Proof

The topology of $\beta X$ is the Zariski topology of the Boolean ring $P X$, with ideals/filters corresponding to closed sets; we thus have a Galois correspondence between filters and closed sets in $\beta X$, so that the poset of filters under inclusion is isomorphic to the poset of open sets of $\beta X$ under inclusion.

Proof

We have that $F = \bigcap (\beta(X) \cap \Phi_X(F))$, because every filter is the intersection of the ultrafilters that contain it. More elementarily (without using AC): we have $(\forall A: P X)\; A \in F \Leftrightarrow prin(\neg A) \notin \Phi_X(F)$.

Proof

By distributivity in the frame $Filt(X)$, we have $\top \leq (F \wedge F') \vee G = (F \vee G) \wedge (F' \vee G)$ iff both $\top \leq F \vee G$ and $\top \leq F' \vee G$. Taking negations on both sides of the “iff” gives the result.

Proof

Since $D$ is inhabited, we have $(colim_d\; \phi(d)) \vee G = colim_d\; (\phi(d) \vee G)$. As $\beta X$ is compact, the frame $Filt(X) = \mathcal{O}(\beta(X))$ is compact in the sense that $\hom(\top, -): Filt(X) \to \mathbf{2}$ preserves filtered colimits, i.e.,

Taking negations on both sides of this “iff” shows that $R_X(colim_{d\in D}\; \phi(d), G)$ holds iff $(\forall_{d\in D})\; R_X(\phi(d), G)$ holds.

Proof

Suppose $(\alpha, \beta)$ and $(\alpha, \beta')$ are funcoids. Then for all $G \in Filt(Y)$ we have $R_X(-, \beta(G)) = R_X(-, \beta'(G))$ since both sides equal $R_Y(\alpha(-), G)$. Hence $\beta(G) = \beta'(G)$ by separability, for all $G$; i.e., $\beta = \beta'$. Similarly, if $(\alpha, \beta)$ and $(\alpha', \beta)$ are funcoids, then $\alpha = \alpha'$. The same argument shows both the functions $\alpha$ and $\beta$ are uniquely determined from the relation $[f]$.

Proof

(Compare Porton, Proposition 6.13.) With the help of Lemma 1, we have

whence $\alpha(F \wedge F') = \alpha(F) \wedge \alpha(F')$ by separability. A similar argument shows that $\alpha(\top) = \top$.

Proof

(Compare Porton, Theorem 6.25.) For a filtered diagram $\phi: D \to Filt(X)$, we have with the help of Lemma 2 the following calculation:

whence $\alpha(colim_d\; \phi(d)) = colim_d\; \alpha(\phi(d))$ by separability.

Proof

$Filt(X)$ is the filtered colimit completion (in the $2$-category of preorders or $\mathbf{2}$-enriched categories) of $(P X)^{op}$; otherwise put, $Filt(X) \cong Lex(P X, \mathbf{2})$. The “Yoneda embedding” $prin: (P X)^{op} \to Lex(P X, \mathbf{2})$, which is the universal map from $(P X)^{op}$ to filtered-cocomplete posets, preserves finite limits. Thus, for any $\alpha: Filt(X) \to Filt(Y)$ that preserves finite limits and filtered colimits, the restriction

preserves finite limits.

In the other direction, because finite limits commute with filtered colimits in $Filt(Y) = Lex(P Y, \mathbf{2})$, the filtered-cocontinuous extension of any left exact $(P X)^{op} \to Filt(Y)$ is also left exact. Thus we have a natural bijection

We also have an isomorphism $Lex(P Y, \mathbf{2}) \cong Rex(P Y, \mathbf{2})^{op}$ that takes $\phi: P Y \to \mathbf{2}$ to $(\lambda B: P Y)(\neg\phi(\neg B))$. Left exact maps $(P X)^{op} \to Lex(P Y, \mathbf{2})$ are thus in bijection with right exact maps $P X \to Rex(P Y, \mathbf{2})$, i.e., with maps $P X \times P Y \to \mathbf{2}$ that preserve finite colimits in each variable, which are topogenies.

Proof

We have $\top \leq \alpha(F) \vee G$ precisely when there is $B \in \alpha(F)$ and $B' \in G$ such that $B \cap B' = \emptyset$. We may write $F$ as a filtered colimit of principal filters $prin(A)$ with $A \in F$; since $\alpha$ preserves filtered colimits, the condition may be rewritten

For such $B, B'$ we have $B' \subseteq \neg B$, so by upward closure of $G$ the condition may be condensed to the equivalent

Coming from the other direction: we have $\top \leq F \vee \beta(G)$ precisely when there is $A \in F$ and $A' \in \beta(G)$ such that $A \cap A' = \emptyset$. Spelling this out, this may be rewritten as

For such $A, A'$ we have $A \subseteq \neg A'$, so the last condition may be condensed down to

which matches (1) since we can swap $B$ and $\neg B$, and we are done.

Proof

(Theorem 1) Combine Proposition 4 with Lemmas 4 and 5.

