Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find an example in \({\mathbb{R}^2}\) to show that equality need not hold in the statement of Exercise 25.

Short Answer

Expert verified

It is proved that equality does not hold.

Step by step solution

01

Set S is affine

Theorem 2states that a set\(S\)is affineif and only if every affine combination of points of\(S\)lies in\(S\). That is,\(S\)is affine if and only if\(S = {\mathop{\rm aff}\nolimits} S\).

02

Show that \(aff\left( {A \cap B} \right) \subset \left( {affA \cap affB} \right)\) need not hold

Consider\(A\)as a line segment through the point\(\left( {0,0} \right)\)to\(\left( {1,0} \right)\)shown below:

\(A = \left\{ {\left( {{\mathop{\rm x}\nolimits} ,y} \right)|y = 0,0 \le {\mathop{\rm x}\nolimits} \le 1} \right\}\)

Consider B as the line segment from the point\(\left( {1,0} \right)\)to\(\left( {2,0} \right)\)shown below:

\(B = \left\{ {\left( {{\mathop{\rm x}\nolimits} ,y} \right)|y = 0,1 \le x \le 2} \right\}\)

Thus, the intersection of\(A\)and\(B\)is only a single point. Therefore it is shown below:

\(\begin{aligned}{}{\mathop{\rm aff}\nolimits} \left( {A \cap B} \right) = A \cap B\\ = \left\{ {\left( {1,0} \right)} \right\}\end{aligned}\)

A line passing through the origin and the point\(\left( {1,0} \right)\)is an affine subset of A, and a line passing through the point\(\left( {1,0} \right)\)and\(\left( {2,0} \right)\)is an affine subset of B. Therefore,\({\mathop{\rm aff}\nolimits} A \cap {\mathop{\rm aff}\nolimits} B\)is a line, which represents\({\mathop{\rm x}\nolimits} - \)axis because the first set is a point and the second is a line. Hence, the\({\mathop{\rm aff}\nolimits} \left( {A \cap B} \right) \subset \left( {{\mathop{\rm aff}\nolimits} A \cap {\mathop{\rm aff}\nolimits} B} \right)\)does not hold.

Thus, it is proved that equality does not hold.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 21–24, a, b, and c are non-collinear points in\({\mathbb{R}^{\bf{2}}}\)and p is any other point in\({\mathbb{R}^{\bf{2}}}\). Let\(\Delta {\bf{abc}}\)denote the closed triangular region determined by a, b, and c, and let\(\Delta {\bf{pbc}}\)be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\)is positive, where\(\overrightarrow {\bf{a}} \),\(\overrightarrow {\bf{b}} \) and\(\overrightarrow {\bf{c}} \)are the standard homogeneous forms for the points.

22. Let p be a point on the line through a and b. Show that\(det\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

Let\({v_1} = \left[ {\begin{array}{*{20}{c}}1\\3\\{ - 6}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{7}}\\3\\{ - {\bf{5}}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{9}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{0}}\\{\bf{9}}\end{array}} \right]\), \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{1.4}\\{{\bf{1}}.{\bf{5}}}\\{ - {\bf{3}}.{\bf{1}}}\end{array}} \right]\), and \({\bf{x}}\left( t \right) = {\bf{a}} + t{\bf{b}}\)for \(t \ge {\bf{0}}\).Find the point where the ray\({\bf{x}}\left( t \right)\)intersects the plane that contains the triangle with vertices\({v_1}\),\({v_{\bf{2}}}\), and\({v_{\bf{3}}}\). Is this point inside the triangle?

Question: Let \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\), \({\bf{n}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{5}}\\{ - {\bf{1}}}\end{array}} \right)\), \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{0}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{4}}\\{\bf{0}}\\{\bf{4}}\end{array}} \right)\), and let H be the hyperplane in\({\mathbb{R}^{\bf{4}}}\) with normal n and passing through p. Which of the points \({{\bf{v}}_{\bf{1}}}\), \({{\bf{v}}_{\bf{2}}}\), and \({{\bf{v}}_{\bf{3}}}\) are on the same side of H as the origin, and which are not?

Question: 25. Let \(p = \left( \begin{array}{l}1\\1\end{array} \right)\). Find a hyperplane \(\left( {f:d} \right)\) that strictly separates \(B\left( {0,3} \right)\) and \(B\left( {p,1} \right)\). (Hint: After finding \(f\), show that the point \(v = \left( {1 - .75} \right)0 + .75p\) is neither in \(B\left( {0,3} \right)\) nor in \(B\left( {p,1} \right)\)).

In Exercises 21-26, prove the given statement about subsets A and B of \({\mathbb{R}^n}\), or provide the required example in \({\mathbb{R}^2}\). A proof for an exercise may use results from earlier exercises (as well as theorems already available in the text).

21. If \(A \subset B\), then B is affine, then \({\mathop{\rm aff}\nolimits} A \subset B\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free