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Chapter 8: Problem 3

In Exercises 3 and \(4,\) determine whether each set is open or closed or neither open nor closed. a. \(\\{(x, y) : y > 0\\}\) b. \(\\{(x, y) : x=2 \text { and } 1 \leq y \leq 3\\}\) c. \(\\{(x, y) : x=2 \text { and } 1 < y < 3\\}\) d. \(\\{(x, y) : x y=1 \text { and } x > 0\\}\) e. \(\\{(x, y) : x y \geq 1 \text { and } x > 0\\}\)

Short Answer

Expert verified
a. Open, b. Closed, c. Open, d. Neither, e. Closed.

Step by step solution

01

Analyze Set (a)

The set \( \{(x, y) : y > 0\} \) contains all points above the x-axis, and it does not include the points on the x-axis where \( y = 0 \). It does not contain its boundary. This makes the set open because there is no point with \( y = 0 \) included.
02

Analyze Set (b)

The set \( \{(x, y) : x=2 \text{ and } 1 \leq y \leq 3\} \) is a closed segment on the line \( x = 2 \). It includes its endpoints because \( y \) takes on values from 1 to 3, including these values, showing that the set is closed.
03

Analyze Set (c)

The set \( \{(x, y) : x=2 \text{ and } 1 < y < 3\} \) is an open interval on the line \( x = 2 \). It does not include the endpoints since \( y \) is strictly between 1 and 3, making the set open.
04

Analyze Set (d)

For the set \( \{(x, y) : xy = 1 \text{ and } x > 0\} \), it represents a hyperbola where \( y = \frac{1}{x} \) for any \( x > 0 \). As this set does not contain any isolated points and is not defined for \( x \leq 0 \), it does not include its 'boundary' at the x-axis; thus, neither open nor closed.
05

Analyze Set (e)

The set \( \{(x, y) : xy \geq 1 \text{ and } x > 0\} \) includes the points where \( xy = 1 \) and all points where \( xy > 1 \), including the hyperbola line \( y = \frac{1}{x} \). Since it includes the boundary line, the set is closed.

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Key Concepts

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

Open Sets
In topology, an open set is a fundamental concept representing a collection of points that does not include its boundary. To visualize this, imagine a cloud without a clear edge. Every point has some room around it to move without touching the edge of the set. For example, considering the set \( \{(x, y) : y > 0\} \), it includes all points above the x-axis but not those on the x-axis itself. Thus, no boundary points are included.
  • An open set allows for every point within it to have a neighborhood fully contained in the set.
  • No boundary points are part of an open set.
  • Examples in a plane include regions not touching a defined edge, like sections above or below an axis without intersecting it.
Closed Sets
Closed sets, unlike open sets, do include their boundary points. Think of a closed set as a region enclosed completely, like a fenced area where every edge is part of the region. In our context, consider the set \( \{(x, y) : x=2 \text{ and } 1 \leq y \leq 3\} \). It includes the line segment between \( y = 1 \) and \( y = 3 \) as well as the endpoints. This indicates it is a closed set.
  • All boundary points are included in a closed set.
  • Every sequence within the set that converges, converges to a point also in the set.
  • Closed sets are often related to bounded regions with clearly defined limits.
Boundary in Topology
The boundary of a set in topology refers to the points that "separate" inside from outside, similar to how a fence defines the perimeter of a yard. These points can be part of either the set or complement of the set. Understanding boundaries help determine whether a set includes such points which impacts its classification as open or closed.
  • A point is a boundary point if every neighborhood around it intersects both the set and its complement.
  • A set is open if it excludes all its boundary points.
  • A set is closed if it includes its boundary points.
Hyperbolas
Hyperbolas are fascinating curves that represent one of the conic sections, resembling two mirrored curves or loops. A standard hyperbola might look like \( xy = c \), where \( c \) is a constant. In the exercise, the set \( \{(x, y) : xy = 1 \text{ and } x > 0\} \) sketches a hyperbola to the right of the y-axis. Unlike ellipses or circles, hyperbolas extend indefinitely and never form a closed loop.
  • Hyperbolas have two distinct branches.
  • They never intersect, confining themselves to opposite quadrants.
  • Often described using asymptotes, illustrating where the curves approach but never reach.
Intervals on the Real Line
Intervals on the real line are straight segments that include or exclude their endpoints. These intervals can either be open or closed depending on their definitions. Open intervals, like \( 1 < y < 3 \), exclude the endpoints, while closed intervals, like \( 1 \leq y \leq 3 \), include them. This classification affects many mathematical operations and properties.
  • Open interval: defined as \( (a, b) \), excludes both \( a \) and \( b \).
  • Closed interval: defined as \( [a, b] \), includes both \( a \) and \( b \).
  • Intervals provide a way to describe a continuous range of values on a number line.

