Can someone clarify the following statement:
Let $H\subset\mathbb{R}$ be a subgroup $\neq 0$. Let $a = \text{inf} \lbrace x\in H \vert x>0 \rbrace$. If $a\notin H$ then there exists a decreasing sequence in $H$ converging to $a$.
What guarantees this decreasing sequence?
Whenever the infimum $a$ of any non-empty subset $S$ of the real numbers that is bounded below has $a\notin S$, one must have $(a,a+\varepsilon)\cap S\neq\emptyset$ for all $\varepsilon>0)$, by the definition of infimum. Now choosing $s_n\in(a,a+2^{-n})\cap S$ for $n\in\Bbb N$ ensures that $\lim_{n\to\infty}s_n=a$.
