If A is a non-empty bounded closed set. Show that sup A $\in$ "closure of A"
To prove this, I tried to reason by contradiction:
Let sup A be not in the closure of A, then by definition of a closed set there is an $\varepsilon > 0$ so that for all a $\in$ A |sup A – a| $\geq \varepsilon$. But by defintion of the supremum it should be that $\forall a\in A$ |sup A – a| $\leq \varepsilon$. This gives us the wanted contradiction, so sup A is indeed an element of the closure of A.
Would this proof be sufficient?
EDIT: say I wanted to proof a similar question:
let $(a_n)_{n\in\mathbb{N}}$ a bounded sequence and define the set A = {$a_n | n\in\mathbb{N}$}. Show that limsup $_{n\rightarrow \infty} (a_n)$ are elements of the closure of A. Would a similar argument work as I used for proving supA in closure of A? since I have tried to do it in a similar way, but I can't really find the contradiction.
By a previous question I know: for $x > limsup _{n\rightarrow \infty}a_n$ we have $\exists N\in\mathbb{N}, \forall n\geq N \;: a_n<x$.
thanks
Best Answer
There is no definition of supremum such that$$(\forall\varepsilon>0)(\forall a\in A):|\sup(A)-a|<\varepsilon.$$What happens is that$$(\forall\varepsilon>0)(\color{red}\exists a\in A):|\sup(A)-a|<\varepsilon\tag1$$(and this is a property of the supremum, rather than its definition). And this is enough, since, as you wrote, if $\sup(A)\notin\overline A$, then there would be some $\varepsilon>0$ such that$$(\forall a\in A):|\sup(A)-a|\geqslant\varepsilon,$$which contradicts $(1)$.