A Point-Free Approach to Canonical Extensions of Boolean Algebras and Bounded Archimedean $\ell$-Algebras

Abstract

Recently W. Holliday gave a choice-free construction of a canonical extension of a boolean algebra B as the boolean algebra of regular open subsets of the Alexandroff topology on the poset of proper filters of B. We make this construction point-free by replacing the Alexandroff space of proper filters of B with the free frame $\mathcal {L}_B$ generated by the bounded meet-semilattice of all filters of B (ordered by reverse inclusion) and prove that the booleanization of $\mathcal {L}_B$ is a canonical extension of B. Our main result generalizes this approach to the category $\varvec{ ba \ell }$ of bounded archimedean $\ell$-algebras, thus yielding a point-free construction of canonical extensions in $\varvec{ ba \ell }$. We conclude by showing that the algebra of normal functions on the Alexandroff space of proper archimedean $\ell$-ideals of A is a canonical extension of $A\in \varvec{ ba \ell }$, thus providing a generalization of the result of Holliday to $\varvec{ ba \ell }$.

Appendix: Technical lemmas

In this appendix we present the technical lemmas used in Sections 5 and 6 to prove that the pair $(D(\mathbb {R}[B]),\alpha )$ is a canonical extension of $A\in \varvec{ ba \ell }$ and $D(\mathbb{R}[B]) \cong N(X_{A}) \cong B(Y_{A})$. We start by recalling that $0 \le u \in A$ is a weak order-unit if $a \wedge u = 0$ implies $a = 0$ for each $a \in A$. It is well known that a strong order-unit is a weak order-unit (see, e.g., [14, Lem. XIII.11.4]). It is easy to see that any positive multiple of a strong order-unit is again a strong order-unit. Thus, every positive multiple of a strong order-unit is a weak order-unit. We will use this in the proof of next lemma.

Lemma A.1

Let $A \in \varvec{ ba \ell }$.

• (1) If $e, f \in {\text {Id}}(A)$ and $0 \le r, s \in \mathbb {R}$, then $re \wedge sf = \min (r, s)(e \wedge f)$.

• (2) Let $a \in A$ and $r, s \in \mathbb {R}$ with $r < s$. If $a \vee r \ge s$, then $a \ge s$.

Proof

(1) Without loss of generality we may assume that $r \le s$. Since $(e - (e\wedge f)) \wedge f = (e \wedge \neg f) \wedge f= 0$, we have $r(e - (e\wedge f))\wedge sf = 0$ by Remark 3.15(9). Therefore,

$$\begin{aligned} 0 \le (re \wedge sf) - r(e \wedge f) \le r(e - (e\wedge f))\wedge sf = 0 \end{aligned}$$

by Remark 3.15(1). Thus, $re \wedge sf = r(e \wedge f) = \min (r,s)(e \wedge f)$.

(2) If $a \vee r \ge s$, then $-s \ge -(a \vee r) = (-a) \wedge (-r)$, so

$$0 \ge s + [(-a) \wedge (-r)] = (s-a) \wedge (s-r).$$

This yields

$$\begin{aligned} 0 = [(s-a) \wedge (s-r)] \vee 0 = [(s-a) \vee 0] \wedge (s-r). \end{aligned}$$

Because $s-r > 0$, it is a weak order-unit. Therefore, $(s-a) \vee 0 = 0$, and so $s-a \le 0$. Thus, $s \le a$.

Lemma A.2

Let $A \in \varvec{ ba \ell }$ and I,J be $\ell$-ideals of A. 

• (1) If $I + J = A$, then there are $a \in I$, $b \in J$ with $0 \le a, b$ and $a + b = 1$.

• (2) $\langle {I}\rangle = A$ implies $I = A$.

• (3) If $I, J \in \mathsf{{Arch}}(A)$, then $I \vee J = \langle {I + J}\rangle$.

• (4) If $I,J \in \mathsf{{Arch}}(A)$ and $I \vee J = A$, then $I + J = A$.

