I'm reading The Reals as Rational Cauchy filter by I. Weiss and I am trying to establish a connection between Dedekind reals and Bourbaki reals.
I'm working with this definition of a Dedekind real a.k.a a Dedekind cut.
Definition 1 A Dedekind real is a pair $(L,U)$ of non-empty subset of $\mathbb{Q}$ such that
- D$_1$. $L$ is a lower cut, i.e. a lower and open set: $a\in L$ iff there exists a rational number $a'$ such that $a<a'$ and $a'\in L$;
- D$_2$. $U$ is an upper cut, i.e. an upper and open set: $b\in U$ iff there exists a rational number $b'\in U$ such that $b'<b$;
- D$_3$. $L,U$ form a two-sided cut, i.e. If $a\in L$ and $b\in U$, then $a<b$.
- D$_4$. $L,U$ are located, i.e. If $a<b$ are rational numbers, then $a\in L$ or $b\in U$;
Detailed discussion of this definition can be found in nLab.
Definition 2 A Bourbaki-MCF real is a minimal Cauchy filter (MCF) $\mathcal{F}$ on $\mathbb{Q}$, i.e.
- C$_1$ $\emptyset\notin \mathcal{F}$;
- C$_2$ if $F_1, F_2 \in \mathcal{F}$, then $F_1\cap F_2 \in \mathcal{F}$;
- C$_3$ if $F_1 \in \mathcal{F}$ and $F_2\subseteq \mathbb{Q}$ such that $F_1\subseteq F_2$, then $F_2 \in \mathcal{F}$;
- C$_4$ $\mathcal{F}$ has the following Cauchy property, for any rational $\varepsilon > 0 $ there exists a rational interval $(a,b) \in \mathcal{F}$ such that $b-a < \varepsilon$;
- C$_5$ $\mathcal{F}$ is minimal, if $\mathcal{G}$ is any subset of $\mathcal{F}$ that satisfies C$_1$-C$_4$ then it is necessarily equal to $\mathcal{F}$, $\mathcal{F}=\mathcal{G}$.
Given a Dedekind real $(L,U)$ we define a filter $$ \mathcal{F}(L,U) = \{ F\subseteq \mathbb{Q}\ |\ (a,b) \subseteq F, a\in L, b\in U\}.$$
I think, this filter is minimal? My attempt at a proof.
Proposition 1. $\mathcal{F}(L,U)$ is a minimal Cauchy filter.
Proof. Using D$_4$, we can prove that for a rational $\varepsilon >0$ there exists $a\in L$, $b\in U$ such that $b-a < \varepsilon$. This proves that $\mathcal{F}(L,U)$ has the Cauchy property. (Details e.g. in in nLab)
We assume that $\mathcal{G} \subseteq \mathcal{F}(L,U)$ is a Cauchy filter. We need to show that $\mathcal{G}=\mathcal{F}(L,U)$. Let $F \in \mathcal{F}(L,U)$ be arbitrarily fixed. From the definition there exists a rational interval $(a,b)$ such that $(a,b)\subseteq F$, $a\in L, b\in U$. Because $L$ is an open lower cut, for $a\in L$ there exists $a'\in L$ such that $a<a'$, and because $U$ is an open upper cut for $b\in U$ there exists $b'\in U$ such that $b'<b$. Let's take a rational $\varepsilon>0$ such that $\varepsilon < \operatorname{min}\{a'-a, b'-a', b-b'\}$
Now, since $\mathcal{G}$ has the Cauchy property there exists a rational interval $(a'',b'')\in \mathcal{G}\subseteq \mathcal{F}(L,U)$ such that $b''-a''<\varepsilon$. Since both $(a', b')\in \mathcal{F}(L,U) $ and $(a'',b'')\in \mathcal{F}(L,U)$ then they have a non-empty intersection as $\mathcal{F}(L,U)$ is a filter, but the idea is now to show that $a<a''$ and $b''<b$.
Because $b'' - a'' < \varepsilon < b'-a'$ either 1) $(a'',b'')\subset (a',b')$ or 2) $a'' < a'$ and $b'' < b'$ or 3) $a' < a''$ and $b' < b''$. If 1) then $(a'', b'')\subset (a,b)$. Assume 2), as $a'\in L$ and $b''\in U$ then $a'<b''$ , so $a' - a'' < b''-a'' < \varepsilon < a' - a$, so $a < a''$ and $b'' < b'<b$. Therefore again $(a'', b'')\subset (a,b)$. The case 3) is analogical to 2).
