# 11.2.3 Dedekind reals are Dedekind complete

We obtained ${\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}$ as the type of Dedekind cuts^{} on $\mathrm{\beta \x84\x9a}$. But we could have instead started
with any archimedean^{} ordered field $F$ and constructed Dedekind cuts on $F$. These would
again form an archimedean ordered field $\stackrel{{\rm B}\u2015}{F}$, the Dedekind completion of $F$,
with $F$ contained as a subfield^{}. What happens if we apply this construction to
${\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}$, do we get even more real numbers? The answer is negative. In fact, we shall prove a
stronger result: ${\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}$ is final.

Say that an ordered fieldΒ $F$ is admissible for $\mathrm{\Xi \copyright}$ when the strict order $$ onΒ $F$ is a map $$.

###### Theorem 11.2.1.

Every archimedean ordered field which is admissible for $\mathrm{\Xi \copyright}$ is a subfield ofΒ ${\mathrm{R}}_{\mathrm{d}}$.

###### Proof.

Let $F$ be an archimedean ordered field. For every $x:F$ define $L,U:\mathrm{\beta \x84\x9a}\beta \x86\x92\mathrm{\Xi \copyright}$ by

$$ |

(We have just used the assumption^{} that $F$ is admissible for $\mathrm{\Xi \copyright}$.)
Then $({L}_{x},{U}_{x})$ is a Dedekind cut. Indeed, the cuts are inhabited and rounded because
$F$ is archimedean and $$ is transitive^{}, disjoint because $$ is irreflexive^{}, and
located because $$ is a weak linear order. Let $e:F\beta \x86\x92{\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}$ be the map $e(x):\beta \x89\u2018({L}_{x},{U}_{x})$.

We claim that $e$ is a field embedding^{} which preserves and reflects the order. First of
all, notice that $e\beta \x81\u2019(q)=q$ for a rational number $q$. Next we have the equivalences,
for all $x,y:F$,

$$ |

so $e$ indeed preserves and reflects the order. That $e\beta \x81\u2019(x+y)=e\beta \x81\u2019(x)+e\beta \x81\u2019(y)$ holds because, for all $q:\mathrm{\beta \x84\x9a}$,

$$ |

The implication^{} from right to left is obvious. For the other direction, if $$ then there merely exists $r:\mathrm{\beta \x84\x9a}$ such that $$, and by taking $s:\beta \x89\u2018q-r$ we get the desired $r$ and $s$. We leave preservation of multiplication^{} by $e$ as
an exercise.
β

To establish that the Dedekind cuts on ${\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}$ do not give us anything new, we need just one more lemma.

###### Lemma 11.2.2.

If $F$ is admissible for $\mathrm{\Xi \copyright}$ then so is its Dedekind completion.

###### Proof.

Let $\stackrel{{\rm B}\u2015}{F}$ be the Dedekind completion of $F$. The strict order on $\stackrel{{\rm B}\u2015}{F}$ is defined by

$$ |

Since $U\beta \x81\u2019(q)$ and ${L}^{\beta \x80\xb2}\beta \x81\u2019(q)$ are elements of $\mathrm{\Xi \copyright}$, the lemma holds as long as $\mathrm{\Xi \copyright}$
is closed under conjunctions^{} and countable^{} existentials, which we assumed from the outset.
β

###### Corollary 11.2.3.

The Dedekind reals are Dedekind complete: for every real-valued Dedekind cut $\mathrm{(}L\mathrm{,}U\mathrm{)}$ there is a unique $x\mathrm{:}{\mathrm{R}}_{\mathrm{d}}$ such that $$ and $$ for all $y\mathrm{:}{\mathrm{R}}_{\mathrm{d}}$.

###### Proof.

By \autoreflem:cuts-preserve-admissibility the Dedekind completion ${\stackrel{{\rm B}\u2015}{\mathrm{\beta \x84\x9d}}}_{\mathrm{\pi \x9d\x96\xbd}}$ of ${\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}$
is admissible for $\mathrm{\Xi \copyright}$, so by \autorefRD-final-field we have an embedding ${\stackrel{{\rm B}\u2015}{\mathrm{\beta \x84\x9d}}}_{\mathrm{\pi \x9d\x96\xbd}}\beta \x86\x92{\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}$, as well as an embedding ${\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}\beta \x86\x92{\stackrel{{\rm B}\u2015}{\mathrm{\beta \x84\x9d}}}_{\mathrm{\pi \x9d\x96\xbd}}$. But these embeddings must be
isomorphisms^{}, because their compositions are order-preserving field homomorphisms which
fix the dense subfieldΒ $\mathrm{\beta \x84\x9a}$, which means that they are the identity^{}. The corollary now
follows immediately from the fact that ${\stackrel{{\rm B}\u2015}{\mathrm{\beta \x84\x9d}}}_{\mathrm{\pi \x9d\x96\xbd}}\beta \x86\x92{\mathrm{\beta \x84\x9d}}_{\mathrm{\pi \x9d\x96\xbd}}$ is an isomorphism.
β

Title | 11.2.3 Dedekind reals are Dedekind complete |

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