For every extended real number $y$ for which $y < x$, $\exists$ $n \geqslant m$ such that $y < a_{n} \leqslant x$

Question

I have to proof this exercise for my math study, and I really don't know how to do it:

Let $(a_{n})_{n=m}^\infty$ be a sequence of real numbers, and let $x$ be the extended real number $x := sup(a_{n})_{n=m}^\infty$. Let $\mathbb{R}^*$ be the set of extended reals.

Prove: For every extended real number $y$ for which $y < x$, $\exists$ $n \geqslant m$ such that $y < a_{n} \leqslant x$

So far, I proved that $a_{n} \leqslant x$ $\forall$ $n \geqslant m$, and:
$M \in \mathbb{R}^*$ is an upper bound of $(a_{n})_{n=m}^\infty\Rightarrow sup(a_{n})_{n=m}^\infty \leqslant M$

Could you please explain me how to prove this? I'm trying this for a couple of days now.

Thanks in advance!

Answer 1

Note that $x-y$ is positive (or infinite). Suppose that there does not exist $a_n$ such that $y<a_n\le x$. Then all $a_n$s would be $\le y$. It follows that $$x=\sup a_n \le y$$ which is impossible.