The Freudenthal and other compactifications of continuous frames

Abstract

The N-star compactifications of frames are the frame-theoretic counterpart of the N-point compactifications of locally compact Hausdorff spaces. A $\pi$-compactification of a frame L is a compactification constructed using a special type of a basis called a $\pi$-compact basis; the Freudenthal compactification is the largest $\pi$-compactification of a rim-compact frame. As one of the main results, we show that the Freudenthal compactification of a regular continuous frame is the least upper bound for the set of all N-star compactifications. A compactification whose right adjoint preserves disjoint binary joins is called perfect. We establish a class of frames for which N-star compactifications are always perfect. For the class of zero-dimensional frames, we construct a compactification which is isomorphic to the Banaschewski compactification and the Freudenthal compactification; in some special case, this compactification is isomorphic to the Stone–Čech compactification.

1 Introduction

The poset of all compactifications of a completely regular frame L is bounded above, the Stone–Čech compactification (in the sense of Banaschewski and Mulvey [8]) being the maximal compactification. If L is a continuous frame, then the poset of all compactifications of L is bounded below, the least compactification (à la Banaschewski [6]) being the minimal compactification. The notion of a $\pi$-compact basis always gives rise to a strong inclusion on a frame. There is a one-to-one correspondence between compactifications and strong inclusions on a frame (this is due to Banaschewski [6]); compactifications corresponding to strong inclusions defined from a $\pi$-compact basis are called $\pi$-compactifications. The Freudenthal compactification of a rim-compact frame is the maximal $\pi$-compactification with a zero-dimensional remainder (as shown by Baboolal [1]), and it is the minimum perfect compactification, by Mthethwa [12]. In [13], a frame is called a J-frame if any binary cover of L by closed sublocales whose intersection is compact has the property that at least one member of the cover is compact. We show that an N-star compactification (in accord with Baboolal [4]) of a J-frame is always perfect. The notion of a rim-perfect compactification was introduced in [14] and it is recalled and discussed herein. For a rim-compact frame, we show that rim-perfect compactifications are larger than all $\pi$-compactifications; in particular, they are larger than the Freudenthal compactification. For a continuous frame L,  let ${\mathfrak {L}}$ be a subcollection of compactifications of L consisting of all the N-star compactifications. We show that members of ${\mathfrak {L}}$ are $\pi$-compactifications and that any upper bound of ${\mathfrak {L}}$ is rim-perfect. The latter is used to deduce that the Freudenthal compactification is the least upper bound for ${\mathfrak {L}}.$ In the last section of this paper, we construct a compactification for zero-dimensional frames which turns out to be isomorphic to the Freudenthal compactification and the Banaschewski compactification; and if the frame at hand is strongly zero-dimensional, this compactification is also isomorphic to the Stone–Čech compactification.

2 Some frame theory

A frame L is a complete lattice which satisfies the infinite distributive law:

$$\begin{aligned} x\wedge \bigvee S=\bigvee \{x\wedge s\mid s\in S\} \end{aligned}$$

for every $x\in L$ and every $S\subseteq L.$ The top element and the bottom element of L will be denoted by 1 and 0,  respectively. A frame homomorphism is a map $h: L\longrightarrow M,$ between the frames L and M,  that preserves finite meets (including 1) and arbitrary joins (including 0). Associated with such a homomorphism is its right adjoint $h_*:M\longrightarrow L$ satisfying $h(x) \le y$ if and only if $x \le h_*(y)$ for all $x\in L$ and $y \in M,$ where $h_*(y) = \bigvee \{x \in L\mid h(x) \le y\}.$ In a frame L,  we have the Heyting operation $\rightarrow$ satisfying the property that, for any $a, x, y\in L$: $a \wedge x \le y$ if and only if $x \le a \rightarrow y,$ making L a complete Heyting algebra. Moreover, $a \rightarrow y = \bigvee \{x \mid a \wedge x \le y\}.$ For any $a,b\in L,$ we say that a is rather below b,  written as $a\prec b,$ if $a\wedge c=0$ and $c\vee b=1,$ for some $c\in L.$ Note that $a\prec b$ if and only if $a^{*} \vee b =1,$ where $a^{*}$ is the pseudocomplement of a,  that is, $a^{*}=a\rightarrow 0=\bigvee \{x\in L \mid x\wedge a=0\}.$ An element $a\in L$ is complemented if $a\vee a^*=1,$ and a frame is zero-dimensional if L is join-generated by complemented elements. A frame L is called regular if $a=\bigvee \{x\in L\mid x\prec a \},$ for all $a\in L.$ We say that a is completely below b in L,  and write $a \prec \prec b,$ if there are $a_r \in L$ where $r\in {\mathbb {Q}}\cap [0,1]$ such that $a_0 = a, a_1 = b$ and $a_r \prec a_s$ for $r < s.$ A frame L is said to be completely regular if $a = \bigvee \{x \mid x \prec \prec a\}$ for every $a \in L.$ A frame L is compact if for any $S\subseteq L$ with $1=\bigvee S,$ there exists a finite $S_0\subseteq S$ such that $1=\bigvee S_0.$ A complete lattice L is called continuous if for $a\in L,$ $a= \bigvee \{x\in L\mid x\ll a\},$ where $x\ll a$ means that for any $S\subseteq L$ with $a\le \bigvee S,$ there exists a finite $F\subseteq S$ such that $x\le \bigvee F.$ If $x\ll a,$ we say that x is well below a. The well below relation interpolates if L is continuous, that is, if $x\ll y$ in L then $x\ll z\ll y$ for some $z\in L.$ A sublocale of a frame L is a subset S of L such that S is closed under arbitrary meets and for each $x \in L$ and each $s \in S,$ $x\rightarrow s \in S.$ The frame ${\uparrow }a={\mathfrak {c}}_L(a)=\{x\in L\mid a\le x\}$ is a sublocale of L that is called the closed sublocale associated with a. The smallest sublocale of a frame L is given by ${{\,\textrm{O}\,}}:=\{1\},$ and of course, the largest sublocale is the entire frame.

For more concepts on the theory of frames, the reader is referred to [10] and [15].

3 Compactifications of frames and strong inclusions

A frame homomorphism $h:M\longrightarrow L$ is dense if for any $x\in M,$ $h(x)=0\Longrightarrow x=0.$ A pair (M, h) is a called a compactification of a frame L if $h:M\longrightarrow L$ is a dense onto frame homomorphism and M is a compact regular frame. The set ${\mathfrak {K}}(L)$ of all compactifications (up to isomorphism) is partially ordered: given two compactifications (M, h) and (N, f) of a frame L,  we say that (M, h) is smaller than (N, f) or (N, f) is larger than (M, h),  written as $(M,h) \le (N,f),$ if there exists a frame homomorphism $\eta :M\longrightarrow N$ such that the following diagram commutes:

If $\eta$ is a frame isomorphism, we say that (M, h) and (N, f) are isomorphic or are equivalent, and write $(M,h)\cong (N, f).$ An upper bound of ${\mathfrak {A}}\subseteq {\mathfrak {K}}(L)$ is a compactification that is larger than all the members of ${\mathfrak {A}}.$ The least upper bound of ${\mathfrak {A}}$ is an upper bound that is smaller than all the other upper bounds. A strong inclusion on a frame L is a binary relation $\triangleleft$ on L such that:

• (1) if $x\le a\triangleleft b\le y,$ then $x\triangleleft y,$

• (2) $\triangleleft$ is a sublattice of $L\times L,$

• (3) $a\triangleleft b$ implies $a\prec b,$

• (4) $a\triangleleft b$ implies $a\triangleleft c \triangleleft b$ for some $c\in L,$

• (5) $a\triangleleft b$ implies $b^*\triangleleft a^*,$ and

• (6) $a=\bigvee \{ x\in L \mid x\triangleleft a\},$ for all $a\in L.$

If $\triangleleft _1$ and $\triangleleft _2$ are strong inclusions on L,  then $\triangleleft _1\subseteq \triangleleft _2\equiv (a\triangleleft _1 b\Longrightarrow a\triangleleft _2 b)$ is a partial order on the set, ${\mathfrak {S}}(L),$ of all strong inclusions on L. Banascheswki [6] showed that there is a one-to-one correspondence between the elements of ${\mathfrak {S}}(L)$ and the elements of ${\mathfrak {K}}(L)$ in the following manner: for any compactification (M, h) of L,  let $h_*:L\longrightarrow M$ be the right adjoint of h. Corresponding to (M, h) is the strong inclusion $\lhd _{M}$ on L defined by $x\lhd _{M} y\equiv {h_*}(x)\prec {h_*}(y).$ Also, if $\lhd$ is a strong inclusion on L,  then we call an ideal $J\subseteq L$ strongly regular with respect to $\lhd ,$ if for each $x\in J,$ there exists $y\in J$ such that $x\lhd y.$ Let $\Im _\lhd L$ be the compact frame consisting of all strongly regular ideals of L with respect to $\lhd .$ We have then a dense frame homomorphism $\Im : \Im _\lhd L\longrightarrow L$ defined by $\Im (J):=\bigvee J$ for all $J\in \Im _\lhd L.$ Thus, $(\Im _\lhd L, \Im )$ is a compactification of L. The right adjoint $k_\lhd :L\longrightarrow \Im _\lhd L$ of $\Im : \Im _\lhd L\longrightarrow L$ is given by $k_\lhd (a)=\{x\in L\mid x \lhd a\}.$ Because of the nature of the one-to-one correspondence between elements of ${\mathfrak {S}}(L)$ and that of ${\mathfrak {K}}(L),$ one has: $(M,h) \le (N,f)$ if and only if $\lhd _M\subseteq \lhd _N,$ where $\lhd _M$ and $\lhd _N$ are the strong inclusions corresponding to compactifications (M, h) and (N, f),  respectively. So, $(M,h)\cong (N, f)$ if and only if $\lhd _M=\lhd _N.$ Equivalently, we have $(M,h)\cong (N, f)$ if and only if $(M,h) \le (N,f)$ and $(N,f)\le (M,h).$

