The title states my question: what aspect of closed makes it attractive for optimization?
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2Open sets may only have a supremum and an infimum, but not maximum or minimum. – Rodrigo de Azevedo Nov 04 '16 at 20:02
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Consider these problems:
$$\begin{cases} \max x \\ \text{s.t.}\\ x \in (0,1) \end{cases}, \begin{cases} \max x \\ \text{s.t.}\\ x \in [0,1] \end{cases}.$$
The first has no solution, while the second has (exactly, $x=1$).
Summarizing, it is better to have closed set since sometimes the optimal value is on the frontier of a set. A closed set does include its frontier. In this way, you don't lose solutions.
the_candyman
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Al right, thank you for your reply! So it has something to do with uniqueness of the solution? Is there more to it? Or is this the only aspect that makes it attractive. – Blackjaguar Nov 04 '16 at 19:55
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@Blackjaguar I did not talk about uniqueness. I just talked about existence of solutions. – the_candyman Nov 04 '16 at 19:57
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@Blackjaguar In general, when you want to solve a problem, you may wish to pose your problem in a way such that you know that there is a solution, doesn't matter how hard it is finding it. You can also take a look to Weierstrass theorem. – the_candyman Nov 04 '16 at 20:00
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This is purely a theoretical consideration. Practically there is nothing wrong with an open set. – LinAlg Nov 04 '16 at 20:42
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Having a restricted and closed (i.e. compact) feasible set and if the (real-valued) objective f is furthermore continous, you are able to apply the Extreme value theorem in order to prove feasibility of your constrained optimization problem.
Jonasson
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