# Relationship between infimum and supremum and closed sets

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 $$O$$?

• 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 Jan 24, 2020 at 20:50
• 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! Jan 24, 2020 at 20:53

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.