4 Certain compactifications of continuous frames

Example 4.1

If L is a completely regular continuous frame, then ${\mathfrak {K}}(L)$ is bounded below and above. This is a consequence of the following general observations:

• (1) Recall from [6] that for a regular continuous frame L,  one has the smallest strong inclusion $\blacktriangleleft$ defined as follows: for $a,b\in L,$ $a\blacktriangleleft b\equiv a\prec b$ and either ${\mathfrak {c}}_L(a^*)$ or ${\mathfrak {c}}_L(b)$ is compact. The least compactification of L,  $(\Im _{\blacktriangleleft } L, \xi ),$ is the compactification corresponding to $\blacktriangleleft ;$ so $\xi (I):=\bigvee I$ for all $I\in \Im _{\blacktriangleleft } L.$ Indeed, $\blacktriangleleft$ is the smallest strong inclusion in the sense that $\blacktriangleleft \subseteq \lhd$ for any strong inclusion $\lhd$ on L. So, $(\Im _{\blacktriangleleft } L,\xi )\le (M,h)$ for any compactification (M, h) of L.

• (2) The operation $\prec \prec$ is a strong inclusion on any completely regular frame L. The Stone–Čech compactification of a completely regular frame L,  in the sense of Banaschewski and Mulvey [8], is given by the dense onto frame homomorphism $\beta :\beta L\longrightarrow L,$ which is defined by $\beta (I):=\bigvee I,$ where $\beta L$ is the collection of all ideals of L which are strongly regular with respect to $\prec \prec .$ That is, $\beta L$ is the set all ideals J with the property that if $x\in J$ then there exists $y\in J$ such that $x\prec \prec y.$ It is always true that $(M,h)\le (\beta L,\beta )$ for any compactification (M, h) of a completely regular frame L.

Definition 4.2

We say that $a\in L$ is a $\pi$-element if ${\mathfrak {c}}_L(a\vee a^{*})$ is compact. A rim-compact frame is a regular frame that is join-generated by $\pi$-elements. For a frame L,  a $\pi$-compact basis for L is a basis B such that:

• (b1) $a, b\in B$ implies that $a\wedge b,$ $a\vee b \in B.$

• (b2) $a\in B$ implies that $a^*\in B.$

• (b3) $a\in B$ implies that a is a $\pi$-element.

Example 4.3

It was shown in [1, Proposition 4.6] that for any $\pi$-compact basis B of a regular frame, one has a strong inclusion on L defined by:

$$\begin{aligned} a \lhd _B b \equiv \ \text {there is}\ c \in B\ \text {such that}\ a \prec c \prec b, \ \text {for all}\ a,b\in L. \end{aligned}$$

(4.1)

Let $\gamma _B L$ be the set of all ideals of L which are strongly regular with respect to $\lhd _B$ and define a map $\gamma _B:\gamma _B L\longrightarrow L$ by $\gamma _B(J):=\bigvee J$ for all $J\in \gamma _B L.$ Then, $(\gamma _B L, \gamma _B)$ is the compactification corresponding to the strong inclusion $\lhd _B$ called the $\pi$-compactification. In particular, for any regular frame L one has the $\pi$-compact basis

$$\begin{aligned} B_F=\{a\in L\mid a\ \text {is a }\pi \text {-element of } L\}. \end{aligned}$$

The Freudenthal compactification is the $\pi$-compactification $(\gamma _{B_{F}} L, \gamma _{B_{F}}).$ To simplify notation, we denote the Freudenthal compactification by $(\gamma L,\gamma ).$ Notice that:

Every $\pi$-compact basis is contained in $B_F,$ and therefore, the Freudenthal compactification is the maximal $\pi$-compactification of a rim-compact frame L. That is, $(\gamma _B L,\gamma _B)\le (\gamma L,\gamma ),$ for any $\pi$-compact basis B of L.

A compactification (M, h) is called perfect [1] if ${h_*} (u\vee u^*)={h_*} (u)\vee {h_*} (u^*)$ for all $u\in L.$ In this case ${h_*}$ also preserves any disjoint binary joins. That is, ${h_*} (u\vee v)={h_*} (u)\vee {h_*} (v)$ whenever $u\wedge v=0$ in L. The Stone–Čech and the Freudenthal compactifications are perfect [1]; the latter being the minimal such for rim-compact frames [12]. In [1, Proposition 3.9], the perfectness of a compactification is characterized in terms of the corresponding strong inclusion as follows:

A compactification (M, h) of a frame L is perfect if and only if its associated strong inclusion $\lhd$ satisfies $x \le y, x \lhd y \vee y^* \Longrightarrow x \lhd y$ for all $x, y \in L.$

Let us recall some nomenclature from [4] for the reader’s convenience:

Definition 4.4

[4]. Let L be a frame and N a positive integer.

• (1) An N-star on L is a collection of N mutually disjoint elements of L,  say $\alpha =\{u_{1}, u_{2},\ldots , u_{N}\},$ such that ${\mathfrak {c}}_{L}\bigg (\bigvee _{u \in \alpha } u\bigg )$ is compact but for each $1 \le i \le N,$ ${\mathfrak {c}}_{L}\bigg (\bigvee _{u \in \alpha \smallsetminus \{u_{i}\}} u\bigg )$ is not compact.

• (2) Let $\alpha = \{u_{1}, u_{2}, \ldots , u_{N}\}$ be an N-star of a frame L. For $u_{i} \in \alpha ,$ let

  $$\begin{aligned} N^{\alpha }_{i} = \bigg \{x \in L \mid {\mathfrak {c}}_{L}\bigg (x \vee \bigvee _{u \in \alpha \smallsetminus \{u_{i}\}} u\bigg ) \ \text {is compact}\bigg \}. \end{aligned}$$

• (3) The relation $\blacktriangleleft _{\alpha }$ on L defined by:

  $$\begin{aligned} a \blacktriangleleft _{\alpha } b\equiv a \prec b\ \text {and for each}\ u_{i} \in \alpha ,\ \text {either}\ a^{*} \in N^{\alpha }_{i}\ \text {or}\ b \in N^{\alpha }_{i} \end{aligned}$$

  is a strong inclusion on L. Let ${\mathfrak {J}}_{\alpha } L$ be the set of all strongly regular ideals with respect to $\blacktriangleleft _{\alpha }$ and $\alpha (I):=\bigvee I$ for all $I\in {\mathfrak {J}}_{\alpha } L.$ The compactification $({\mathfrak {J}}_{\alpha } L,\alpha )$ is called the N-star compactification of L,  and its right adjoint is denoted by $k_\alpha .$

Remark 4.5

Note that $N_{i}^{\alpha }$ in (2) of the definition above is a proper filter of L,  see [4, Lemma 3.2].

Lemma 4.6

[4, Lemma 3.3]. Let L be a regular continuous frame having an N-star $\alpha = \{u_{1}, u_{2}, \ldots , u_{N}\}$ and suppose that $1 \le s \le t \le N.$ Then $x \in N_{s}^{\alpha } \cap \cdots \cap N_{t}^{\alpha }$ if and only if ${\mathfrak {c}}_{L}\bigg (x \vee (u_{1} \vee \cdots \vee u_{s - 1}) \vee (u_{t+1} \vee \cdots \vee u_{N})\bigg )$ is compact. In particular for $s = 1, t = N,$ then $x \in \bigcap _{i = 1}^{N} N_{i}^{\alpha }\Longleftrightarrow {\mathfrak {c}}_{L}(x)$ is compact.

An N-star compactification is the frame-theoretic counterpart of the N-point compactification. Conditions under which 2-star compactifications are perfect were studied in [11].

Now let us recall the definition of a Wallman compactification of a frame:

Definition 4.7

[2]. Let L be a frame.

