The Sierpinski Locale

In this blog post we are going to prove the existence of a locale $\mathbb{S}$ with an open subset $U_0 \subset \mathbb{S}$ which has the property that for every locale $X$ the map $$ \Hom(X, \mathbb{S}) \to \Omega(X), \qquad \chi \mapsto \chi^*(U_0) $$ is a bijection. In the language of frames, this means that the frame $\Omega(\mathbb{S})$ is the free frame on one generator. We are going to work with intuitionistic reasoning.

Classically, it is not too hard to show that $\mathbb{S}$ is given by the locale associated with the Sierpinski space, i.e. with the frame of open subsets $\Omega(\mathbb{S}) = \{\{\}, \{0\}, \{0, 1\}\}$ and $U_0 = \{0\}$. In the absence of the law of excluded middle however, the construction of $\mathbb{S}$ is more delicate.

Free $\wedge$-Semilattices

A $\wedge$-semilattice is a partially ordered set $L$ which admits a top element $\top$ and in which any two elements $a, b \in L$ admit an infimum $a \wedge b$. (Equivalently, every finite family of elements—including the empty family—admits an infimum.) A morphism of $\wedge$-semilattices is a monotone map which preserves the top elements and pairwise infima. As a first step towards the construction of the free frame on one generator, we give a description of the free $\wedge$-semilattice on one generator.

Proposition. Consider the ordered set $\{0, 1\}$ as a $\wedge$-semilattice. Given any $\wedge$-semilattice $L$ and an element $a \in L$, there exists a unique morphism $f \colon \{0, 1\} \to L$ such that $f(0) = a$.

Proof. Since any such morphism must satisfy $f(0) = a$ and $f(1) = \top$ uniqueness follows immediately. Conversely, defining $f \colon \{0, 1\} \to L$ by $f(0) := a$ and $f(1) := \top$, we need to show that $f$ is a morphism of $\wedge$-semilattices. To show that $f(x \wedge y) = f(x) \wedge f(y)$, one needs to consider 4 cases in each of which the required equality follows easily. For example, if $x = 0$ and $y = 1$, then $f(0 \wedge 1) = f(0) = a = a \wedge \top = f(0) \wedge f(1)$. ∎

This is enough to continue with the construction of the Sierpinski locale, however it is interesting to remark that there exists a free $\wedge$-semilattice on an arbitrary set of generators. In fact, for the proof, it is easier to think about $\vee$-semilattices.

Let $M$ be a set. A subset $A \subset M$ is called finitely enumerated if there exists an $n \in \mathbb{N}$ such and a map $\sigma \colon \{1, \dots n\} \to M$ with image $A$. For example, the empty subset is finitely enumerated (take $n$ = 0) and the union of two finitely enumerated subsets is finitely enumerated. It is clear that the set of finitely enumerated subsets of $M$ forms a $\vee$-semilattice with join given by union.

Let $L$ be a $\vee$-semilattice. Note that by induction on the number of elements necessary to enumerate a finitely enumerated subset it follows that every finitely enumerated subset $A \subset L$ admits a supremum $\bigvee_{a \in A} a$.

Proposition. Let $M$ be a set. Consider the $\vee$-semilattice $F(M)$ of finitely enumerated subsets of $M$ and the map $u \colon M \to F(M)$, $a \mapsto \{a\}$. If $L$ is any $\vee$-semilattice and $f \colon M \to L$ is a map, then there exists a unique morphism $\widetilde{f} \colon F(M) \to L$ such that $\widetilde{f} \circ u = f$.

Proof Sketch. If $A \in F(M)$ is given by $A = \{a_1, \dots, a_n\}$, then $A = \{a_1\} \cup \dots \cup \{a_n\} = u(a_1) \vee \dots \vee u(a_n)$ in $F(M)$ and hence $$ \widetilde{f}(A) = \widetilde{f}(u(a_1)) \vee \dots \vee \widetilde{f}(u(a_n)) = f(a_1) \vee \dots \vee f(a_n) = \bigvee_{b \in f(A)} b. $$ This shows uniqueness. Conversely, defining $\widetilde{f}(A) := \bigvee_{b \in f(A)} b$, it is not hard to show that $\widetilde{f}$ satisfies all the requirements (note that $f(A)$ is finitely enumerated, so the join is defined by our remarks above). ∎

Corollary. Let $M$ be a set. Consider the $\wedge$-semilattice $F^{\wedge}(M)$ of finitely enumerated subsets of $M$ with the partial ordering given by $A \le B \colon\iff B \subset A$. Note that the infimum of two subsets is given by the union and the top elemet is given by the empty set. Consider the map $u \colon M \to F^{\wedge}(M)$, $a \mapsto \{a\}$. If $L$ is any $\wedge$-semilattice and $f \colon M \to L$ is a map, then there exists a unique morphism $\widetilde{f} \colon F(M) \to L$ such that $\widetilde{f} \circ u = f$. ∎

In the case of the one-point set $\{*\}$, there are only two finitely enumerated subsets, namely $\{\}$ and $\{*\}$ and we recover $F^{\wedge}(\{*\}) \cong \{0, 1\}$.

