A set that is bounded below possess a subsequence that converges to its infimum.

Question

Assume $S \subset \mathbb{R}$ is nonempty and bounded below. Show that there exists a sequence $(a_n)_{ n\in \mathbb{N}}\in S^{\mathbb{N}}$ of elements of $S$ converging to $\inf(S)$.

Answer 1

Let $s$ denote the infimum of $S$. For each $\epsilon > 0$, the quantity $s+\epsilon$ cannot be a lower-bound for $S$. Hence, there exists (see the note below if not clear) $x_\epsilon \in S$ such that $$ s\leq x_\epsilon < s+\epsilon. $$ Given $n \in \mathbb{N}$ and using $\epsilon = 1/n$ in the above, we obtain $x_n \in S$ such that $$ s \leq x_n < s + \frac{1}{n}. $$ By the squeeze theorem, it follows that $x_n \to s$ as $n \to \infty$.

Note: If $s+\epsilon$ is not a lower bound for $S$, this means that the statement $$ s+\epsilon \leq x, \quad \forall x \in S $$ is false. Put otherwise, there must exist $x \in S$ for which the above fails, i.e. such that $x < s+ \epsilon$. But, because $s$ is a lower bound for $S \ni x$, we still have $s \leq x$.