Show that if $\mathrm{inf}S = 0$, then $S^{−1}$ is not bounded above.
If $\mathrm{inf}S = 0$ then there exists $L$ such that $L$ is a lower bound of $S$ and for each lower bound $l$, of $S$, $L \geqslant l$.
my idea is : proof by contradiction. assume there is upper bound for $S^{−1}$. so there must be least upper bound for $S^{−1}$ which we call $u^*$. so there is $\varepsilon>0$ such that $u^*-\varepsilon = 1/s$. Rearranging, we have $s = \frac{1}{u^*-\varepsilon}$. if we choose $\varepsilon=u^*+1$ we get a negative $s$ which is a contradiction. However I think my proof is sketchy or incorrect. So far we have only learned the least upper bound property, upper/lower bounds and Archimedean property to solve this problem.