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I am required to prove that the Axiom of Completeness is equivalent to the Cut Property of Real Numbers.

The Following in my attempt so far

Proof. $(\Rightarrow)$. By hypothesis $A,B$ are non-empty sets bounded above and below respectively. Now define the set $H = \{h\in\mathbf{R}\ |\ h\ \text{ is a lower bound for }B\}$ and note that $A\subseteq H$. The axiom of completeness assures us that $\sup B$ exists and by exercise $\textbf{1.3.3}$ it is apparent that $\sup H = \inf B$. Now Let $c = \sup H = \inf B$ and assume that $a\in A$ and $b\in B$ respectively, then $b\geq c$ and since $a\in H$ it follows that $a\leq c$.

$(\Leftarrow).$ Now let $E\subseteq\mathbf{R}$ such that $E\neq\varnothing$ and is bounded above $\dots\dots$


I dont know where should i start for the converse in particular i am not sure how to partition $\mathbf{R}$ given the set $E$. Any hints and only hints would be appreciated.

The full statement of the cut property and exercise $\textbf{1.3.3}$ is as follows.

If $A$ and $B$ are nonempty, disjoint sets with $A\cup B = \mathbf{R}$ and $a < b$ for all $a\in A$ and $b\in B$, then there exists $c\in\mathbf{R}$ such that $x\leq c$ whenever $x\in A$ and $x\ge c$ whenever $x\in B$.

$\textbf{1.3.3}$ Let $A$ be non-empty and bounded below, and define $B = \{b\in\mathbf{R}|b\text{ is a lower bound for A}\}$, then $\sup B = \inf A$.

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  • $\begingroup$ A side comment: your proofs will be clearer if you explain to the reader ahead of time which direction you're doing. Taking a quick glance at what you've written, I'm not sure what you've done and what you're still doing. $\endgroup$ – goblin Jul 10 '18 at 13:37
  • $\begingroup$ @goblin My apologies $\endgroup$ – Atif Farooq Jul 10 '18 at 13:38
  • $\begingroup$ You have nothing to apologize for; I was giving practical advice, not trying to be critical. $\endgroup$ – goblin Jul 10 '18 at 13:39
  • $\begingroup$ @goblin Is it clear now or should i edit it again ? $\endgroup$ – Atif Farooq Jul 10 '18 at 13:39
  • $\begingroup$ Yep, that's a lot better, $\endgroup$ – goblin Jul 10 '18 at 13:40

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