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Q27E

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Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Question: 27. Give an example of a closed subset \(S\) of \({\mathbb{R}^{\bf{2}}}\) such that \({\rm{conv}}\,S\) is not closed.

The set is \(S = \left\{ {\left( {x,y} \right):{x^2}{y^2} = 1,\,\,y > 0} \right\}\).

See the step by step solution

Step by Step Solution

Step 1: Assume subset  \(S\) such that \({\rm{conv }}S\) is not closed

One of the possible sets is \(S = \left\{ {\left( {x,y} \right):{x^2}{y^2} = 1,\,\,y > 0} \right\}\). This set is not closed as the equation \({x^2}{y^2} = 1,\,\,\,y > 0\) is of a hyperbola in an upper half-plane that is shown below:

Step 2: Check whether the assumed set \(S\)is in \({\mathbb{R}^{\bf{2}}}\)

The hyperbola \(xy = 1\) opens upwards; that is, it satisfies all values of \(x\) and for \(y > 0\).

So, the set \(S\) is in \({\mathbb{R}^{\bf{2}}}\).

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