Solution for Exercise 05 from Chapter 01 in Baby Rudin

Books/BabyRudin/Chapter01/ex05.gapry.tex

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+\subsection*{Exercise 05 (Gapry)}
+\begin{flushleft}
+For all $x$ in the set $A$, $-x$ is in the set $-A$.
+Let's claim $y = -x$, we know that for all $y \in -A$, $y \le sup(-A)$ is true. 
+We have the following inferences:
+\begin{flalign*}
+          y &\le  sup(-A) &\\
+    \iff -x &\le  sup(-A) &\\
+\implies  x &\ge -sup(-A)
+\end{flalign*}
+It means $-sup(-A)$ is one of the lower bound of the set $A$. \\
+\vspace{10px}
+Let's claim $B$ is a set of all lower bound of the set $A$. \\
+Obviously, for all $b$ in the set $B$, $b \le x$ is true for all $x \in A$. \\
+\vspace{10px}
+Since $b \le x \implies -b \ge -x$ is true, it follows $-b$ will greater than or equal to all element in the set $-A$. It proves $-b \ge sup(-A)$ is true. \\
+\vspace{10px}
+Since $-b \ge sup(-A) \implies b \le -sup(-A)$ is true, it follows $-sup(-A)$ is the upper bound of $b$. \\
+\vspace{10px}
+We prove the $x \ge -sup(-A)$ and $b \le -sup(-A)$ are both true, hence $-sup(-A)$ is the greatest lower bound of the set $A$, that is $-sup(-A) = inf(A)$.
+\end{flushleft}
+