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A closed set $O$ is a subset of the real numbers such that $\exists x \in O, \forall \epsilon>0$ s.t. $(x-\epsilon,x+\epsilon)$ is not a subset of $O$.

Are the only two $x \in O$ s.t. $(x-\epsilon,x+\epsilon)$ is not a subset of $O$, the infimum/minimum or the supremum/maximum of that set $O$?

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  • $\begingroup$ Your definition is wrong. An open set $A$ is a set such that for each $x\in A$ we have some open interval $J$ st $x\in J\subset A$. A closed set is the complement of an open one. For instance, both the void set and all the real numbers are both closed and open. A bounded closed set always contain it supremum and infimum, but there can be more points that aren't in the interior of the set: take the Cantor's set for example $\endgroup$
    – Arararararagi-kun
    Jan 24, 2020 at 20:50
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    $\begingroup$ There is a saying in many books, sets are not like doors: sets can be open or closed, they can also be open and closed! They can even be neither! $\endgroup$
    – mathematics2x2life
    Jan 24, 2020 at 20:53

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That's not what a closed set is. You just defined what it means for a set to be not open. A set is closed if the complement is open. But certainly neither is required. For example, take $r\in\mathbb{Q}$. Then for all $\epsilon>0$, the set $(r-\epsilon,r+\epsilon)$ is not a subset of $\mathbb{Q}.$ I think you see open and closed as opposites rather than describing a relationship between the set and its complement.

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