Exercise 1.5

Question

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Exercise 5: Let $A$ be a nonempty set of real numbers, which is bounded below. Let $-A$ be the set of all numbers $-x$ where $x \in A$. Prove that:
  $$ \inf{A} = -\sup(-A) $$

ghostofgarborg · Accepted Answer

Assume $\alpha$ is the greatest lower bound of $A$. If $x \in (-A)$ then $-x \in A$, so $\alpha \leq -x$, and therefore $ - \alpha \geq x$. This implies that $-\alpha$ is an upper bound for $(-A)$. If $ \beta < -\alpha$ then $-\beta > \alpha$, and there is an $x \in A$ such that $x < -\beta$. Then $-x \in -A$, and $-x > \beta$. This shows that $-\alpha$ is the least upper bound of $(-A)$, and we are done.

Amit Awasthi · Answer

\begin{proof}
      Since $A$ is bounded below, $\inf A$ exists, and implies $\inf A \le x$ for all $x \in A$. Then $-\inf A \ge -x$ for all $x \in A$ and so $-\inf A$ is an upper bound of $-A$.
      \par Now prove this is the \emph{least} upper bound. Assume there is an upper bound $y$ of $-A$ such that $-x \le y < -\inf A$ for all $x \in A$. This implies $x \ge y > \inf A$, which is a contradiction.
\end{proof}

Exercise 1.5

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Exercise 1.5

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