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Most popular questions from this chapter

Let \(T\) be a tetrahedron in "standard" position, with three edges along the three positive coordinate axes in \(\mathbb{R}^{3}\) , and suppose the vertices are \(a \mathbf{e}_{1}, b \mathbf{e}_{2}, c \mathbf{e}_{3},\) and \(0,\) where \(\left[\mathbf{e}_{1} \quad \mathbf{e}_{2} \quad \mathbf{e}_{3}\right]=I_{3} .\) Find formulas for the barycentric coordinates of an arbitrary point \(\mathbf{p}\) in \(\mathbb{R}^{3}\) .

Suppose a B茅zier curve is translated to \(\mathbf{x}(t)+\mathbf{b} .\) That is, for \(0 \leq t \leq 1,\) the new curve is $$ \begin{aligned} \mathbf{x}(t)=(1-t)^{3} \mathbf{p}_{0} &+3 t(1-t)^{2} \mathbf{p}_{1} \\ &+3 t^{2}(1-t) \mathbf{p}_{2}+t^{3} \mathbf{p}_{3}+\mathbf{b} \end{aligned} $$ Show that this new curve is again a B茅zier curve. [Hint: Where are the new control points?]

Prove the given statement about subsets \(A\) and \(B\) of \(\mathbb{R}^{n} .\) A proof for an exercise may use results of earlier exercises. If \(A \subset B\) and \(B\) is convex, then \(\operatorname{conv} A \subset B\)

Let \(\mathbf{v}_{1}=\left[\begin{array}{l}{1} \\ {0}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{l}{1} \\ {2}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{l}{4} \\ {2}\end{array}\right], \mathbf{v}_{4}=\left[\begin{array}{l}{4} \\ {0}\end{array}\right],\) and \(\mathbf{p}=\left[\begin{array}{l}{2} \\ {1}\end{array}\right] .\) Confirm that \(\mathbf{p}=\frac{1}{3} \mathbf{v}_{1}+\frac{1}{3} \mathbf{v}_{2}+\frac{1}{6} \mathbf{v}_{3}+\frac{1}{6} \mathbf{v}_{4}\) and \(\mathbf{v}_{1}-\mathbf{v}_{2}+\mathbf{v}_{3}-\mathbf{v}_{4}=\mathbf{0}\) Use the procedure in the proof of Caratheodory's Theorem to express \(\mathbf{p}\) as a convex combination of three of the \(\mathbf{v}_{i}\) 's. Do this in two ways.

Given points \(\mathbf{p}_{1}=\left[\begin{array}{l}{1} \\\ {0}\end{array}\right], \mathbf{p}_{2}=\left[\begin{array}{l}{2} \\\ {3}\end{array}\right],\) and \(\mathbf{p}_{3}=\left[\begin{array}{r}{-1} \\\ {2}\end{array}\right]\) in \(\mathbb{R}^{2}\) let \(S=\operatorname{conv}\left\\{\mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3}\right\\} .\) For each linear functional \(f,\) find the maximum value \(m\) of \(f\) on the set \(S,\) and find all points \(\mathbf{x}\) in \(S\) at which \(f(\mathbf{x})=m .\) $$ \begin{array}{ll}{\text { a. } f\left(x_{1}, x_{2}\right)=x_{1}-x_{2}} & {\text { b. } f\left(x_{1}, x_{2}\right)=x_{1}+x_{2}} \\ {\text { c. } f\left(x_{1}, x_{2}\right)=-3 x_{1}+x_{2}}\end{array} $$
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