I cannot understand why we say "The Completeness axiom" (that $\mathbb{R}$ has least upper bound property) when it can be proved from Dedekind's construction?
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Does this answer your question? Proving that $\mathbb{R}$ satisfies the Least Upper Bound property. – Unknown May 30 '20 at 08:27
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We could describe axiomatically $\Bbb R$ as a Dedekind-complete (A) ordered field (B) $(R,+,\cdot,<)$. It turns out that for any two such ordered fields $(R,+,\cdot,<)$ and $(R',\oplus,\odot,\prec)$ there is exactly one homomorphism of fields $f:(R,+,\cdot)\to (R',\oplus,\odot)$, and that said homomorphism is also an isomorphism of ordered fields.
In that sense, Dedekind completeness is an axiom, although, in any explicit construction of a model of the field of real numbers within set theory, the fact that the "candidate" $\Bbb R$ has the least-upper-bound property is a theorem.
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1Expanding on your second paragraph, Dedekind constructs a model of the reals, not the reals. – Eric Towers May 30 '20 at 08:43
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