Originally Posted by

**Jhevon** here is a sketch of the proof

if S has a minimum, then inf(S) = min{S} and so is in the set

now suppose S does not have a minimum, and let x = inf(s)

then there is some $\displaystyle \epsilon > 0$ so that $\displaystyle (x + \epsilon )\in S$. let $\displaystyle s_1 = x + \epsilon$. then by the denseness of $\displaystyle \mathbb{R}$, there is some $\displaystyle s_2 \in \mathbb{R}$, so that $\displaystyle x < s_2 < s_1$ and $\displaystyle s_2 \in S$. continue this logic by induction to create a sequence $\displaystyle \{ s_n \}$ that converges to $\displaystyle x$