Proof

(1) Since $I + J = A$, there are $x \in I$ and $y \in J$ with $1 = x + y$. Since $y \in J$, we have $1 + J = x + J \le |x| + J$. Set $a = 1 \wedge |x|$. Then $0 \le a \le 1$ and $a \in I$ because $x \in I$, so $|x| \in I$. Therefore, $a + J = (1 + J) \wedge (|x| + J) = 1 + J$. Thus, $b := 1-a \in J$. Clearly $a + b = 1$ and $0 \le b$ since $a \le 1$.

(2) Suppose $\langle {I}\rangle = A$. Then $1 \in \langle {I}\rangle$, so $(n\cdot 1-1)^+ \in I$ for each $n \ge 1$ by Proposition 4.8. In particular, $(2\cdot 1-1)^+ \in I$. Thus, $1 \in I$, and so $I = A$.

(3) Since $I + J \subseteq I \vee J$ and $I \vee J$ is archimedean, we have $\langle {I+J}\rangle \subseteq I \vee J$. On the other hand, $\langle {I+J}\rangle$ is an archimedean $\ell$-ideal which contains both I and J, so it contains $I \vee J$. Thus, $I \vee J = \langle {I + J}\rangle$.

(4) This follows from (2) and (3).

Let L be a frame and B its free boolean extension. We recall from Section 2 that for $a \in L$, we write $a^*$ for the pseudocomplement of a in L. On the other hand, we write $\lnot a$ for the complement of a in B.

Lemma A.3

Let $A \in \varvec{ ba \ell }$ and B be the free boolean extension of $\mathsf{{Arch}}(A)$.

• (1) If $I \in \mathsf{{Arch}}(A)$, then $x_I = \bigvee \{ x_{\lnot J} \mid J \vee I = A\}$.

• (2) If $0 \le f \in D(\mathbb {R}[B])$, then there are $0 \le r_I \in \mathbb {R}$ with $f = \bigvee \{r_I x_{\lnot I} \mid I \in \mathsf{{Arch}}(A) \}$.

• (3) Let $0 \le f \in D(\mathbb {R}[B])$ and $0 \le t \in \mathbb {R}$. If $f = \bigvee \{ r_I x_{\lnot I} \mid I \in \mathsf{{Arch}}(A) \}$ with $r_I \ge 0$, then $f + t = \bigvee \{ (t + r_I)x_{\lnot I} \mid I \in \mathsf{{Arch}}(A)\}$.

Proof

(1) Since $\mathsf{{Arch}}(A)$ is a regular frame (see Theorem 4.4),

$$\begin{aligned} I = \bigvee \{ K \mid K \prec I\} = \bigvee \{ K \mid K^* \vee I = A\}. \end{aligned}$$

We show that $I = \bigvee \{ \lnot J \mid J \vee I = A\}$. The right-to-left inclusion is clear. For the left-to-right inclusion it is sufficient to show that if $K^* \vee I = A$, then $K \le \lnot J$ for some J with $J \vee I = A$. But if we set $J = K^*$, then $K \subseteq K^{**} = J^* \le \lnot J$. Thus, $I = \bigvee \{ \lnot J \mid J \vee I = A\}$, and hence $x_I = \bigvee \{ x_{\lnot J} \mid J \vee I = A\}$.

(2) Each element of $D(\mathbb {R}[B])$ is a join from $\mathbb {R}[B]$. A nonnegative element of $\mathbb {R}[B]$ can be written in the form $r_1 x_{b_1} + \cdots + r_n x_{b_n} = r_1 x_{b_1} \vee \cdots \vee r_n x_{b_n}$ for some $0 \le r_i \in \mathbb {R}$ and $b_1, \dots , b_n \in B$ with $b_i \wedge b_j = 0$ whenever $i \ne j$ (see Remark 3.14). Since each $b \in B$ is a finite join of elements of the form $J \wedge \lnot I$ with $I, J \in \mathsf{{Arch}}(A)$, we may write a nonnegative element of $\mathbb {R}[B]$ as a join of elements of the form $r(x_J \wedge x_{\lnot I})$. Thus, by (1), if $0 \le f \in D(\mathbb {R}[B])$, we may write f as a join of elements of the form $rx_{\lnot I}$ with $I \in \mathsf{{Arch}}(A)$.

