Let $\phi\neq A\subset \mathbb{R}$ that is bounded from below. Show that there exists a decreasing sequence $(s_n)$ with $s_n \in A$ for all $n$ such that $\lim s_n = \inf A$.
My attempt:
Since $\phi \neq A$ is bounded below, then by the completeness axioms, $\inf A$ exists. Let $s=\inf A$. By the definition of $\inf A$,
$s< s+\frac{1}{n}$ for all $n\in \mathbb{N}$
which means that we can find a sequence $s_n$ in $A$ such that
$s\leq s_n < s+\frac{1}{n}$. Hence $|s_n-s|<\frac{1}{n}\to 0$, so $\lim s_n=s=\inf A$.
My issue is how to prove that $(s_n)$ is decreasing? Any help would be appreciated. Thank you!
Edit:
consider $s_n=s+\frac{1}{2n}$. We have
$s+\frac{1}{2(n+1)}=s_{n+1}\leq s_n=s+\frac{1}{2n}$
which shows that $s_n$ is decreasing