Proof

(This proof is a reformulation of Porton’s proof, and uses the axiom of choice, specifically the fact that every proper filter is contained in an ultrafilter, which is slightly weaker than AC.) As in the proof of Proposition 3, the poset of funcoids is isomorphic to the poset $Lex((P X)^{op}, Filt(Y))$, or equivalently $Lex(P X, Filt(Y))$ since negation provides an isomorphism $(P X)^{op} \cong P X$. Now $Filt(Y) = \mathcal{O}(\beta Y)$ (cf. Proposition 1) is a sober spatial frame, so that (abbreviating $\mathcal{O}(\beta Y)$ to $\mathcal{O}$) there is a frame embedding

and this induces an embedding

that preserves finite meets and arbitrary joins (note this embedding is a left adjoint because we have an adjunction $j \dashv j_\ast$ of left exact maps, and $Lex(P X, -)$ will preserve that adjunction). The codomain is a power of the frame $Lex(P X, \mathbf{2}) \cong Filt(X)$ and is thus a frame itself; the domain $Lex(P X, Filt(Y))$, being closed under finite meets and arbitrary joins, is a subframe, as was to be shown.

Second proof

We prove a more general fact: that if $L$ is any frame, then $Lex((P X)^{op}, L)$ is also a frame. Then apply this to $L = Filt(Y)$.

Indeed, a frame $L$ is precisely a lex-total poset, i.e., a poset with the property that the Yoneda-Dedekind embedding $y: L \to [L^{op}, \mathbf{2}]$ has a left exact left adjoint $\sigma: [L^{op}, \mathbf{2}] \to L$. Applying the functor $Lex((P X)^{op}, -)$ to this adjunction $\sigma \dashv y$ (both left exact maps), we get an induced adjunction with right adjoint

Since $Filt(X)$ is a frame, so is $[L^{op}, Filt(X)]$. We thus have realized $Lex((P X)^{op}, L)$ as the category of algebras (or poset of fixed points) of a left exact monad (aka a nucleus) on a frame $[L^{op}, Filt(X)]$. By a well-known result in locale theory (see for example Mac Lane-Moerdijk, section IX.4, Proposition 3), this implies $Lex((P X)^{op}, L)$ is a frame, as was to be shown.

Proof

(See also this MO discussion.) This poset is isomorphic to the poset of join-preserving maps $\psi: L \coloneqq L_1 \otimes \ldots \otimes L_n \to \mathbf{2}$. This tensor product $L$ is a distributive lattice by Proposition 8. The poset of such $\psi$ is dual to the poset of ideals in $L$, or to the poset of filters in $L^{op}$, where $L^{op}$ is also a distributive lattice. Then apply Corollary 1.

Proof

This poset is dual to the frame of ideals of the distributive lattice $P X_1 \otimes \ldots \otimes P X_n$ which, by Proposition 9, is the coproduct $\sum_i P X_i$ in the category of Boolean algebras. By Stone duality, the poset of ideals in the Boolean algebra $\sum_i P X_i$ is isomorphic to the topology of its spectrum $\beta X_1 \times \ldots \times \beta X_n$.

Proof

If $f: X \to Y$ is a function and $C$ is an endotopogeny on $Y$, then there is an induced endotopogeny $f^\ast C \coloneqq (\beta(f) \times \beta(f))^{-1}(C) \subseteq \beta X \times \beta X$ by pulling back. This $f^\ast C$ is the largest endotopogeny on $X$ that renders $f$ continuous, and is the initial structure that lifts $f: X \to Y$ seen as a source diagram.

More generally, given a source diagram $f_i: X \to X_i$ of sets where $X_i$ have given endotopogenies $C_i$, the initial lift is the meet of the endotopogenies $f_i^\ast C_i$ on $X$. This shows that the forgetful functor is topological. The case for topogenic spaces is wholly similar.

Proof

If $F, F' \subseteq L$ are two filters, then their meet may be calculated as

and their join as

The inclusion $(F \wedge G) \vee (F \wedge H) \leq F \wedge (G \vee H)$ holds in any lattice, and we have

which completes the proof.

Proof

Certainly finite meets distribute over arbitrary joins in $[L, \mathbf{2}]$. Since the inclusion $Lex(L, \mathbf{2}) \hookrightarrow [L, \mathbf{2}]$ is closed under finite meets and filtered joins, the result follows.

Proof

The preceding lemmas show that $F \wedge -$ preserves finite joins and filtered joins, and therefore arbitrary joins.

Proof

Certainly each element $a \otimes b$ is idempotent, since $a \otimes b = (a + a) \otimes b = a \otimes b + a \otimes b$.

Proof

Note that if $b \leq c$, then $a b \leq a c$ since $a c = a(b + c) = a b + a c$. If the unit $1$ is maximal, it follows that $a b \leq a 1 = a$ and similarly $a b \leq b$. On the other hand, if $x \leq a$ and $x \leq b$, then $x = x x \leq a b$. This shows $a b$ is the meet of $a$ and $b$.

