Question:20.In Exercises 17–20, prove the given statement about subsets \(A\) and \(B\) of \({\mathbb{R}^n}\) . A proof for an exercise may use results of earlier exercises.

20. a. \({\rm{conv}}\,\left( {{\rm{A}} \cap {\rm{B}}} \right) \subset \left( {\left( {{\rm{conv}}\,A} \right)\,\, \cap \left( {{\rm{conv}}\,B} \right)} \right)\),

b. Find an example in \({\mathbb{R}^2}\) to show that equality need not hold in part (a).

Short Answer

Expert verified
  1. It is shown that \({\rm{conv}}\,\left( {{\rm{A}} \cap {\rm{B}}} \right) \subset \left( {\left( {{\rm{conv}}\,A} \right)\,\, \cap \left( {{\rm{conv}}\,B} \right)} \right)\).
  2. A square with \(A\) be the two adjacent corners of the square, and B be the other two corners of the square.

Step by step solution

01

Step 1:Use the result of Exercise 18

The combinations of A must contain combinations of \(A\)and\(B\). So, \(\,\left( {A \cap B} \right) \subset A\) .

Similarly,a convex combination of \(A\) must containsevery convex combination of \(A\) and\(B\);that is, \({\rm{conv}}\,\left( {{\rm{A}} \cap \,B} \right) \subset {\rm{conv}}\,A\)or \({\rm{conv}}\,\left( {{\rm{A}} \cap \,B} \right) \subset {\rm{conv}}\,B\).

02

Step 2:Draw a conclusion

If \({\rm{conv}}\,\left( {{\rm{A}} \cap \,B} \right) \subset {\rm{conv}}\,A\) is true, then \({\rm{conv}}\,\left( {{\rm{A}} \cap {\rm{B}}} \right) \subset \left( {\left( {{\rm{conv}}\,A} \right)\,\, \cap \left( {{\rm{conv}}\,B} \right)} \right)\) must also be true as a convex combination of points ofAandconvex combination of points of \(B\)must contain some or all points of convex combinations of \(A\)and\(B\).

03

 Assume an example in \({\mathbb{R}^2}\) as a requirement

Consider a square and assume \(A\) be the two diagonally opposite corners of the square, whereas B be the other diagonally opposite corners of the square.

Then, \({\rm{conv}}\,A\,\, \cap {\rm{conv}}\,B\) represents the diagonals of the square and contains their intersection point, whereas \({\rm{conv}}\,\left( {A\,\, \cap \,B} \right)\)should be an empty set as \(A\,\, \cap \,B\) is empty.

So, \({\rm{conv}}\,\left( {{\rm{A}} \cap {\rm{B}}} \right) \subset \left( {\left( {{\rm{conv}}\,A} \right)\,\, \cap \left( {{\rm{conv}}\,B} \right)} \right)\)is followed.

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