Say there exists a series {A}, which has an upper bound, meaning it has a supremum, call is "a" ( sup(A) = a ). My question is, is {A}≤a or {A}<a? My confusion arises from the fact that we know that if "a" is included in {A} then it is also its maximum, meaning {A}≤a, does that still stand for the supremum? I know that if there is a x∈A, st x≤a than a is the maximum of {A}, so consequently if sup(A) would have the same definition, meaning {A}≤a, wouldn't it be a maximum too? I would really appreciate some help to clear up this confusion, thank you.
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The difference is that $\max A\in A$ (when it exists), whereas it is not always true that $\sup A\in A$. Take $A=]-1,1[$ for instance. Then $\sup A=1\notin A$, whereas $A$ has no maximum element. – Anthony Nov 12 '20 at 21:42
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https://math.stackexchange.com/q/18605/42969 – Martin R Nov 12 '20 at 21:44
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so if sup(A)=/= max(A) then {A}≤ sup(a) is still true? – user844179 Nov 12 '20 at 21:45
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1If a subset $S$ of $\mathbb R$ has a maximum element then necessarily the supremum of $S$ is equal to the maximum of $S$. However, even if a bounded set $S$ has no maximum it still has a supremum. For example, consider the set of all rational numbers whose square is less than $2$. It has no maximum but it has a supremum. – littleO Nov 12 '20 at 22:02
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If a set $A$ has an upper bound, then $A$ has a least upper bound (in $\mathbb{R}$). Then for every $a \in A$, the supremum $\sup A$ is an upper bound of $A$ so $a \leq \sup A$. So, to abuse notation a bit, $A \leq \sup A$.
The supremum is meant to generalize the idea of a maximum. You can show that if the set contains an upper bound, then supremum and maximum coincide. In particular, on finite sets the maximum and the supremum are precisely the same.