• (1) $B \subseteq L$ is called a Wallman basis of L if:

  • (a) B is a bounded sublattice of L,  i.e., $0, 1\in B$ and if $a, b\in B$ then $a\wedge b,$ $a\vee b \in B.$

  • (b) For every $a \in L,$ $a =\bigvee \{b\mid b \in B, b\prec _B a\},$ where $b\prec _B a$ means that there exists $c \in B$ such that $b \wedge c = 0$ and $c \vee a = 1.$

  • (c) For every $a, b \in B$ such that $a \vee b = 1$ there exists $c, d \in B$ such that $c \wedge d = 0$ and $a \vee c = b \vee d = 1.$

• (2) Let L be a regular frame and B a Wallman basis for L. Define $\lhd _\omega$ by

  $$\begin{aligned} a \lhd _\omega b \iff \ \text {there exists} \ c \in B \ \text {such that}\ a \prec _{B} c \prec _{B} b. \end{aligned}$$

  Then $\lhd _\omega$ is a strong inclusion on L. The compactification $({\mathfrak {J}}_{\omega } L,\omega ),$ where $\omega (I):=\bigvee I$ for all $I\in \Im _{\lhd _\omega } L,$ corresponding to $\lhd _\omega ,$ is called the Wallman compactification.

• (3) Note that $a \prec _{B} c \Longrightarrow a \prec c$ whenever B is a Wallman basis.

Lemma 4.8

[2, Proposition 3.7]. Let L be a non-compact regular continuous frame. Then $(\Im _{\blacktriangleleft } L,\xi )$ is the Wallman compactification of L with the Wallman basis $B_1 = \{b \in L : \ \text {either} \ {\mathfrak {c}}_{L}(b)\ \text {or}\ {\mathfrak {c}}_{L}(b^{*})\ \text {is compact}\}.$

5 Freudenthal and N-star compactifications

Recall that the Freudenthal compactification is defined for the class of rim-compact frames and note that the Freudenthal compactification exists for all regular continuous frames because:

Lemma 5.1

Any regular continuous frame is rim-compact.

Proof

Suppose that L is a regular continuous frame and let $a\in L.$ Then $a=\bigvee \{x\in L\mid x\ll a\}.$ But, $x\ll a$ and $a\le 1$ implies that $x\ll 1.$ So ${\mathfrak {c}}_L(x^*)$ is compact, by [3, Proposition 3.3]. Hence ${\mathfrak {c}}_L(x\vee x^*)$ is compact, and so $a=\bigvee \{ x\in L\mid {\mathfrak {c}}_L(x\vee x^*)\ \text {is compact}\}.$ That is, L is rim-compact. $\square$

Definition 5.2

Let $\alpha = \{u_{1}, u_{2},\ldots , u_{N}\}$ be an N-star on a regular continuous frame L. The N-star basis of L is the set

$$\begin{aligned} B_{\alpha } = \{ a \in L \mid a \in N_{i}^{\alpha } \ \text {or} \ a^{*} \in N_{i}^{\alpha } \ \text {for each }1 \le i \le N\}. \end{aligned}$$

Remark 5.3

Note that if we consider the 1-star

$$\begin{aligned} N_1=\{x\in L\mid {\mathfrak {c}}_{L}(x)\ \text {is compact}\}, \end{aligned}$$

then

$$\begin{aligned} B_1&= \{b \in L \mid \ \text {either} \ b\in N_1\ \text {or}\ b^{*}\in N_1\}\\&= \{b \in L \mid \ \text {either} \ {\mathfrak {c}}_{L}(b)\ \text {or}\ {\mathfrak {c}}_{L}(b^{*})\ \text {is compact}\}. \end{aligned}$$

Hence, $B_1$ is the 1-star basis corresponding to the least compactification $(\Im _{\blacktriangleleft } L,\xi ).$ So, all the results about $B_\alpha$ that we establish in this paper are true for the special case of $B_1.$

Notation 5.4

For the remainder of this section, unless stated otherwise, L will be a regular continuous frame, $\alpha = \{u_{1}, u_{2},\ldots , u_{N}\}$ will be an N-star on L and $B_\alpha$ will be the associated N-star basis. As the terminology suggests, indeed, $B_\alpha$ is a basis, a special one:

Lemma 5.5

$B_{\alpha }$ is a $\pi$-compact basis.

Proof

First, we argue that $B_{\alpha }$ is a basis for L. Since L is continuous, for $a \in L$ we have that $a = \bigvee \{b\mid b\ll a\}.$ Now, $b \ll a \le 1$ implies $b \ll 1.$ Thus, ${\mathfrak {c}}_{L}(b^{*})$ is compact by [3, Proposition 3.3]. So, $b^{*} \in \bigcap _{i = 1}^{N} N_{i}^{\alpha }$ by Lemma 4.6. That is, $b^{*} \in N_{i}^{\alpha }$ for every i. But this implies that $b \in B_{\alpha }$ by the definition of $B_{\alpha }.$ It is known that if L is regular, then $b \ll a \Longrightarrow b \prec a$ (for example, see [15, Lemma VII.5.2.1]). As a result, $a = \bigvee \{b\mid b \prec a, b \in B_{\alpha }\}.$ Now we show that this basis $B_{\alpha }$ is a $\pi$-compact basis:

• (b1) Let $a, b \in B_{\alpha }.$ For each i,  we have the following possible scenarios:

  • (i) $a, b \in N_{i}^{\alpha }$: then $a \vee b \in N_{i}^{\alpha }$ since $a \vee b \ge a$ and $N_{i}^{\alpha }$ is a filter. Also $a \wedge b \in N_{i}^{\alpha }$ since both a and b are in $N_{i}^{\alpha }$ and it is a filter. Thus $a \vee b \in B_{\alpha }$ and $a \wedge b \in B_{\alpha }.$

  • (ii) $a \in N_{i}^{\alpha }$ and $b^{*} \in N_{i}^{\alpha }$: since $a \in N_{i}^{\alpha },$ then $a \vee b \in N_{i}^{\alpha }.$ Also, $a \wedge b \le b$ implies that $b^{*} \le (a \wedge b)^{*}.$ Since $b^{*} \in N_{i}^{\alpha },$ then $(a \wedge b)^{*} \in N_{i}^{\alpha }.$ Thus $a \wedge b \in B_{\alpha }.$

  • (iii) $a^{*} \in N_{i}^{\alpha }$ and $b \in N_{i}^{\alpha }$: Similar argument as in (ii).

  • (iv) $a^{*} \in N_{i}^{\alpha }$ and $b^{*} \in N_{i}^{\alpha }$: this implies that $(a \vee b)^{*} = a^{*} \wedge b^{*} \in N_{i}^{\alpha },$ hence $a \vee b \in B_{\alpha }.$ Also, $b^{*} \le (a \wedge b)^{*}$ implies that $(a \wedge b)^{*} \in N_{i}^{\alpha },$ and thus, $a \wedge b \in B_{\alpha }.$

• (b2) Suppose that $a \in B_{\alpha }.$ Then either $a \in N_{i}^{\alpha }$ or $a^{*} \in N_{i}^{\alpha }$ for each i. If $a \in N_{i}^{\alpha },$ then $a^{**} \in N_{i}^{\alpha },$ since $a \le a^{**}$ and $N_{i}^{\alpha }$ is a filter. Thus, either $a^{**} \in N_{i}^{\alpha }$ or $a^{*} \in N_{i}^{\alpha }.$ Hence, $a^{*} \in B_{\alpha }.$

• (b3) Take $a \in B_{\alpha }.$ Then $a \in N_{i}^{\alpha }$ or $a^{*} \in N_{i}^{\alpha }$ for each i. If $a \in N_{i}^{\alpha }$ then $a \vee a^{*} \ge a$ implies that $a \vee a^{*} \in N_{i}^{\alpha },$ since $N_{i}^{\alpha }$ is a filter. Similarly, if $a^{*} \in N_{i}^{\alpha }$ then $a \vee a^{*} \in N_{i}^{\alpha }.$ So, $a \vee a^{*} \in N_{i}^{\alpha }$ for all $1 \le i \le N.$ That is, $a \vee a^{*} \in \bigcap _{i = 1}^{N} N_{i}^{\alpha }.$ It follows by Lemma 4.6 that ${\mathfrak {c}}_{L}(a \vee a^{*})$ is compact. $\square$

Recall that $a \blacktriangleleft _{\alpha } b$ if and only if $a \prec b$ and for each $u_{i} \in \alpha ,$ either $a^{*} \in N^{\alpha }_{i}$ or $b \in N^{\alpha }_{i}.$ The strong inclusion $\blacktriangleleft _{\alpha }$ can now be defined using elements of the $\pi$-compact basis $B_{\alpha }$ as follows:

Theorem 5.6

$\blacktriangleleft _{\alpha }=\lhd _{B_{\alpha }}$ for any N-star $\alpha$ on L. In particular,  $({\mathfrak {J}}_{\alpha } L,\alpha )$ is a $\pi$-compactification with a $\pi$-compact basis $B_\alpha .$

