If $ M $ is an upper bound of $ A $ and $ \exists $ $ \{ a_n \} \in A :\lim \sup a_n = M $ prove that $ \sup A = M $

Question

I'm having some difficulty proving this proposition :

Let $ A \subset \Bbb R $ with $ A \neq \emptyset $. If $ M $ is an upper bound of $ A $ and $ \exists $ $ \{ a_n \} \in A :\lim\limits \sup a_n = M $ show that $$ \sup A = M $$

Intuitively this proposition makes perfect sense to me but I have no idea how to prove it rigorously.

Answer 1

$$ limsup_A(a_n) \leq sup_A(a_n)$$ $$ sup_A(a_n) \leq M $$ $$ limsup_A(a_n) = M $$ combining these leads to $$ M = limsup_A(a_n) \leq sup_A(a_n) \leq M$$

FYI: What you wrote is not what you meant. Writing $$ \exists \{a_n\} \in A : limsup(a_n) = M$$ means that there exists a Set {a_n} within A. Thats not what you mean. You mean that there exists element a_n in A $$ \exists a_n \in A$$ If you meant there exists a series , then you write $$ (a_n)_{n \in \mathbb{N}} $$

and the existens Quantor is redundant in that case.