(3) By Remark 3.14(2,3), $x_I + x_{\lnot I} = x_I \vee x_{\lnot I} = 1$. Therefore, by Remark 3.15(1),

$$\begin{aligned} f + t = \bigvee \{ t + r_I x_{\lnot I} \mid I \in \mathsf{{Arch}}(A) \} = \bigvee \{ (t + r_I)x_{\lnot I} + tx_I \mid I \in \mathsf{{Arch}}(A) \}. \end{aligned}$$

Because $x_{\lnot I} \wedge x_I = 0$, Remark 3.15(9) implies $(t + r_I)x_{\lnot I} \wedge tx_I = 0$, so

$$(t + r_I)x_{\lnot I} + tx_I = (t + r_I)x_{\lnot I} \vee tx_{I}.$$

 Thus, by (1),

$$\begin{aligned} f + t= & {} \bigvee \{ (t + r_I)x_{\lnot I} \vee tx_I \mid I \in \mathsf{{Arch}}(A) \} \\= & {} \bigvee \{ (t + r_I)x_{\lnot I} \vee tx_{\lnot J} \mid I, J \in \mathsf{{Arch}}(A), I \vee J = A\}. \end{aligned}$$

Now, $t \le t + r_{J}$, so $tx_{\lnot J} \le (t+r_{J})x_{\lnot J}$. Consequently, $f + t = \bigvee \{ (t + r_I)x_{\lnot I} \mid I \in \mathsf{{Arch}}(A)\}$.

Remark A.4

Let $A\in \varvec{ ba \ell }$. For $S\subseteq A$ we let $[{S}]$ be the $\ell$-ideal of A generated by S. It is well known (see, e.g., [34, p. 96]) that

$$\begin{aligned}{}[{S}] = \{ x \in A : |x| \le \sum_{i=1}^{k} n_{i} |a_{i}|\; \text{for some}\; n_1, \ldots, n_{k} \ge 1\; \text{and}\; a_{1}, \ldots, a_{k} \in S\}. \end{aligned}$$

If $S = \{a\}$, we write $[{a}]$ for $[{S}]$.

Let $A \in \varvec{ ba \ell }$. We recall that $X_A = \mathsf{{Arch}}(A) \setminus \{A\}$, that $X_A$ is ordered by inclusion, and that $X_A$ is viewed as an Alexandroff space.

Lemma A.5

Let $A \in \varvec{ ba \ell }$.

• (1) If $a, b \in A$ with $a < b$ and $b - a \in \mathbb {R}$, then $\langle {b^+, a^-}\rangle = A$.

• (2) If $I,J \in X_A$ with $I \not \subseteq J$, then there is $K \in X_A$ with $J \subseteq K$ and $K + I = A$.

Proof

(1) Set $I =\langle {b^+, a^-}\rangle$. Since $0 \le a^+ \le b^+$, we have $a^+ \in I$, so $a = a^+ - a^- \in I$. Also, as $0 \le b^- \le a^-$, we have $b^- \in I$, and so $b \in I$. Thus, $b - a \in I$. Since $b-a$ is a nonzero real number, it is a unit in A, and hence $I = A$.

(2) Since $I \not \subseteq J$ there is $a \in I$ with $a \notin J$. Because J is archimedean, by Proposition 4.8, there is $n\ge 1$ with $(n|a|-1)^+ \notin J$. Let $K = J \vee \langle {(n|a|-1)^-}\rangle$. Then $J \subseteq K$, and $I \vee K = A$ by (1). We show that $K \in X_A$. Otherwise $1 = x + y$ with $0 \le x, y$, $x \in J$, and $y \in \langle {(n|a|-1)^-}\rangle$ by Lemma A.2(1,4). We claim that $y (n|a| - 1)^+ = 0$. To see this, we set $b = n|a|-1$. Since $y \in \langle {b^-}\rangle$, we have $(y-1/p)^+ \in [{b^-}]$ for each $p \ge 1$ by Proposition 4.8. Therefore, by Remark A.4, for each p there is m with $(y-1/p)^+ \le mb^-$. Thus, $0 \le (y-1/p)^+ b^+ \le mb^- b^+ = 0$ by Remark 3.15(7), and so $(y - 1/p)^+ b^+ = 0$. Because $y-1/p \le (y-1/p)^+$, we have

