If a finite nonempty set of real numbers has a maximum then it has a minimum, prove.

Question

so as the title says. I first proved that all finite nonempty sets of real numbers have maximum, but I cannot see how from this I can prove that it also has a minimum.

Any advice would be appreciated.

Answer 1

You can express your set as $A = \{x_1, x_2, \ldots, x_n\}$. Denote $\{-x_1, \ldots, -x_n\}$ by $-A$. By condition, there exists $i \in \{1, 2, \ldots, n\}$ such that $-x_i = \max(-A)$. Try to show that $x_i = \min A$.

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Details: $-x_i = \max(-A)$ means that $-x_i \geq -x_k$ for all $k \in \{1,2, \ldots, n\}$, which in turn is equivalent to say that $x_i \leq x_k$ for all $k \in \{1,2, \ldots, n\}$, which exactly means $x_i = \min A$.

Answer 2

Hint If $A$ is your set, what can you say about the maximum of $$B= \{ -x |x \in A \} ?$$