Proof

Certainly $P \otimes Q$ is a commutative monoid in the symmetric monoidal category of join-semilattices. By the lemma it suffices to show that the unit ${\top_P} \otimes {\top_Q}$ is maximal and that multiplication is idempotent.

First we show $\top_P \otimes \top_Q$ is the maximal element of $P \otimes Q$. Indeed,

and similarly $\top_P \otimes b \leq \top_P \otimes \top_Q$, so $a \otimes b \leq \top_P \otimes \top_Q$. Since any element of $P \otimes Q$ is a join $\sum_i a_i \otimes b_i$ of elements $a_i \otimes b_i$ which all have $\top_P \otimes \top_Q$ as upper bound, any element has $\top_P \otimes \top_Q$ as upper bound.

Second we show multiplication is idempotent. Clearly $a \otimes b$ is idempotent for any $a \in P$ and $b \in Q$, by idempotency of $a$ and $b$. More generally

which is bounded above by $\sum_i a_i \otimes b_i$ since $a_i a_j \leq a_i$ and $b_i b_j \leq b_i$, but is also bounded below by $\sum_i a_i a_i \otimes b_i b_i = \sum_i a_i \otimes b_i$. This completes the proof.

Proof

We must show that if $P, Q$ are Boolean algebras, then every element in the distributive lattice $P \otimes Q$ has a complement (necessarily unique by distributivity). For $a \in P$ and $b \in Q$, let $\neg a$ and $\neg b$ denote their complements in $P$ and $Q$. Then an easy calculation shows that the complement of $a \otimes b$ in $P \otimes Q$ is $\neg a \otimes b + a \otimes \neg b + \neg a \otimes \neg b$. Moreover, any element in $P \otimes Q$ is a finite join of elements of the form $a_i \otimes b_i$, so we just need the auxiliary lemma that the join $x + y$ of two complemented elements in a distributive lattice is also complemented. Indeed, $\neg x \cdot \neg y$ is that complement (for instance, $1 \leq x + y + (\neg x)(\neg y)$ since $1 = (x + \neg x)(y + \neg y) = x y + x(\neg y) + (\neg x)y + (\neg x)(\neg y)$ and $x y + x(\neg y) + (\neg x)y = x + (\neg x)y \leq x + y$).

Proof

For a subset $A \subseteq X$, let $[A]$ denote the corresponding basic clopen of $\beta(X)$:

It suffices to show that the image $\beta(f)([A])$ is open in $\beta(Y)$; in fact we will show $\beta(f)([A]) = [f(A)]$. For $U \in \beta(X)$ we have

and

so a necessary condition for $V \in \beta(f)([A])$ is $A \cap f^{-1}(B) \neq \emptyset$ for all $B \in V$. Note that $A \cap f^{-1}(B) \neq \emptyset$ iff $f(A \cap f^{-1}(B)) \neq \emptyset$; using Frobenius reciprocity, this is the same as saying $f(A) \cap B \neq \emptyset$. But $f(A) \cap B \neq \emptyset$ for all $B \in V$ if and only if $f(A) \in V$ (the “if” is clear, and moreover either $f(A) \in V$ or $\neg f(A) \in V$, where the latter clearly negates $f(A) \cap B \neq \emptyset$ for all $B \in V$, and so the “only if” must hold as well).

We have thus shown that $V \in \beta(f)([A])$ implies $V \in [f(A)]$. In the other direction, if $V \in [f(A)]$, then $f(A) \cap B \neq \emptyset$ for all $B \in V$, which as we saw is equivalent to $A \cap f^{-1}(B) \neq \emptyset$ for all $B \in V$. Hence the sets $A \cap f^{-1}(B)$ generate a filter on $X$, which may be extended to an ultrafilter $U$ on $X$, for which $U \in [A]$ and $V \subseteq \beta(f)(U)$ by construction, whence $V = \beta(f)(U)$ by maximality of ultrafilters. Thus $V \in [f(A)]$ implies $V \in \beta(f)([A])$.

Proof

Put $X = \sum_j X_j$ in $Set$. Each inclusion $\beta(X_j) \hookrightarrow \beta(X)$ induced by the inclusion $i_j: X_j \to X$ is open, by Lemma 10, and hence their union is open. The subspace topology on the union is the same as the disjoint sum topology. In the category $Tych$ of Tychonoff spaces, where coproducts are formed the same as they are in $Top$, an inclusion $i: X \to Y$ is dense if and only if it is epic. But it is obvious that $\sqcup_j \beta(X_j) \to \sum_j \beta(X_j)$ is epic in $Tych$: if $Z$ is Tychonoff and $i: Z \to \bar{Z}$ its Stone-Cech compactification, then a map $f: X \to Z$ is uniquely determined by a map $i f: X \to \bar{Z}$ in $CompHaus$, which in turn is uniquely determined by the family of their restrictions to the summands $\beta(X_j)$, or by the restriction along the inclusion of the disjoint sum. This completes the proof.