Exercise 1.4

Question

Exercise 4: Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$,  and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$.

ghostofgarborg · Accepted Answer

Let $x \in E$. By definition of lower and upper bounds, $\alpha \leq x \leq \beta$.

Amit Awasthi · Answer

\begin{proof}
      Assume $\alpha > \beta$. Since $E 
e \emptyset$, there is some $\gamma \in E$ such that $\gamma \ge \alpha$. This implies $\gamma \ge \alpha > \beta$, which is a contradiction since $\beta$ is an upper bound.
\end{proof}

Exercise 1.4

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Exercise 1.4

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