Proof

We start by showing that $\lhd _{B_{\alpha }}\subseteq \blacktriangleleft _{\alpha }.$ Suppose $x \lhd _{B_\alpha } a.$ Then there exists $y \in B_{\alpha }$ such that $x \prec y \prec a.$ If $y \in N_{i}^{\alpha },$ then $x \blacktriangleleft _{\alpha } y \prec a,$ so $x \blacktriangleleft _{\alpha } a.$ If $y^{*} \in N_{i}^{\alpha },$ then $x \prec y \blacktriangleleft _{\alpha } a,$ thus we also have $x \blacktriangleleft _{\alpha } a.$

Now we show that $\blacktriangleleft _{\alpha }\subseteq \lhd _{B_{\alpha }}.$ Let $x, a \in L$ and suppose that $x \blacktriangleleft _{\alpha } a.$ Then $x \prec a$ and either $x^{*} \in N_{i}^{\alpha }$ or $a \in N_{i}^{\alpha }.$ If ${\mathfrak {c}}_{L}(x^{*})$ or ${\mathfrak {c}}_{L}(a)$ is compact, then $x \blacktriangleleft a.$ By Lemma 4.8, we have that there exists $y \in B_1 = \{ b \in L \mid {\mathfrak {c}}_{L}(b) \ \text {or}\ {\mathfrak {c}}_{L}(b^{*})\ \text {is compact}\}$ such that $a \prec _{B} y \prec _{B} a.$ Therefore, $a \prec y \prec a.$ Since either ${\mathfrak {c}}_{L}(y)$ or ${\mathfrak {c}}_{L}(y^{*})$ is compact, either $y \in N_{i}^{\alpha }$ for all i or $y^{*} \in N_{i}^{\alpha }$ for all i,  by Lemma 4.6. It follows that $y \in B_{\alpha },$ and whence $x \lhd _{B_{\alpha }} a.$ So, from this end, assume that neither ${\mathfrak {c}}_{L}(x^{*})$ nor ${\mathfrak {c}}_{L}(a)$ is compact. It follows by Lemma 4.6 that $x^{*} \not \in \bigcap _{i = 1}^{N} N_{i}^{\alpha }$ and $a \not \in \bigcap _{i = 1}^{N} N_{i}^{\alpha }.$ We can, without loss of generality, re-order $N_{i}^{\alpha }$ such that:

• (1) For some t,  $a \in N_{i}^{\alpha }$ for every i such that $1 \le i \le t-1,$ whereas $a \not \in N_{i}^{\alpha }$ whenever $t \le i \le N,$ and

• (2) $\exists k \le t - 1$ such that $x^{*} \in N_{i}^{\alpha }$ for every i where $k \le i \le N,$ while $x^{*} \not \in N_{i}^{\alpha }$ whenever $1 \le i \le k -1.$

We have $a \in \bigcap _{i = 1}^{t - 1} N_{i}^{\alpha },$ so ${\mathfrak {c}}_{L}\bigg (a \vee \bigvee _{j = t}^{N}u_{j}\bigg )$ is compact, by Lemma 4.6. Similarly, since $x^{*} \in \bigcap _{i = k}^{N} N_{i}^{\alpha },$ then ${\mathfrak {c}}_{L}\bigg (x^{*} \vee \bigvee _{j = 1}^{k-1}u_{j}\bigg )$ is compact. This implies that ${\mathfrak {c}}_{L}\bigg (x^{*} \vee \bigvee _{j = 1}^{t-1}u_{j}\bigg )$ is compact since it is contained in ${\mathfrak {c}}_{L}\bigg (x^{*} \vee \bigvee _{j = 1}^{k-1}u_{j}\bigg )$: indeed, $k- 1 \le t - 1 \Longrightarrow \bigvee _{j = 1} ^{k - 1} u_{j} \le \bigvee _{j = 1} ^{t - 1} u_{j} \Longrightarrow x^{*} \vee \bigvee _{j = 1} ^{k - 1} u_{j} \le x^{*} \vee \bigvee _{j = 1} ^{t - 1} u_{j}\Longrightarrow {\mathfrak {c}}_{L}\bigg (x^{*} \vee \bigvee _{j = 1}^{t-1}u_{j}\bigg ) \subseteq {\mathfrak {c}}_{L}\bigg (x^{*} \vee \bigvee _{j = 1}^{k-1}u_{j}\bigg ).$ Since $\blacktriangleleft _{\alpha }$ is a strong inclusion, then $x \blacktriangleleft _{\alpha } a \Longrightarrow x \prec a \Longrightarrow x^{*} \vee a = 1.$ By continuity of L,  we can write $a = \bigvee \{z\in L\mid z\ll a\}.$ In the following calculation, the last step follows by the compactness of ${\mathfrak {c}}_{L}\bigg (x^{*} \vee \bigvee _{j = 1}^{t-1}u_{j}\bigg )$:

$$\begin{aligned} x^{*} \vee a = 1&\Longrightarrow x^{*} \vee \bigvee \{z\in L\mid z\ll a\} = 1 \\&\Longrightarrow x^{*} \vee \bigvee \{z\in L\mid z\ll a\} \vee \bigvee _{j = 1} ^{t - 1} u_{j}= 1 \\&\Longrightarrow x^{*} \vee z_{0} \vee \bigvee _{j = 1} ^{t - 1} u_{j} = 1,\quad \text {for some}\ z_{0} \ll a. \end{aligned}$$

So, $x \prec \big (\bigvee _{j = 1} ^{t - 1} u_{j}\big ) \vee z_{0}.$

Again, by continuity of L,  we can write $x^{*} = \bigvee \{w\in L\mid w\ll x^*\}.$ We now have

The last step follows by compactness of ${\mathfrak {c}}_{L}\bigg (a \vee \bigvee _{j = t}^{N}u_{j}\bigg ).$ Since L is regular, by [15, Lemma VII.5.2.1], we have $w_{0} \ll x^{*} \Longrightarrow w_{0} \prec x^{*} \Longrightarrow w_{0}^{*} \vee x^{*} = 1 \Longrightarrow x \prec w_{0}^{*}.$ The latter together with $x \prec \big (\bigvee _{j = 1} ^{t - 1} u_{j}\big ) \vee z_{0}$ implies that $x \prec \bigg (\big (\bigvee _{j = 1} ^{t - 1} u_{j}\big ) \vee z_{0} \bigg ) \wedge w_{0}^{*}.$ Let $y = \bigg (\big (\bigvee _{j = 1} ^{t - 1} u_{j}\big ) \vee z_{0} \bigg ) \wedge w_{0}^{*}.$ Then $x \prec y.$

Claim: $y \prec a.$ Let $v=\big (\bigvee _{j = t}^{N} u_{j}\big ) \vee w_{0}.$ By $(\dag ),$ $a\vee v=1.$ Note that $y = \bigg ( \big (\bigvee _{j = 1} ^{t - 1} u_{j} \big ) \vee z_{0} \bigg ) \wedge w_{0}^{*} = \bigg ( \big (\bigvee _{j = 1} ^{t - 1} u_{j} \big ) \wedge w_{0}^{*} \bigg ) \vee (z_{0} \wedge w_0^{*}).$ The frame law and the fact that elements of $\alpha$ are pairwise disjoint gives us:

$$\begin{aligned} \left( \left( \bigvee _{j = 1} ^{t - 1} u_{j} \right) \wedge w_{0}^{*} \right) \wedge v&= \left( \left( \bigvee _{j = 1} ^{t - 1} u_{j} \right) \wedge w_{0}^{*} \right) \wedge \left( \left( \bigvee _{j = t}^{N} u_{j} \right) \vee w_{0} \right) \\&= \left( \left( \bigvee _{j = 1} ^{t - 1} u_{j} \right) \wedge w_{0}^{*} \wedge \left( \bigvee _{j = t}^{N} u_{j}\right) \right) \vee \left( \left( \bigvee _{j = 1} ^{t - 1} u_{j} \right) \wedge w_{0}^{*} \wedge w_{0} \right) \\&= 0. \end{aligned}$$

Thus, $\bigg (\big (\bigvee _{j = 1} ^{t - 1} u_{j} \big ) \wedge w_{0}^{*} \bigg ) \wedge v = 0$ and $a \vee v = 1.$ Hence we have $\bigg (\big (\bigvee _{j = 1} ^{t - 1} u_{j} \big ) \wedge w_{0}^{*} \bigg ) \prec a.$ Furthermore, again by [15, Lemma VII.5.2.1], $z_0 \ll a \Longrightarrow z_0 \prec a .$ But $z_{0} \wedge w_{0}^{*} \le z_{0},$ so $z_{0} \wedge w_{0}^{*} \prec a.$ It follows that $y=\bigg ( \big (\bigvee _{j = 1} ^{t - 1} u_{j} \big ) \wedge w_{0}^{*} \bigg ) \vee (z_{0} \wedge w_0^{*}) \prec a.$ We now show that, for all i,  either $y \in N_{i}^{\alpha }$ or $y^{*} \in N_{i}^{\alpha }.$

Case 1: $1 \le i \le t-1$:

