Supremum of a set to the power of -1

Question

$$ A\subseteq(0,\infty) $$ $$ \sup(A)=3, \\ \inf(A)=1 $$ $$ B=\left\{a^{-1}:a\in A\right\} $$

How would I go about formally proving that 1 is the supremum of B?

Answer 1

Since we are considering inverses of real numbers it should be clear that $1 = \inf(A) $would be a good choice for an upper bound for $B$. It is clear that the supremum will be positive since, for example, take the inverse of any element in $A \subset (0,\infty)$, then we must have that $\sup(B) >$ than that. To prove this is the least upper bound, suppose there was $0< x < 1$ $(\implies \frac{1}{x} > 1)$ such that $x > a^{-1}=\frac {1}{a}$ for all $a \in A$. But then $a > \frac{1}{x}$ so that $1 < \frac{1}{x} < a$ since $x$ is fixed, contradicting that $1$ is the infimum of $A$.