Free Frames

$\newcommand{\downset}{{\,\downarrow}}$Let $L$ be a $\wedge$-semilattice. A set $U \subset L$ is called a down set if whenever $l, m \in L$ such that $m \in U$ and $l \le m$, then also $l \in U$. It is easy to see that intersections and unions of down sets are again down sets and hence the set of all down sets on $L$ is a topology on $L$ and, in particular, a frame. We denote this frame by $G(L)$. Note that the map $$ u \colon L \to G(L), \qquad l \mapsto l\downset := \{ m \in L \mid m \le l \} $$ is a morphism of $\wedge$-semilattices. Indeed, given $l, l' \in L$, we have $$ m \in (l \wedge l')\downset \iff m \le l \wedge l' \iff m \le l \text{ and } m \le l' \iff m \in l\downset \cap l'\downset $$ and hence $(l \wedge l')\downset = l\downset \cap l'\downset$.

Proposition. Let $L$ be a $\wedge$-semilattice. If $F$ is any frame and $f \colon L \to F$ is a morphism of $\wedge$-semilattices, then there exists a unique morphism of frames $\widetilde{f} \colon G(L) \to F$ such that $\widetilde{f} \circ u = f$.

Proof. Note that if $U \in G(L)$, then $U = \bigvee_{l \in L} l\downset$ because $U$ is a down set. Hence, if $\widetilde{f}$ is such a morphism, then $$ \begin{split} \widetilde{f}(U) &= \widetilde{f}(\bigcup_{l \in U} l\downset) \\ &= \bigvee_{l \in U} \widetilde{f}(l\downset) \\ &= \bigvee_{l \in U} f(l). \end{split} $$ This shows uniqueness of $\widetilde{f}$. Conversely, we define $\widetilde{f}$ by setting $\widetilde{f}(U) := \bigvee_{l \in U} f(l)$ for $U \in G(L)$.

If $\{U_i\}_{i \in I}$ is a family of elements $U_i \in G(L)$, then $$ \widetilde{f}(\bigcup_{i \in I} U_i) = \bigvee_{l \in \bigcup_{i \in I} U_i} f(l) = \bigvee_{i \in I} \bigvee_{l \in U_i} f(l) = \bigvee_{i \in I} \widetilde{f}(U_i). $$ Furthermore, we have $\top \ge \widetilde{f}(L) = \bigvee_{l \in L} f(l) \ge f(\top) = \top$ and $$ \begin{split} f(U) \wedge f(V) &= \bigvee_{l \in U} f(u) \wedge \bigvee_{m \in V} f(m) \\ &= \bigvee_{l \in U, m \in V} f(l) \wedge f(m) \\ &= \bigvee_{l \in U, m \in V} f(l \wedge m) \\ &\le \bigvee_{k \in U \cap V} f(k) \\ &= \widetilde{f}(U \cap V) \\ &\le \widetilde{f}(U) \wedge \widetilde{f}(V) \end{split} $$ for $U, V \in G(L)$. The first $\le$ is because $U$ and $V$ are down sets, so $l \wedge m \in U \cap V$ whenever $l \in U$ and $m \in V$. The second $\le$ holds because $\widetilde{f}$ is easily shown to be monotone. This shows that $\widetilde{f}$ is a morphism of frames.

It remains to be shown that $\widetilde{f} \circ u = f$. For $l \in L$ we have $f(l) \le \bigvee_{m \in l\downset} f(m)$ because $l \in l\downset$ and $\bigvee_{m \in l\downset} f(m) \le f(l)$ because $f(m) \le f(l)$ for all $m \in l\downset$. Hence $$ \widetilde{f}(u(l)) = \widetilde{f}(l\downset) = \bigvee_{m \in l\downset} f(m) = f(l) $$ which concludes the proof. ∎

The Sierpinski Locale

For the frame $G(\{0, 1\})$ of downward-closed sets of $\{0, 1\}$ with the element $U_0 := u(\{0\}) = \{0\}\downset = \{0\} \in G(\{0, 1\})$ we know now that the map of sets $$ \Hom(G(\{0, 1\}), F) \to \Hom_{\text{}}(\{0, 1\}, F) \to F, \qquad f \mapsto f \circ u \mapsto (f \circ u)(0) = f(U_0) $$ is bijective, where the first $\Hom$-set is a $\Hom$-set of frames and the second one is a $\Hom$-set of $\wedge$-semilattices.

Setting $\mathbb{S}$ to be the locale with frame of opens given by $\Omega(\mathbb{S}) := G(\{0, 1\})$, this means that for any locale $X$ the map $$ \Hom(X, \mathbb{S}) \to \Omega(X), \qquad \chi \mapsto \chi^*(U_0) $$ is bijective.

As a special case of the universal property, we can describe the points of $\mathbb{S}$. Recall, that the set of points of a locale $X$ is defined as $\Hom(\mathbb{1}, X)$ where $\mathbb{1}$ is the terminal locale with frame of opens $\Omega(\mathbb{1}) = \Omega := P(\{*\})$ given by the set of truth values, i.e. the power set of a one-point set. We have that the map $$ \Hom(\mathbb{1}, \mathbb{S}) \to \Omega(\mathbb{1}) = \Omega, \qquad p \mapsto p^*(U_0) $$ is bijective.

Since the frame of $\mathbb{S}$ was defined as the frame of open subsets of a topology on $\{0, 1\}$, the locale $\mathbb{S}$ is by construction spacial. In particular, it is completely determined by the sober topological space given by the set of points $\Hom(\mathbb{1}, \mathbb{S}) \cong \Omega$ together with the induced topology. I am not 100% sure what this topology is, it is claimed here that it is the topology of upper sets, while it is claimed here that it is the Scott topology. I might think about this question at some point.