Exercise 1.3.1

Question

\begin{enumerate}[label=(\alph*)] 
  \item Write a formal definition in the style of Definition 1.3.2 for the \emph{infimum} or \emph{greatest lower bound} of a set.
  \item Now, state and prove a version of Lemma 1.3.8 for greatest lower bounds.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We have $i = \inf A$ if and only if
    \begin{enumerate}[label=(\roman*)] 
      \item Lower bound, $a \ge i$ for all $a \in A$
      \item Greatest lower bound, If $b$ is a lower bound on $A$ then $b \le i$
     \end{enumerate}
  \item Suppose $i$ is a lower bound for $A$, it is the greatest lower bound if and only if for all $\epsilon>0$, there exists an $a\in A$ such that $i + \epsilon > a$. \par
    First suppose $i = \inf A$, then for all $\epsilon > 0$, $i+\epsilon$ cannot be a lower bound on $A$ because (ii) implies all lower bounds $b$ obey $b \le i$, therefore there must be some $a \in A$ such that $i+\epsilon > a$.

Second suppose for all $\epsilon > 0$ there exists an $a \in A$ such that $i+\epsilon > a$. In other words $i+\epsilon$ is not a lower bound for all $\epsilon$, which is the same as saying every lower bound $b$ must have $b \le i$ implying (ii).
   \end{enumerate}

Exercise 1.3.1

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Exercise 1.3.1

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