infimum and supremum of subsets question

Question

Let $B \subseteq \mathbb{R}_{+}$ such that B is non-empty. consider $B^{-1} = \left \{b^{-1} : b\in B \right \}$.
Show that if $B^{-1}$ is unbounded from above, then $\inf\left(B\right)=0$

How can i prove that? tnx!

Answer 1

B^−1 is unbounded from above hence for every n in N, there exist b in B such that 1/b>n then 1/n > b for every n. Hence inf B =0(note infimum exist because B bounded below by 0)

Answer 2

Hint: if not, then $\inf B =\delta>0$. Therefore $$ b \geq \delta>0 $$ for every $b \in B$, and hence$ \frac{1}{b} \leq \frac{1}{\delta}$. In particular, $B^{-1}$ is bounded from above. Contradiction.