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From wiki, $\mathbf R$ has the following properties:

  1. Least upper bound property
  2. Nested intervals theorem
  3. Monotone convergence theorem
  4. Bolzano–Weierstrass theorem
  5. Cauchy completeness

Moreover, for a general ordered field, (1)-(4) are equivalent, while (5) is weaker than them. However, the proofs of (5) that I have seen all rely on (1)-(4). Can you give a proof of (5) that does not rely on (1)-(4)?

JSCB
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    How do you define the real numbers? – user49640 Jun 14 '17 at 15:03
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    It kind of depends on how you're defining $\Bbb{R}$, right? If you define $\Bbb{R}$ as the completion of $\Bbb{Q}$ with respect to the usual Archimedean absolute value, completeness is essentially free. – sharding4 Jun 14 '17 at 15:04
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    No. Because without 1-4 $\mathbb R$ has no meaning. The reason we have $\mathbb R$ and are able to refer to it is because we prove an extension of $\mathbb Q$ exists that has 1-4. We call it $\mathbb R$. It's a bit like proving kangaroos are marsupials without being allowyed to use any definitions of "placenta" or marsupials lack thereof. – fleablood Jun 14 '17 at 15:12
  • Any paper treating the definition of $\mathbb{R}$ will prove the least upper bound property. If you want a definition that gives Cauchy completeness easily, the completion of $\mathbb{Q}$ via Cauchy-sequences is probably the quickest, even though it is a little bit subtle. – nombre Jun 14 '17 at 21:46

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5) is weaker than the others in the sense that 5) should be replaced by 5)$^\prime$ Cauchy completeness and the Archimedean property. This means that to prove completeness you do need to use one of the properties 1)-4), while to prove any of the properties 1)-4) you also need the Archimedean property and not just completeness. For a complete non Archimedean field see complete-non-archimedean

Gio67
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