Since ${\mathfrak {c}}_{L}(\bigvee _{i = 1}^{N}u_{j})$ is compact, then $\bigvee _{j = 1}^{t-1}u_{j} \in N_{i}^{\alpha },$ for $1 \le i \le t-1.$ But $N_{i}^{\alpha }$ is a filter, so $\big (\bigvee _{j = 1}^{t-1}u_{j}\big ) \vee z_{0} \in N_{i}^{\alpha },$ for every $1 \le i \le t-1.$ Also, $w_{0} \ll x^{*} \le 1 \Longrightarrow w_{0} \ll 1 \Longrightarrow {\mathfrak {c}}_{L}( w_{0}^{*})\ \text {is compact}\ \Longrightarrow w_{0}^{*} \in N_{i}^{\alpha } \ \text {for every } i.$ Since $N_{i}^{\alpha }$ is a filter, then $\bigg (\big (\bigvee _{j = 1} ^{t - 1} u_{j}\big ) \vee z_{0} \bigg ) \wedge w_{0}^{*} \in N_{i}^{\alpha }.$ That is, $y \in N_{i}^{\alpha }$ whenever $1 \le i \le t -1.$

Case 2: $t \le i \le N$:

Firstly, $z_{0} \ll a \le 1 \Longrightarrow z_{0} \ll 1 \Longrightarrow {\mathfrak {c}}_{L}(z_{0}^{*})\ \text {is compact}\ \Longrightarrow z_{0}^{*} \in N_{i}^{\alpha } \ \text {for every }i.$ Now, $\bigvee _{j = t}^{N} u_{j} \in N_{i}^{\alpha }$ for $t \le i \le N,$ since ${\mathfrak {c}}_{L}\bigg (\bigvee _{i = 1}^{N}u_{j}\bigg )$ is compact. Now, $(\bigvee _{j = t}^{N} u_{j} ) \wedge \big (\bigvee _{j = 1}^{t-1} u_{j}\big ) = 0\Longrightarrow \big (\bigvee _{j = 1}^{t-1} u_{j}\big )^{*} \ge (\bigvee _{j = t}^{N} u_{j}).$ But $N_{i}^{\alpha }$ is a filter, so $\big (\bigvee _{j = 1}^{t-1} u_{j}\big )^{*} \in N_{i}^{\alpha }$ for $t \le i \le N.$ Again, $N_{i}^{\alpha }$ is closed under finite meets, so for $t \le i \le N,$ $\big (\bigvee _{j = 1}^{t-1} u_{j}\big )^{*} \wedge z_{0}^{*} \in N_{i}^{\alpha }.$ Since $y = \bigg ( \big (\bigvee _{j = 1} ^{t - 1} u_{j}\big ) \vee z_{0}\bigg ) \wedge w_{0}^{*} \le \big (\bigvee _{j = 1}^{t-1} u_{j}\big ) \vee z_{0},$ then $\big (\bigvee _{j = 1}^{t-1} u_{j}\big )^{*} \wedge z_{0}^{*} \le y^{*}.$ So, $N_{i}^{\alpha }$ being a filter, we get $y^{*} \in N_{i}^{\alpha }$ for $t \le i \le N.$ In conclusion, $x \blacktriangleleft _{\alpha } a$ implies that there exists $y \in B_{\alpha }$ such that $x \prec y \prec a;$ that is, $x \lhd _{B_\alpha } a.$ $\square$

Recall from [13] that a frame L is called a J-frame if whenever $a\wedge b= 0$ in L and ${\mathfrak {c}}_L(a\vee b)$ is compact, then ${\mathfrak {c}}_L(a)$ or ${\mathfrak {c}}_L(b)$ is compact. The following result is of independent interest:

Proposition 5.7

If L is a J-frame,  then $({\mathfrak {J}}_{\alpha } L,\alpha )$ is a perfect compactification.

Proof

We are going to argue that $({\mathfrak {J}}_{\alpha } L,\alpha )$ is isomorphic to the Freudenthal compactification $(\gamma L,\gamma ).$ It is always true that $B_\alpha \subseteq B_{F},$ so $\lhd _{B_\alpha }\subseteq \lhd _{B_{F}}.$ We show the other inclusion. Suppose $x \lhd _{B_{F}} y.$ Then there exists $w \in B_{F}$ such that $x \prec w \prec y.$ Thus, ${\mathfrak {c}}_{L}(w \vee w^{*})$ is compact. Since L is a J-frame, either ${\mathfrak {c}}_{L}(w)$ or ${\mathfrak {c}}_{L}(w^{*})$ is compact. It follows by Lemma 4.6 that $w \in N_{i}^{\alpha }$ or $w^{*} \in N_{i}^{\alpha }$ for all i. Therefore $x \lhd _{B_\alpha } y.$ Thus, $\lhd _{B_\alpha }=\lhd _{B_{F}}.$ But we have seen in Proposition 5.6 that $\blacktriangleleft _{\alpha } = \lhd _{B_\alpha },$ so $\blacktriangleleft _{\alpha }=\lhd _{B_{F}}.$ Therefore $({\mathfrak {J}}_{\alpha } L, \alpha )\cong (\gamma L,\gamma ).$ Since $(\gamma L,\gamma )$ is perfect by [1, Proposition 4.10], then $({\mathfrak {J}}_{\alpha } L,\alpha )$ is perfect. $\square$

In the three lemmas below, L is a rim-compact frame with a $\pi$-compact basis B,  and $k_{B}$ is the right adjoint of the join map $\gamma _B:\gamma _B L\longrightarrow L$ corresponding to $\lhd _{B}.$

Lemma 5.8

[1, Lemma 4.5]. If $w \in L$ and $u \in B$ with $w \vee u = 1,$ then there exists $v \in B$ such that $v \prec u$ and $w \vee v = 1.$

Lemma 5.9

$k_{B}(u) \vee k_{B}(v) = k_{B}(u \vee v)$ for any $u, v \in B.$

Proof

Let $u, v \in B.$ Since $u \le u \vee v$ and $v \le u \vee v,$ then $k_{B}(u) \vee k_{B}(v) \subseteq k_{B}(u \vee v).$ To prove that $k_{B}(u \vee v) \subseteq k_{B}(u) \vee k_{B}(v),$ let $x \in k_{B}(u \vee v) .$ Then $x \lhd _{B} u \vee v.$ By the interpolation property of the strong inclusion, there exists $z \in L$ such that $x \lhd _{B} z \lhd _{B} u \vee v.$ Then $x \prec z \prec u \vee v.$ Find $w \in L$ such that $z \wedge w= 0$ and $w \vee u \vee v = 1.$ Since $u, v \in B,$ by applying Lemma 5.8 to u and to v,  we can get $u', v' \in B$ such that $u' \prec u,$ $v' \prec v$ and $w \vee u' \vee v' = 1.$ By Lemma 5.8 again, there exists $u'', v'' \in B$ such that $u'' \prec u',$ $v'' \prec v'$ and $w \vee u'' \vee v'' = 1.$ It follows from the latter that $w^{*} \prec (u'' \vee v'').$ But $z \wedge w = 0$ implies $z \le w^{*}.$ Thus, $z \prec (u'' \vee v'').$ So, we can write $z = (z \wedge u'') \vee (z \wedge v'').$ Then $z \wedge u'' \le u'' \prec u' \prec u$ and $z \wedge v'' \le v'' \prec v' \prec v.$ Hence, $z \wedge u'' \prec u' \prec u$ and $z \wedge v'' \prec v' \prec v.$ Since $u', v' \in B,$ then by definition of $\lhd _{B},$ we have $z \wedge u'' \lhd _{B} u$ and $z \wedge v'' \lhd _{B} v.$ Therefore, $z \in k_{B}(u) \vee k_{B}(v).$ Now, $x \lhd _{B} z$ implies that $x \in k_{B}(u) \vee k_{B}(v).$ It follows that $k_{B}(u \vee v) \subseteq k_{B}(u) \vee k_{B}(v).$ $\square$

Let us note the general observation that if $h: M \longrightarrow L$ is a dense and onto frame homomorphism, then $h_*(a^*) = (h_*(a))^*$ for all $a \in L,$ and $h(x^*) = h(x)^*,$ for all $x \in M$ (for example, see [1] for proofs). We apply the former to the right adjoint $k_B$ of the dense onto frame homomorphism $\gamma _B:\gamma _B L\longrightarrow L$ in the result below.