We conclude that $(a'',b'')\subset (a,b) \subseteq F$, by the filter property of $\mathcal{G}$ this implies that $F\in\mathcal{G}$, and that $\mathcal{F}(L,U) \subseteq\mathcal{G}$. This proves that $\mathcal{F}(L,U)$ is minimal. $\blacksquare$.
Therefore the function $(L,U) \mapsto \mathcal{F}(L,U)$ is a mapping from Dedekind reals to Bourbaki-MCF reals.
Edit on 2023-01-27
I think the following defines a mapping on opposite direction.
Given a minimal Cauchy filter $\mathcal{F}$, we can define two sets \begin{align*} L(\mathcal{F}) &= \{\operatorname{inf} F\ |\ F\in \mathcal{F}\} \\ U(\mathcal{F}) &= \{\operatorname{sup} F\ |\ F\in \mathcal{F}\} \end{align*} which define a Dedekind cut of $\mathbb{Q}$.
Proposition 2. $(L(\mathcal{F}), U(\mathcal{F}))$ is a Dedekind real.
Proof Infima and suprema do not necessarily exist in $\mathbb{Q}$. But from the Cauchy property, $L(\mathcal{F})$ and $U(\mathcal{F})$ are non-empty because there exists an interval $(a,b)\in \mathcal{F}$ for some rational $\varepsilon > 0$. So some $a = \operatorname{inf} (a,b) \in L(\mathcal{F}), b = \operatorname{sup} (a,b) \in U(\mathcal{F})$.
Let's prove that $L(\mathcal{F})$ is a lower subset. If $a' < \operatorname{inf} F$ for some $F\in \mathcal{F}$, then $a' < a \in F$ for some $a \in F$, then $F\subseteq (a', a) \cup F$. Thus $(a', a) \cup F \in \mathcal{F}$ and $\operatorname{inf} (a',a) \cup F = a'$.
Let's show that $L(\mathcal{F})$ is open. The filter $\mathcal{F}$ is minimal Cauchy so it is round, i.e. for every $F\in L(\mathcal{F})$ there exists a rational number $\varepsilon > 0$ such that any interval $(a,b) \in \mathcal{F}$ of length $b-a<\varepsilon$ satisfies $(a,b)\subseteq F$. (See Proposition 3.6 The Reals as Rational Cauchy filter by I. Weiss.) If $a \in L(\mathcal{F})$ then $a= \operatorname{inf} F$ for some $F \in \mathcal{F}$. From roundness we get a rational $\varepsilon > 0$ such that if $b'-a' < \varepsilon$ and $(a',b') \in \mathcal{F}$ then $(a',b')\subseteq F$. Now, since $\mathcal{F}$ is Cauchy then we can find such an interval $(a'', b'') \in \mathcal{F}$ such that $b''-a'' <\varepsilon/2$. Let take now interval $(a''-\varepsilon/4,b''+\varepsilon/4)$ which has length $b''-a'' + \varepsilon/2 < \varepsilon$ and $(a''-\varepsilon/4,b''+\varepsilon/4) \in \mathcal{F}$. Therefore we conclude that $(a''-\varepsilon/4,b''+\varepsilon/4) \subseteq F$. From this we get $a \leqslant a''-\varepsilon/4 < a''= \operatorname{inf}(a'',b'')$ and $L(\mathcal{F})$ is open. $U(\mathcal{F})$ is an upper cut by similar reasoning.
If $a = \operatorname{inf}A $ and $b = \operatorname{sup}B $ for some $A,B\in \mathcal{F}$ then we can again apply the roundness to the set $A\cap B$ in a similar fashion as in the proof of openness to get that $a<b$.
Finally $L(\mathcal{F})$ and $U(\mathcal{F})$ are located. If $a < b$, then we show that $(a,+\infty)\in \mathcal{F}$ or $(-\infty,b) \in \mathcal{F}$. Let us suppose this is not the case therefore any $F\in \mathcal{F}$ cannot be a subset of $(a,+\infty)$ because this would mean $(a,+\infty)\in \mathcal{F}$, so $F\subseteq (-\infty,a])$. Similarly, it cannot be a subset of $(-\infty, b)$, so $F\subseteq [b,+\infty)$, but $(-\infty,a]$ and $[b,+\infty)$. Only the empty set is a subset of both, but $\emptyset\notin\mathcal{F}$. Therefore $(a,+\infty)\in \mathcal{F}$ or $(-\infty,b) \in \mathcal{F}$, and thus $a\in L(\mathcal{F})$ and $b\in U(\mathcal{F})$. $\blacksquare$.
