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Question:28. Give an example of a compact set\(A\)and a closed set\(B\)in\({\mathbb{R}^2}\)such that\(\left( {{\rm{conv}}\,A} \right) \cap \left( {{\rm{conv}}\,B} \right) = \emptyset \)but\(A\)and\(B\)cannot be strictly separated by a hyperplane.

Short Answer

Expert verified

The sets are \(A = \left\{ {\left( {x,y} \right):\left| x \right| < 1,\,\,y = 0} \right\}\), and \(B = \left\{ {\left( {x,y} \right):{x^2}{y^2} = 1,\,\,y > 0} \right\}\).

Step by step solution

01

Assume set \(A\)and \(B\) as per required condition

One of the possible sets is \(B = \left\{ {\left( {x,y} \right):{x^2}{y^2} = 1,\,\,y > 0} \right\}\) and \(A = \left\{ {\left( {x,y} \right):\left| x \right| < 1,\,\,y = 0} \right\}\). This set is in \({\mathbb{R}^2}\).

02

Check whether \(\left( {{\rm{conv}}\,A} \right) \cap \left( {{\rm{conv}}\,B} \right) = \emptyset \)

The convex combination of both sets is null as the set\(B\)opens only in half-plane, whereas the set\(A\)lies within\( - 1 < x < 1\).

Thus, satisfy the condition\(\left( {{\rm{conv}}\,A} \right) \cap \left( {{\rm{conv}}\,B} \right) = \emptyset \).

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