Lemma
120
[22S]Let \(A⊆ X\) be a not empty set. We recall these properties of the supremum.
If \(A\) has maximum \(m\) then \(m=\sup A\).
Let \(s∈ X\). We have \(s=\sup A\) if and only if
for every \(x∈ A\) we have \(x≤ s\).
for every \(x∈ X\) with \(x{\lt}s\) there exists \(y∈ A\) with \(x{\lt} y\).
This last property is of very wide use in the analysis!
The proof is left as a (useful) exercise.
Solution
1