I'm stuck on a problem where I have the following,
$A \subset \mathbb R$. Then there exists sequence ${a_{n}}$ and ${b_{n}}$ of $A$ such that $$\lim_{n \to \infty}{a_{n}}=\sup(A),\ \ \ \ \lim_{n \to \infty}{b_{n}}=\inf(A).$$
I have a small outline of the proof that I have done however I'm not exactly sure if this is the right way to go,
I want to find a sequence $x_n$ in $\mathbb R$ such that $x_n<\sup(A)$ for each $n$ and $x_n\to a$ and if $\alpha=\sup(A)<\infty$ I let $x_n=\alpha-\frac{1}{n}$ and if $\sup(A)=\infty$ then I'll let $x_n=n$. I also would follow this same pattern for the $\inf(A)$ since $\sup(A)=-\inf(-A)$.
Or would it be better to show that $a_n$ is increasing in $A$ and $b_n$ is increasing in $A$ and since both sequences are part of $A$ we know they are bounded both above and below so $sup(A)$ and $inf(A)$ exists so I can use that to prove the initial proof.
Any help is welcomed.