Lemma 5.10

For any $x,y\in L,$ $x \lhd _{B} y$ implies $k_{B}(x) \prec k_{B}(y).$

Proof

Let $x,y\in L.$ Then:

$$\begin{aligned} x \lhd _{B} y&\Longrightarrow \exists z \in L \ \text {such that } x \lhd _{B} z \lhd _{B} y \\&\Longrightarrow \exists v, w \in B \ \text {such that}\ x \prec v \prec z \prec w \prec y\\&\Longrightarrow x \prec v \prec w \prec y. \end{aligned}$$

Then $v^{*} \vee w = 1,$ which implies that $k_{B}(v^{*} \vee w) = k(1)=L=1_{\gamma _{B}L},$ where $1_{\gamma _{B}L}$ is the top element of $\gamma _{B}L.$ Since B is a $\pi$-compact basis, $v^{*}, w \in B,$ and so, by Lemma 5.9, we get $k_{B}(v^{*}) \vee k_{B}(w) = k_{B}(v^{*} \vee w) = 1_{\gamma _{B}L}.$ That is, $(k_{B}(v^{}))^* \vee k_{B}(w)= 1_{\gamma _{B}L}.$ Therefore $k_{B}(v) \prec k_{B}(w).$ But $k_{B}(x) \le k_{B}(v) \prec k_{B}(w) \le k_{B}(y),$ which implies $k_{B}(x) \prec k_{B}(y).$ $\square$

Definition 5.11

[14] A compactification (M, h) of a frame L is said to be rim-perfect if for any $\pi$-elements $u, v \in L,$ $u^{*} \vee v^{*} = 1\Longrightarrow {h_*} (u^{*}) \vee {h_*} (v^{*}) = 1.$

Remark 5.12

A word on rim-perfectness and $\pi$-compactifications is in order:

• (1) Since any compactification (M, h) corresponds to the strong inclusion $\lhd _M$ defined by: for any $x,y\in L,$ $x\lhd _M y \equiv h_*(x)\prec h_*(y),$ we can rephrase rim-perfectness in terms of strong inclusions as follows: a compactification (M, h) with the corresponding strong inclusion $\lhd _M$ is rim-perfect if and only if for any $\pi$-elements $u,v\in L,$ $u \prec v^{*}$ implies $u \lhd v^{*}.$ We shall employ this strategy in all our results where we must show rim-perfectness.

• (2) In particular, to show that a certain $\pi$-compactification $(\gamma _B L,\gamma _B)$ is rim-perfect, it is enough to prove that for any $\pi$-elements $x,y\in L,$ $x \prec y^{*}$ implies $x \lhd _{B} y^{*}.$ Indeed, if we apply Lemma 5.10 to $x \lhd _{B} y^{*},$ we get rim-perfectness.

Example 5.13

The Freudenthal compactification is rim-perfect: First, recall that $B_F$ consists of all $\pi$-elements of a rim-compact frame L. Now, if $a,b\in B_F$ and $a \prec b^{*},$ then $a^{*} \vee b^{*} = 1,$ and so, by Lemma 5.8, there exists $c\in B_F$ such that $c\prec b^*$ and $a^*\vee c=1.$ That is, $a\prec c\prec b^*,$ so $a\lhd _{B_F} b^*.$ It follows that $(\gamma L,\gamma )$ is rim-perfect by the remark above. Note that this argument fails for any other $\pi$-compactification, since the $\pi$-compact basis corresponding to such a $\pi$-compactification does not contain all $\pi$-elements of the frame.

Next, we show that all $\pi$-compactifications are smaller than rim-perfect compactifications.

Proposition 5.14

Let (M, h) be a rim-perfect compactification of L. If B is a $\pi$-compact basis of L,  then $(\gamma _B L,\gamma _B)\le (M,h).$

Proof

Let $\lhd$ be the strong inclusion associated with (M, h) and $\lhd _{B}$ be the strong inclusion corresponding to $(\gamma _B L,\gamma _B),$ with a $\pi$-compact basis B. We have

$$\begin{aligned} x \lhd _{B} y&\Longrightarrow \exists z \in L \ \text {such that}\ x \lhd _{B} z \lhd _{B} y \quad (\text {since }\lhd _{B}\text { is a strong inclusion})\\&\Longrightarrow \exists w, v \in B \ \text {such that}\ x \prec w \prec z \prec v \prec y \\&\Longrightarrow x \prec w \prec z \prec v^{**} \prec y \\&\Longrightarrow x \prec w \prec v^{**} \prec y. \end{aligned}$$

Recall that every member of B is $\pi$-element. Now, since (M, h) is rim-perfect and $w, v^{*} \in B,$ then $w \prec v^{**}\Longrightarrow w \lhd v^{**}.$ Hence $x \prec w \lhd v^{**} \prec y.$ Thus, $x \lhd y,$ and it follows that $\lhd _{B} \subseteq \lhd .$ $\square$

In particular, the Freudenthal compactification is smaller than any rim-perfect compactification, and this will be important in our main result, so we formally record it:

Corollary 5.15

If (M, h) is a rim-perfect compactification,  then $(\gamma L,\gamma ) \le (M,h).$

Recall from the introduction that ${\mathfrak {L}}$ is the subset of ${\mathfrak {K}}(L)$ consisting of N-star compactifications of a regular continuous frame L,  ordered by the usual partial order of compactification, $\le ,$ that was described in Section 3 of this paper. We now show that any upper bound of ${\mathfrak {L}}$ is rim-perfect.

Theorem 5.16

If (M, h) is an upper bound of ${\mathfrak {L}},$ then (M, h) is rim-perfect.

Proof

First, let $\lhd _M$ be the strong inclusion corresponding to (M, h). Since (M, h) is larger than any N-star compactification, then $\blacktriangleleft _\alpha \subseteq \lhd _M,$ for any N-star $\alpha$ on L. We use Remark 5.12 to show that (M, h) is rim-perfect. So, let $x,y\in L$ be such that ${\mathfrak {c}}_{L}(x \vee x^{*})$ and ${\mathfrak {c}}_{L}(y \vee y^{*})$ are compact and assume that $x \prec y^{*}.$ We need to show that $x \lhd _M y^{*}.$ Suppose that ${\mathfrak {c}}_{L}(x)$ or ${\mathfrak {c}}_{L}(y)$ are compact. Now,

$$\begin{aligned} B_{1} = \{z \in L \mid {\mathfrak {c}}_{L}(z) \ \text {or}\ {\mathfrak {c}}_{L}(z^{*})\ \text {is compact}\} \end{aligned}$$

is the $\pi$-compact basis corresponding to the least compactification $({\mathfrak {J}}_{\blacktriangleleft } L, \xi ),$ by Lemma 5.5 and Theorem 5.6, and $x \in B_{1}$ or $y \in B_{1}.$ This implies that either $x^{*} \in B_{1}$ or $y^{*} \in B_{1},$ since $B_{1}$ is a $\pi$-compact basis. Let us assume that $x^{*} \in B_{1}.$ Then

$$\begin{aligned} x \prec y^{*}&\Longrightarrow x^{*} \vee y^{*} = 1 \\&\Longrightarrow \exists w \in B_{1}, w \prec x^{*}, w \vee y^{*} = 1 \quad (\text {by Lemma}~5.8)\\&\Longrightarrow x \le x^{**} \prec w^{*} \prec y^{*} \\&\Longrightarrow x \prec w^{*} \prec y^{*}. \end{aligned}$$

Since $w \in B_{1},$ then $w^{*} \in B_{1}.$ Thus, by Theorem 5.6, we have $x \blacktriangleleft y^{*}.$ But $\blacktriangleleft$ is the smallest strong inclusion, so $\blacktriangleleft \subseteq \lhd _M,$ and whence $x \lhd _M y^{*}.$ By a similar argument, if we assume that $y^{*} \in B_{1},$ we get $x \lhd _M y^{*}.$

Now, let us assume that ${\mathfrak {c}}_{L}(x)$ and ${\mathfrak {c}}_{L}(y)$ are not compact. We have the following cases to consider:

Case 1: ${\mathfrak {c}}_{L}(x \vee y)$ is compact.

Since $x \prec y^{*},$ then $x \wedge y = 0.$ Then $\vartheta = \{x, y\}$ is a 2-star for L. Therefore, we have the 2-star compactification $({\mathfrak {J}}_{\vartheta } L, {\vartheta }),$ with

$$\begin{aligned}&N_{1}^{\vartheta } = \{z \in L : {\mathfrak {c}}_{L}(z \vee y) \ \text {is compact}\} \\&N_{2}^{\vartheta } = \{z \in L : {\mathfrak {c}}_{L}(x \vee z) \ \text {is compact}\}. \end{aligned}$$

Notice that $y^{*} \in N_{1}^{\vartheta },$ because ${\mathfrak {c}}_{L}(y \vee y^{*})$ is compact. Similarly, $x^{*} \in N_{2}^{\vartheta }.$ We now have $x \prec y^{*}$ and either $x^{*} \in N_{i}^{\vartheta }$ or $y^{*} \in N_{i}^{\vartheta }$ for each $i= 1, 2.$ So, $x \blacktriangleleft _{\vartheta } y^{*},$ by Theorem 5.6. Since (M, h) is an upper bound of N-star compactifications, then $\blacktriangleleft _{\vartheta }\subseteq \lhd _M,$ and so $x \lhd _M y^{*}.$

Case 2: ${\mathfrak {c}}_{L}(x \vee y)$ is not compact.