Therefore the function $\mathcal{F} \mapsto (L(\mathcal{F}),U(\mathcal{F}))$ is a mapping from Bourbaki-MCF reals to Dedekind reals.
This leads me to thinking that the definition of the Bourbaki reals can be simplified as the essential part of the $\mathcal{F}(L,U)$ is its filter base given by open intervals. Open intervals of $\mathbb{Q}$ can be identified with pairs of rationals such that $a<b$. We can define a partial order and a meet amongst such pairs by
\begin{align*}(a,b) &\sqsubseteq (a',b') \iff a'\leqslant a \text{ and }b\leqslant b', \bot \sqsubseteq (a,b) \sqsubseteq \top \\ (a,b) & \sqcap (a',b')= \begin{cases}(\operatorname{max}\{a,a'\}, \operatorname{min}\{b,b'\}) &\text{if } \operatorname{max}\{a,a'\} < \operatorname{min}\{b,b'\} \\ \bot &\text{if } \operatorname{max}\{a,a'\} \geqslant \operatorname{min}\{b,b'\} \end{cases}\end{align*}
This defines a meet-semilattice on $\mathscr{I}(\mathbb{Q})=\{(a,b)\in\mathbb{Q}\times \mathbb{Q}\ | \ a<b\} \cup \{\bot, \top\}$. We can now consider filters of open intervals identified with filters on $(\mathscr{I}(\mathbb{Q}), \sqsubseteq, \sqcap)$ instead of filters on $\mathbb{Q}$ (i.e. filters on $(\mathscr{P}(\mathbb{Q}), \subseteq, \cap)$).
Definition 3 A Bourbaki-OIF real is a non-empty subset $B$ of $\mathbb{Q}\times\mathbb{Q}$, representing a filter of open intervals (OIF), that satisfies the following conditions
- B$_1$.if $(a,b)\in B$, then $a<b$, i.e. $B\subseteq \mathscr{I}(\mathbb{Q})\setminus\{\bot\}$;
- B$_2$ if $(a,b), (a',b') \in B$, then $(a,b)\sqcap (a',b') \in B$;
- B$_3$ if $(a,b) \in B$ and $a'<b'$ such that $(a,b)\sqsubseteq (a',b')$, then $(a',b') \in B$;
- B$_4$ $B$ has the following Cauchy property, for any rational $\varepsilon > 0 $ there exists $(a,b) \in B$ such that $b-a < \varepsilon$;
- B$_5$ $B$ is minimal, if $C$ is any subset of $B$ that satisfies B$_1$-B$_4$ then it is necessarily equal to $B$, $B=C$.
If $(L,U)$ is a Dedekind real, then $L\times U$ is a Bourbaki-OIF real in the sense of the definition 3, $(L,D)\mapsto \mathbb{B}(L,U) = L\times U$.
If $B$ is a Bourbaki-OIF real, then $(\operatorname{proj}_1(B), \operatorname{proj}_2(B))$ is a Dedekind real, $B\mapsto \mathbb{D}(B)=(\operatorname{proj}_1(B), \operatorname{proj}_2(B))$.
The identity $(\mathbb{D}\circ \mathbb{B})(L,U) = (L,U)$ is trivial. The identity $(\mathbb{B}\circ \mathbb{D})(B) = B$ requires an observation that $B\subseteq \operatorname{proj}_1(B) \times \operatorname{proj}_2(B))$ but $\operatorname{proj}_1(B) \times \operatorname{proj}_2(B)$ is minimal therefore $B= \operatorname{proj}_1(B) \times \operatorname{proj}_2(B)$.
So a real numbers is a set of pairs of rational numbers or a pair of subsets of rational numbers and we can move betweens these two interpretations. I think this approach has the benefit that the order is easier to define for Dedekind reals while arithmetic operations are easier to define for Bourbaki reals.
A complete construction of reals as subsets of $\mathbb{Q}\times\mathbb{Q}$ using this idea: The Reals as Sets of Rational Intervals