$$(y-1/p)b^+ \le (y-1/p)^+ b^+ = 0,$$

so $yb^+ \le (1/p) b^+$, which yields $pyb^+ \le b^+$. Since this is true for all $p \ge 1$, it follows that $yb^+ = y(|n|a-1)^+ = 0$ as A is archimedean. This verifies the claim. Therefore, $(n|a| - 1)^+ = (n|a| - 1)^+ (x + y) = (n|a| - 1)^+ x$, and so $(n|a| - 1)^+ \in J$, which is a contradiction. Thus, $K \in X_A$.

Lemma A.6

Let $A \in \varvec{ ba \ell }$, B be the free boolean extension of $\mathsf{{Arch}}(A)$, $I \in X_A$, and $0 \le f,g \in D(\mathbb {R}[B])$. Suppose that whenever $tx_{\lnot I} \le f$ for $0 \le t \in \mathbb{R}$, there is $K \in X_A$ with $I \subseteq K$ and $tx_{\lnot K} \le g$. Then $f \le g$.

Proof

Suppose that $f \not\le g$. Then $0 \not\le g - f$, so $(g-f)^{-} \ne 0$. By Lemma 5.7(1), there are $r>0$ and $I \in X_{A}$ with $rx_{\lnot I} \le (g-f)^{-}$. By Remark 3.15(7), 

$$0 \le rx_{\lnot I}(g-f)^+ \le (g-f)^+ (g-f)^- = 0$$

which yields $rx_{\lnot I}(g-f)^{+} = 0$. Therefore,

$$0 = rx_{\lnot I}[(g-f)\vee 0] = r(gx_{\lnot I} - fx_{\lnot I}) \vee 0.$$

Since $0 < r$, we have $g x_{\lnot I} -f x_{\lnot I} \le 0$, and hence $g x_{\lnot I} \le f x_{\lnot I}$. Consequently, because $r x_{\lnot I} \le (g-f)^{-}$ and $x_{\lnot I}$ is an idempotent,

$$r x_{\lnot I} \le (g-f)^{-} x_{\lnot I} = [(f-g)\vee 0] x_{\lnot I} = (f x_{\lnot I} -g x_{\lnot I} )\vee 0 = f x_{\lnot I} -g x_{\lnot I}.$$

This gives $(r+g) x_{\lnot I} = r x_{\lnot I} + g x_{\lnot I} \le f x_{\lnot I}$. Note that if $J \supseteq I$, then $x_{\lnot J} \le x_{\lnot I}$. Therefore, multiplying the equation above by $x_{\lnot J}$ and using Definition 3.10, we have

$$(r+g)x_{\lnot J} \le fx_{\lnot J} \textrm{ for each } J \in X_{A}, J \supseteq I.$$

(A.1)

We use this to obtain a contradiction to$f \not\le g$. Since $0 \le f$, we have

$$r x_{\lnot I} \le (g-f)^{-} = (f-g)^{+} \le f.$$

By hypothesis, there is $K_{1} \in X_{A}$ with $K_{1} \supseteq I$ and $rx_{\lnot K_1} \le g$, so $rx_{\lnot K_1} \le gx_{\lnot K_1}$. Thus, as $(r+g)x_{\lnot K_{1}} \le fx_{\lnot K_{1}}$ by Equation (A.1), we have $2rx_{\lnot K_{1}} \le fx_{\lnot K_{1}} \le f$. From this the hypothesis yields $K_{2} \in X_{A}$ with $K_2 \supseteq K_1$ and $2rx_{\lnot K_2} \le g$. Using Equation (A.1) again yields $3rx_{\lnot K_2} \le f$. Continuing, for each $n \ge 1$ there is $K_{n} \in X_{A}$ with $(n+1)rx_{\lnot K_n} \le f$. Multiplying this by $x_{\lnot K_n}$ gives $(n+1)rx_{\lnot K_n} \le fx_{\lnot K_n}$. Since $D(\mathbb{R}[B])$ is bounded, there is $m > 0$ with$f \le m$. Then $fx_{\lnot K_n} \le mx_{\lnot K_n}$, and so $(n+1)rx_{\lnot K_n} \le mx_{\lnot K_n}$. By Remark 3.13(1), this forces $(n+1)r \le m$ for each $n \ge 1$. The obtained contradiction shows that $f \le g$.