If ${\mathfrak {c}}_{L}(x^{*})$ or ${\mathfrak {c}}_{L}(y^{*})$ is compact, then $x \prec y^{*} \Longrightarrow x \blacktriangleleft y^{*} \Longrightarrow x \lhd _M y^{*},$ and we are done. Now assume that ${\mathfrak {c}}_{L}(x^{*})$ and ${\mathfrak {c}}_{L}(y^{*})$ are not compact. Let $v_{1} = x,$ $v_{2} = y$ and $v_{3} = x^{*} \wedge y^{*}.$ We argue that $\nu = \{v_{1}, v_{2}, v_{3}\}$ is a 3-star on L. First, observe that $\nu$ is a set of three pairwise disjoint elements in L. Next, we have ${\mathfrak {c}}_{L}(v_{1} \vee v_{2}) = {\mathfrak {c}}_{L}(x \vee y)$ is not compact by assumption. Note that $x \prec y^{*}$ implies that $x \le y^{*}$ and $y \le x^{*}.$ Now, ${\mathfrak {c}}_{L}(v_{2} \vee v_{3}) = {\mathfrak {c}}_{L}(y \vee (x^{*} \wedge y^{*})) = {\mathfrak {c}}_{L}((y \vee x^{*}) \wedge (y \vee y^{*})) = {\mathfrak {c}}_{L}(x^{*} \wedge (y \vee y^{*})),$ so ${\mathfrak {c}}_{L}(v_{2} \vee v_{3})$ cannot be compact, otherwise ${\mathfrak {c}}_{L}(x^{*})$ will be compact, which would contradict our assumption. Similarly, ${\mathfrak {c}}_{L}(v_{1} \vee v_{3}) = {\mathfrak {c}}_{L}(y^{*} \wedge (x \vee x^{*})).$ Therefore ${\mathfrak {c}}_{L}(v_{1} \vee v_{3})$ cannot be compact, since ${\mathfrak {c}}_{L}(y^{*})$ is not compact. Finally, we have that

$$\begin{aligned} {\mathfrak {c}}_{L}(v_{1} \vee v_{2} \vee v_{3})&= {\mathfrak {c}}_{L}(x \vee y \vee (x^{*} \wedge y^{*})) \\&= {\mathfrak {c}}_{L}((x \vee y \vee x^{*}) \wedge (x \vee y \vee y^{*}) ) \\&= {\mathfrak {c}}_{L}((x \vee x^{*}) \wedge ( y \vee y^{*}) ) \quad (\text {since}\ x \le y^{*}, y \le x^{*})\\&= {\mathfrak {c}}_{L}(x \vee x^{*})\vee {\mathfrak {c}}_{L}(y \vee y^{*}). \end{aligned}$$

Therefore, ${\mathfrak {c}}_{L}(v_{1} \vee v_{2} \vee v_{3})$ is compact, since ${\mathfrak {c}}_{L}(x \vee x^{*})$ and ${\mathfrak {c}}_{L}(y \vee y^{*})$ are compact. Thus, we have a 3-star compactification $({\mathfrak {J}}_{\nu } L, \nu )$ with

$$\begin{aligned}&N_{1}^{\nu } = \{z \in L : {\mathfrak {c}}_{L}(z \vee v_{2} \vee v_{3}) \ \text {is compact}\} \\&N_{2}^{\nu } = \{z \in L : {\mathfrak {c}}_{L}(v _{1} \vee z \vee v_{3}) \ \text {is compact}\}\\&N_{3}^{\nu } = \{z \in L : {\mathfrak {c}}_{L}(v _{1} \vee v_{2} \vee z) \ \text {is compact}\}. \end{aligned}$$

Now, $y^{*} \in N_{1}^{\nu },$ since

$$\begin{aligned} {\mathfrak {c}}_{L}(y^{*} \vee v_{2} \vee v_{3})&= {\mathfrak {c}}_{L}(y^{*} \vee y \vee (x^{*} \wedge y^{*})) \\&= {\mathfrak {c}}_{L}((y^{*} \vee y \vee x^{*}) \wedge (y^{*} \vee y \vee y^{*})) \\&= {\mathfrak {c}}_{L}((y^{*} \vee x^{*}) \wedge ( y \vee y^{*})) \quad (\text {since}\ y \le x^{*}) \\&= {\mathfrak {c}}_{L}( y \vee y^{*})\quad (\text {since}\ y^{*} \vee x^{*} = 1), \end{aligned}$$

and ${\mathfrak {c}}_{L}( y \vee y^{*})$ is compact. Similarly, $x^{*} \in N_{2}^{\nu }$ because ${\mathfrak {c}}_{L}(v _{1} \vee x^{*} \vee v_{3}) = {\mathfrak {c}}_{L}(x \vee x^{*})$ is compact. Also, $x^{*} \in N_{3}^{\nu }$ because ${\mathfrak {c}}_{L}(v _{1} \vee v_{2} \vee x^{*}) = {\mathfrak {c}}_{L}(x \vee y \vee x^{*}) = {\mathfrak {c}}_{L}(x \vee x^{*})$ is compact. Thus, $x \prec y^{*}$ and either $x^{*} \in N_{i}^{\nu }$ or $y^{*} \in N_{i}^{\nu }$ for each $i = 1, 2, 3.$ Hence, by Theorem 5.6, $x \blacktriangleleft _{\nu } y^{*}.$ Since (M, h) is an upper bound of N-star compactifications, $\blacktriangleleft _{\nu }\subseteq \lhd _M.$ This implies that $x \lhd _M y^{*}.$ Therefore, in all the cases, $x \prec y^{*}$ implies $x \lhd _M y^{*},$ so (M, h) is rim-perfect. $\square$

Now, to the main result of this section:

Theorem 5.17

The Freudenthal compactification is the least upper bound of ${\mathfrak {L}}.$

Proof

On the one hand, any N-star compactification is a $\pi$-compactification by Proposition 5.6, and since $(\gamma L,\gamma )$ is the largest $\pi$-compactification, then $(\gamma L,\gamma )$ is an upper bound for ${\mathfrak {L}}.$ On the other hand, if (M, h) is an upper bound for ${\mathfrak {L}},$ then (M, h) is rim-perfect by Proposition 5.16. So, $(\gamma L,\gamma ) \le (M,h),$ by Corollary 5.15. $\square$

6 When Freudenthal agrees with Banaschewski and Stone–Čech

For any frame L,  as usual, let BL denote the Boolean part of L. That is,

$$\begin{aligned} BL= \{ b \in L \mid b \ \text {is complemented}\}. \end{aligned}$$

Let L be zero-dimensional. Let us denote the frame of all ideals of the lattice BL by $\beta _0 L.$ It is a compact and zero-dimensional frame, and the map $\beta _0:\beta _0 L \longrightarrow L$ defined by $\beta _0 (I):=\bigvee I$ is a dense onto frame homomorphism. The pair $(\beta _0 L,\beta _0 )$ is called the Banaschewski compactification. It is the largest zero-dimensional compactification of a zero-dimensional frame, see [5] (the reader would note that $\beta _0 L$ is sometimes denoted by $\zeta L$ in the literature). We shall construct a compactification which is isomorphic to the Banaschewski compactification and the Freudenthal compactification for the class of zero-dimensional frames.

Proposition 6.1

Let L be a zero-dimensional frame. Then : 

• (1) BL is a $\pi$-compact basis for L.

• (2) L is rim-compact.

Proof

(1) Since L is a zero-dimensional frame, then L has a basis T consisting of complemented elements. Thus, $T\subseteq BL,$ and this makes BL a basis for L. Let us now argue that BL satisfies all the conditions of a $\pi$-compact basis:

• (b1) Let $a, b \in BL.$ Then $(a \vee b) \vee (a \vee b)^{*} = (a \vee b) \vee (a^{*} \wedge b^{*}) = (a \vee b \vee a^{*}) \wedge (a \vee b \vee b^{*}) = 1.$ Also, $(a \wedge b) \vee (a \wedge b)^{*} \ge (a \wedge b) \vee (a^{*} \vee b^{*}) = (a^{*} \vee b^{*} \vee a) \wedge (a^{*} \vee b^{*} \vee b) = 1.$ Hence, $a\vee b, a\wedge b\in BL.$

• (b2) It is clear that if $a \in BL,$ then $a^{*} \in BL.$

• (b3) For $b \in BL,$ we have $b\vee b^{*} = 1,$ so ${\mathfrak {c}}_{L}(b \vee b^{*}) = {{\,\textrm{O}\,}}$ is compact. Therefore, b is a $\pi$-element in L.

(2) Any zero-dimensional frame is regular. This together with the fact that elements of BL are all $\pi$-elements of L (by (b3) of Part (1)) implies that L is rim-compact. $\square$

As mentioned in Example 4.3, for any rim-compact frame L and any $\pi$-compact basis B of L,  one has the strong inclusion $\lhd _B$ defined by:

$$\begin{aligned} a \lhd _B b \equiv \ \text {there is}\ c \in B\ \text {such that}\ a \prec c \prec b,\quad \text {for all}\ a,b\in L. \end{aligned}$$

(6.1)

For a zero-dimensional frame L,  we denote the compactification corresponding to the strong inclusion $\lhd _{BL}$ by $(\gamma _0 L,\gamma _0).$ Notice that if L is a Boolean frame, then $L=BL,$ so $\lhd _{B_F}=\lhd _{BL},$ and therefore $(\gamma L,\gamma )\cong (\gamma _0 L,\gamma _0).$ We shall show that this remains valid for all zero-dimensional frames.

