$E$ bounded above, show that there is a sequence $(a_n) \subset E$ such that $a_n \to \sup E$

Question

Let $E$ be an infinite set of real numbers which is bounded above. Show that there exists a sequence $( a_n ) \subset E$ such that $\lim a_n = \sup E$ if $\sup E \not \in E$.

My attempt:

Let $(a_n)\subset E$ such that it converges to some $L$ such that $L\not\in E$. Let $L=\sup(E)$. This implies that $(a_n)$ is bounded above by $\sup(E)$ and that $\lim_{n\to \infty} (a_n)= L = \sup(E)$.

Answer 1

It's not very clear what you are asking. But anyway, since $E$ is bounded above, $\sup E<\infty $. Denote it $\ell$. By definition, $$\forall \varepsilon>0, \exists a\in E: \ell-\varepsilon<a<\ell.$$ How can you transform this expression to get a sequence $(a_n)\subset E$ that converge to $\ell$ ?

Answer 2

Hint: by definition of supremum, for each $n\ge1$ there exists an element $y\in E$ such that $y>\sup E-\frac1n$.