We arrive at our final auxiliary lemma, Item (2) of which has the most involved proof.

Lemma A.7

Let $A \in \varvec{ ba \ell }$, $0 \le a \in A$, and $I \in X_A$. Set $s_{I} = \sup \{r \mid (a-r)^{-} \in I \}$. 

• (1) $(a - s_I)^- \in I$.

• (2) $rx_{\lnot I} \le \alpha (a)$ iff $(a-r)^- \in I$.

• (3) $s_I = \sup \{ r \mid rx_{\lnot I} \le \alpha (a) \}$ and $I \vee \langle {(a - s_I)^+}\rangle \ne A$.

Proof

(1) If $(a-r)^- \in I$, then $a + I \ge r + I$ in A/I by Remark 4.7(2). We use this to show that $(a - s_I)^- \in I$. For each $n \ge 1$ there is r with $(a-r)^- \in I$ and $s_I - 1/n \le r$. Therefore, $(s_I - 1/n) + I \le r + I \le a + I$, and so $(s_I - a) + I \le 1/n + I$. Since this is true for all n, we have $(s_I-a) + I \le 0 + I$ as A/I is archimedean. Thus, $s_I + I\le a + I$. Applying Remark 4.7(2) again yields $(a-s_I)^- \in I$.

(2) If $(a-r)^- \in I$, then $rx_{\lnot I} \le \alpha (a)$ by Lemma 5.7(1). Conversely, suppose that $rx_{\lnot I} \le \alpha (a)$. The result is clear if $r \le 0$ since then $(a-r)^- = 0 \in I$, so assume $r > 0$. By (1) and Lemma 5.7(1), we may write $\alpha (a) = \bigvee \{ s_J x_{\lnot J} \mid J \in \mathsf{{Arch}}(A)\}$, where $s_J$ is given in (1). To show $(a-r)^- \in I$ we then need to show $r \le s_I$. We have $rx_{\lnot I} \le \bigvee \{ s_J x_{\lnot J} \mid J \in \mathsf{{Arch}}(A) \}$ and $rx_{\lnot I} \le r$. Therefore,

$$\begin{aligned} rx_{\lnot I}\le & {} \bigvee \{ s_Jx_{\lnot J} \mid J \in \mathsf{{Arch}}(A)\} \wedge r \\= & {} \bigvee \{ s_Jx_{\lnot J} \wedge r \mid J \in \mathsf{{Arch}}(A)\} \\= & {} \bigvee \{ \min (s_J, r)x_{\lnot J} \mid J \in \mathsf{{Arch}}(A) \} \end{aligned}$$

by Remark 3.15(2) and Lemma A.1(1). To simplify notation set $t_J = \min (s_J, r)$. Then $(a - t_J)^- \le (a - s_J)^-$, so $(a - t_J)^- \in J$. From Remark 3.14(3) we get

$$\begin{aligned} r = rx_I \vee rx_{\lnot I} \le rx_I \vee \bigvee \{ t_Jx_{\lnot J} \mid J \in \mathsf{{Arch}}(A)\} \le r \end{aligned}$$

since the join is bounded by r, so equality holds throughout. Fix $\varepsilon > 0$. Then

$$\begin{aligned} r &= \left( rx_I \vee \bigvee \{ t_Jx_{\lnot J} \mid J \in \mathsf{{Arch}}(A)\}\right) \vee (r - \varepsilon ) \\ &= rx_I \vee \bigvee \{ t_Jx_{\lnot J} \vee (r - \varepsilon ) \mid J \in \mathsf{{Arch}}(A)\}. \end{aligned}$$