Proposition 6.2

If L is a zero-dimensional frame,  then $(\gamma _0 L,\gamma _0)$ is a perfect compactification.

Proof

Let $a, b \in L.$ It suffices to show that if $a \le b$ and $a \lhd _{BL} b \vee b^{*},$ then $a \lhd _{BL} b.$ Now, $a \lhd _{BL} b \vee b^{*}$ implies that there is a complemented $t \in L$ such $a \prec t \prec b \vee b^{*}.$ But $a \prec b \vee b^{*}$ implies $a^{*} \vee b \vee b^{*} = 1,$ and since $b^{*} \le a^{*},$ then $a^{*} \vee b = 1.$ Therefore $a \prec b.$ We also have that $a \prec t,$ thus $a \prec t \wedge b.$ Let $w = t \wedge b.$ We claim that $w \prec b.$ Indeed, since $t \prec b \vee b^{*},$ then there exists $s \in L$ such that $t \wedge s = 0$ and $b \vee b^{*} \vee s = 1.$ Therefore

$$\begin{aligned} w \wedge (b^{*} \vee s) = (w \wedge b^{*}) \vee (w \wedge s) = (t \wedge b \wedge b^{*}) \vee (t \wedge b \wedge s) = 0 \vee 0 = 0. \end{aligned}$$

Thus $w \prec b.$ It follows that $a \prec w \prec b.$ It remains to show that w is complemented. Let us first show that $t \vee t^{*} \le w \vee w^{*}.$ On the one hand, $w = t \wedge b \le t$ implies that $t^{*} \le w^{*},$ so $t^{*} \le w^{*} \vee w.$ On the other hand, $t \prec b \vee b^{*}$ implies that $t = (t \wedge b) \vee (t \wedge b^{*}) = w \vee (t \wedge b^{*}) \le w \vee b^{*} \le w \vee w^{*}.$ The latter inequality follows by the fact that $w \le b$ implies $b^{*} \le w^{*}.$ Therefore, $t \vee t^{*} \le w \vee w^{*}.$ But $t \vee t^{*} = 1,$ since t is complemented. Thus, $w \vee w^{*} = 1.$ In summary, $w \in BL$ and $a \prec w \prec b.$ That is, $a \lhd _{BL} b,$ as required. $\square$

Recall that a Stone frame is a compact zero-dimensional frame. Since $\gamma _{0}L$ is compact, it follows by the result below that it is a Stone frame. In the proof of the following result we use the Banaschewski–Brümmer characterization (see [7, Lemma 1.1]) which states that a compactification (M, h) of a frame L is such that M is zero-dimensional if and only if the associated strong inclusion $\lhd _M$ satisfies the condition:

$$\begin{aligned} a\lhd _M b\Longrightarrow \exists c\in L\ \text {such that}\ a \le c\lhd _M c \le b. \end{aligned}$$

Proposition 6.3

If L is zero-dimensional,  then $\gamma _{0}L$ is zero-dimensional.

Proof

Suppose $a\lhd _{BL} b$ in L. Then $a\prec c\prec b,$ for some $c\in BL.$ Since a complemented elemented is rather below itself, then $a\prec c \prec c\prec c\prec b.$ Thus, $a\le c \lhd _{BL}c\prec b,$ and so $\gamma _{0}L$ is zero-dimensional. $\square$

All zero-dimensional compactifications of a zero-dimensional frame L are smaller than $\gamma _{0}L$:

Proposition 6.4

Let L be a zero-dimensional frame. If (M, h) is a zero-dimensional compactification of L,  then $(M,h) \le (\gamma _0 L,\gamma _0).$

Proof

Let (M, h) be a zero-dimensional compactification and $\lhd _M$ bet the strong inclusion associated with (M, h). To show that $(M,h) \le (\gamma _0 L,\gamma _0),$ we shall argue that $\lhd _M \subseteq \lhd _{BL}.$ Take $x, y\in L$ and suppose that $x \lhd _M y.$ Since a strong inclusion interpolates, there exists $w \in L$ such that $x \lhd _M w \lhd _M y.$ That is, $h_*(x) \prec h_*(w) \prec h_*(y),$ where $h_*$ is the right adjoint of h. Being zero-dimensional, M has a basis B of complemented elements, we can write $h_*(w) = \bigvee \{b_{i}\in B\mid b_{i} \le h_*(w)\}.$ From $h_*(x) \prec h_*(w),$ we get $h_*(x)^{*} \vee h_*(w) = 1_{M},$ therefore $h_*(x)^{*} \vee \bigvee \{b_{i}\in B\mid b_{i} \le h_*(w)\} = 1_{M}.$ Using compactness of M we get that $h_*(x)^{*} \vee \bigvee _{i=1}^{n} \{b_{i}\in B\mid b_{i} \le h_*(w)\} = 1_{M}.$ Since each $b_i$ is complemented in M and a finite join of complemented elements is complemented, then $b=\bigvee _{i=1}^{n} \{b_{i}\in B\mid b_{i} \le h_*(w)\}$ is complemented. We now have $b \le h_*(w)$ with $h_*(x)^{*} \vee b = 1_{M}.$ Thus, $h_*(x) \prec b \le h_*(w) \prec h_*(y).$ This implies that $h_*(x) \prec b \prec h_*(y).$ Hence, $x \prec h(b) \prec y.$ By complementedness of b in M,  we get $h(b) \vee h(b^{*}) = h(b \vee b^{*}) = h(1_{M}) = 1_{L},$ so h(b) is complemented in L. So, $h(b)\in BL$ and $x \prec h(b) \prec y.$ That is, $x \lhd _{BL} y.$ $\square$

Zero-dimensionality of $\beta _0 L$ and the result above implies that $(\beta _0 L, \beta _0) \le (\gamma _0 L,\gamma _0).$ It follow by the maximality of $(\beta _0 L, \beta _0)$ that:

Corollary 6.5

If L is zero-dimensional,  then $(\beta _0 L, \beta _0)\cong (\gamma _0 L,\gamma _0).$

Let $h:M\longrightarrow L$ be a compactification with the right adjoint $h_*: L\longrightarrow M.$ Note that $h_*[L]=\{h_*(a)\mid a\in L\}$ is a sublocale of M,  and that L and $h_*[L]$ are isomorphic as locales. In [9], the remainder of L in M is defined to be the supplement of $h_*[L]$ in M. That is:

$$\begin{aligned} M\smallsetminus h_*[L]=\bigvee \{S\in \mathcal {S}(M)\mid S\cap h_*[L]={{\,\textrm{O}\,}}\}, \end{aligned}$$

where $\mathcal {S}(M)$ is the lattice of all sublocales of M. We identify $h_*[L]$ with L and write $M\smallsetminus L$ for the remainder.

Definition 6.6

[14]. A sublocale S of a locale L is zero-dimensionally embedded in L if L has a basis B such that ${\mathfrak {c}}_L(b\vee b^*) \cap S= {{\,\textrm{O}\,}},$ for each $b\in B.$

If a sublocale S of a frame L is zero-dimensionally embedded in L,  then S is zero-dimensional. If L is a zero-dimensional frame with a basis B of complemented elements, then $S\cap {\mathfrak {c}}_L(b\vee b^*)=S\cap {\mathfrak {c}}_L(1)={{\,\textrm{O}\,}},$ for any $b\in B.$ Thus, every sublocale of a zero-dimensional frame L is zero-dimensionally embedded in L. In particular, the remainder $\gamma _{0}L\smallsetminus L$ is zero-dimensionally embedded in $\gamma _{0}L.$ To see that $(\gamma _0 L,\gamma _0)$ is also isomorphic to the Freudenthal compactification for a zero-dimensional frame L,  we employ [14, Proposition 3.26] which states that:

If (M, h) is any compactification of L with a zero-dimensionally embedded remainder, then $(M,h)\cong (\gamma L,\gamma )$ if and only if (M, h) is perfect.

This combined with Proposition 6.2 yields:

Corollary 6.7

If L is zero-dimensional,  then $(\gamma _0 L,\gamma _0) \cong (\gamma L,\gamma ).$

Banaschewski and Brümmer [7] studied completely regular frames with the property that $\beta L$ is zero-dimensional and called such frames the strongly zero-dimensional frames. For these frames, we have:

Corollary 6.8

If L is a strongly zero-dimensional frame,  then

$$\begin{aligned} (\beta L, \beta )\cong (\beta _0 L, \beta _0)\cong (\gamma _0 L,\gamma _0)\cong (\gamma L,\gamma ). \end{aligned}$$