they disprove statements or theories.
In our exercise, the set \( \text{A} \) provided serves as a counterexample to the claim that the convex hull of a closed set is always closed.
By examining the convex hull of \( A = \{ (x, \frac{1}{x}) ; x > 0 \} \cup \{ (0,0) \} \), we see that the hull contains the points along \( y = 0 \) for \( x \geq 0 \).
However, even though (0,0) is a limit point, other points like \( (x, \frac{1}{x}) \) do not converge to zero, indicating the hull is not closed.

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Most popular questions from this chapter

Let \(X\) be a normed space, \(M \subset X\) and \(\varepsilon>0 .\) We say that \(A \subset M\) is an \(\varepsilon=n e t\) in \(M\) if for every \(x \in M\) there is \(y \in A\) such that \(\|x-y\|<\varepsilon\). We say that \(A \subset M\) is \(\varepsilon\) -separated in \(M\) if \(\|x-y\| \geq \varepsilon\) for all \(x \neq y \in A\). Clearly, a maximal \(\varepsilon\) -separated subset of \(M\) in the sense of inclusion is an \(\varepsilon\) -net in \(M\). Let \(X\) be an \(n\) -dimensional real normed space. Show that if \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is an \(\varepsilon\) -net in \(B_{X}\), then \(N \geq \varepsilon^{-n} .\) On the other hand, there is an \(\varepsilon\) -net \(\left\\{x_{j}\right\\}_{j=1}^{N}\) for \(B_{X}\) with \(N \leq\left(\frac{2}{c}+1\right)^{n}\). Hint: Fix a linear isomorphism \(T\) of \(X\) onto \(\mathbf{R}^{n}\). For a compact subset \(C\) of \(X\), we define the volume of \(C\) by \(\operatorname{vol}(C)=\lambda(T(C))\), where \(\lambda\) is the Lebesgue measure on \(\mathbf{R}^{n}\). Let \(B(x, r)\) be the closed ball centered at \(x\) with radius \(r\). We have \(B_{X} \subset \bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon\right)\), so $$ \begin{aligned} \operatorname{vol}\left(B_{X}\right) & \leq \operatorname{vol}\left(\bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon\right)\right) \leq \sum_{j=1}^{N} \operatorname{vol}\left(B\left(x_{j}, \varepsilon\right)\right) \\ &=N \cdot \operatorname{vol}\left(\varepsilon B_{X}\right)=N \cdot \varepsilon^{n} \cdot \operatorname{vol}\left(B_{X}\right) \end{aligned} $$ Thus \(N \geq \varepsilon^{-n}\). On the other hand, if \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is a maximal \(\varepsilon\) -separated set in \(B_{X}\), then \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is an \(\varepsilon\) -net in \(B_{X}\). Since \(B\left(x_{j}, \varepsilon / 2\right) \cap B\left(x_{i}, \varepsilon / 2\right)=\emptyset\) if \(i \neq j\) and \(B\left(x_{j}, \varepsilon / 2\right) \subset(1+\varepsilon / 2) B_{X}\), we have $$ \begin{aligned} N \cdot(\varepsilon / 2)^{n} \cdot \operatorname{vol}\left(B_{X}\right) &=\operatorname{vol}\left(\bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon / 2\right)\right) \\ & \leq \operatorname{vol}\left((1+\varepsilon / 2) B_{X}\right)=(1+\varepsilon / 2)^{n} \operatorname{vol}\left(B_{X}\right) . \end{aligned} $$ Thus \(N \leq\left(\frac{1+\varepsilon / 2}{\varepsilon / 2}\right)^{n}\).

Show that \(\overline{\operatorname{span}}(L)=\overline{\operatorname{span}(L)}\) and \(\overline{\operatorname{conv}}(M)=\overline{\operatorname{conv}(M)}\). Hint: By the definition, \(\overline{\operatorname{span}}(L)\) is the intersection of all closed subspaces containing \(L\), and \(\overline{\operatorname{span}(L)}\) is one such subspace. From this, one inclusion follows. The other inclusion follows similarly because \(\operatorname{span}(L)\) is the intersection of all subspaces containing \(L .\)

Let \(A\) be an open set in a normed space \(X\). Show that \(\operatorname{conv}(A)\) is open. Hint: Given \(\left\\{x_{i}\right\\}_{i=1}^{n} \in A\) and \(\left\\{\lambda_{i}\right\\}_{i=1}^{n}\) such that \(\lambda_{i} \geq 0\) for every \(i\) and \(\sum \lambda_{i}=1\), note that if \(\lambda_{1}>0\) and \(O_{1}\) is an open set containing \(x_{1}\) and contained in \(A\), then \(\operatorname{conv}(A)\) contains the open set \(\lambda_{1} O_{1}+\lambda_{2} x_{2}+\ldots+\lambda_{n} x_{n}\)

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 \((1,1, \ldots) \notin c_{0}\).

Let \(Y\) be a closed subspace of a normed space \(X\). Show that if \(Y\) and \(X / Y\) are both Banach spaces, then \(X\) is a Banach space. Note: A property \(\mathcal{P}\) is said to be a three-space property if the following holds: Let \(Y\) be a closed subspace of a space \(X\). If \(Y\) and \(X / Y\) have \(\mathcal{P}\), then \(X\) has \(\mathcal{P}\) (see, e.g., [CaGo]). Thus, the property of being complete is a three-space property in the class of normed linear spaces. Hint: If \(\left\\{x_{n}\right\\}\) is Cauchy in \(X\), there is \(x \in X\) such that \(\hat{x}_{n} \rightarrow \hat{x}\). There are \(\left\\{y_{n}\right\\}\) in \(Y\) such that \(\left\\{x_{n}-x-y_{n}\right\\} \rightarrow 0\). Thus \(\left\\{y_{n}\right\\}\) is Cauchy, so \(y_{n} \rightarrow y\) and \(x_{n} \rightarrow x+y\).
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