Prove that the l.u.b of A is an element of A if A is closed.

Question

Prove that The Lower upper Bound of A is an element of A if A is closed. Please help, it would be much appreciated.

Answer 1

Let $s:= \sup A$. To each $n \in \mathbb N$ there is $a_n \in A$ such that

$$s-\frac{1}{n} <a_n \le s.$$

Hence $a_n \to s$. Since $A$ is closed, we have $s \in A$.