If $t_J \le r - \varepsilon$, then $t_J x_{\lnot J} \le r - \varepsilon$. Therefore,

$$\begin{aligned} r &= rx_I \vee \bigvee \{ t_Jx_{\lnot J} \vee (r - \varepsilon ) \mid J \in \mathsf{{Arch}}(A)\} \\ &= \left( rx_I \vee \bigvee \{t_Jx_{\lnot J} \mid r - \varepsilon < t_J \}\right) \vee (r-\varepsilon ). \end{aligned}$$

Thus, we have $rx_I \vee \bigvee \{ t_Jx_{\lnot J} \mid r - \varepsilon < t_J \} = r$ by Lemma A.1(2). Multiplying both sides by $r^{-1}$ yields $x_I \vee \bigvee \{ r^{-1}t_Jx_{\lnot J} \mid r - \varepsilon < t_J \} = 1$. Consequently, $x_{\lnot I} \le \bigvee \{ r^{-1}t_Jx_{\lnot J} \mid r - \varepsilon < t_J\}$ by Remark 3.13(2), and hence $rx_{\lnot I} \le \bigvee \{ t_Jx_{\lnot J} \mid r - \varepsilon < t_J\}$. Let $S = \{ J \in \mathsf{{Arch}}(A) \mid r - \varepsilon < t_J \}$. We then have $rx_{\lnot I} \le \bigvee \{ t_J x_{\lnot J} \mid J \in S\}$ and $(a - (r-\varepsilon ))^- \in J$ for each $J \in S$ since $(a - (r - \varepsilon ))^- \le (a - t_J)^-$ and $(a - t_J)^- \in J$. Because $t_J \le r$ for each J by definition and $r > 0$, we have $rx_{\lnot I} \le \bigvee \{ rx_{\lnot J} \mid J \in S\}$, so $x_{\lnot I} \le \bigvee \{ x_{\lnot J} \mid J \in S\}$. Let $K=\bigwedge_{\mathsf{Arch}(A)} S$. Then $K = \bigcap S$. If $J \in S$, then $K \subseteq J$, so $\neg J \le \neg K$ in B, and hence $x_{\neg J} \le x_{\neg K}$. Therefore, $x_{\neg I} \le x_{\neg K}$. This yields $\neg I \le \neg K$, so $\bigcap S=K \subseteq I$. As $(a-(r-\varepsilon ))^- \in J$ for each $J \in S$, we see that $(a-(r-\varepsilon ))^- \in I$. Since this is true for all $\varepsilon$, we have $a + I \ge (r - \varepsilon ) + I$ for each $\varepsilon$, so $a + I \ge r + I$ because I is archimedean. Thus, $(a-r)^- \in I$.

(3) We write $s = s_I$ for convenience. The first part of the statement follows from (1) and (2). Suppose that $I \vee \langle {(a-s)^+}\rangle = A$. Then $I + [{(a-s)^+}] = A$ by Lemma A.2(4). By Lemma A.2(1) and Remark A.4, there are $0 \le x,y$ with $x \in I$, $y \le n(a-s)^+$ for some n, and $x + y =1$. Then $1/n - y/n = x/n \in I$ and $1/n - y/n \ge 1/n - (a-s)^+$. Therefore, $1/n - y/n \ge (1/n - (a-s)^+) \vee 0$, so $(1/n - (a-s)^+)^+ \in I$. Using Items (3), (1), (2), and (4) of Remark 3.15, we have

$$\begin{aligned} \left( 1/n - (a-s)^+\right) ^{+}&= \left( 1/n - ((a-s)\vee 0)\right)^{+} = \left( 1/n + ((s-a) \wedge 0)\right) ^+ \\ &= \left( (s + 1/n - a) \wedge 1/n\right) \vee 0 \\ &= \left( (s + 1/n - a) \vee 0\right) \wedge 1/n \\ &= (a - (s + 1/n))^- \wedge 1/n. \end{aligned}$$

Thus, $(a - (s + 1/n))^- \wedge 1/n \in I$. Since $A/I \in \boldsymbol{\mathit{ba}\ell}$ and $1/n$ is a weak order-unit, $(a-(s+1/n))^- \in I$. This is a contradiction to the definition of $s = s_I$. Thus, $I \vee \langle {(a-s_I)^+}\rangle \ne A$.