Constructive subtleties about the Sierpinski Space

Question

Classically, $\newcommand{\Hom}{\mathrm{Hom}}\newcommand{\IS}{\mathbb{S}}\newcommand{\P}{\mathscr{P}}\newcommand{\upset}{\!\uparrow}$ the Sierpinski space is given by the set $\mathbb{S} = \{0, 1\}$ together with the topology of upwards-closed sets, i.e. $\Omega(\mathbb{S}) = \{\{\}, \{1\}, \{0, 1\}\}$. Together with the open subset $U_0 := \{1\} \in \Omega(\mathbb{S})$ it satisfies the following universal property: for every topological space $T$, the canonical map $$ \Hom(T, \IS) \to \Omega(T), \qquad \chi \mapsto \chi^*(U_0) \label{1}\tag{$*$} $$ is a bijection.

From now on, let's work without the law of excluded middle. I know for abstract reasons (if necessary, I can elaborate) that there still exists a topological space $\mathbb{S}$ together with an open subset $U_0 \in \Omega(\mathbb{S})$ such that the map above is a bijection. Moreover, the set of points is given by the set of truth values $\Omega$ (the set of all subsets of a one-point set, i.e. $\Omega = \P(\{*\})$). Indeed, letting $T$ be the discrete topological space with underlying set $\{*\}$ and topology $\P(\{*\})$, the bijectivity of \eqref{1} tells us that the set of points of $\mathbb{S}$ is in bijection with $\Omega$.

My question: what is the topology $\tau$ on $\Omega$ such that $\mathbb{S} = (\Omega, \tau)$ satisfies the universal property above?

It is claimed on the nlab that $\tau$ is the Alexandrov topology of upwards-closed sets of $\Omega$, together with $U_0 = \{\top\}$. I'm trying to prove that it satisfies the universal property. It is easy to see that if there exists an inverse map to \eqref{1}, it must be given by $$ \Omega(T) \to \Hom(T, \mathbb{S}), \qquad U \mapsto \chi_U, $$ where $\chi_U \colon T \to \Omega$ is the characteristic function satisfying $\chi_U(t) = \top \iff t \in U$. (If we think of $\Omega$ as $\P(\{*\})$, then $\chi_U$ is given by $\chi_U(t) = \{x \in \{*\} \mid t \in U\}$.) What must be shown however is that the function $\chi_U$ is continuous with respect to the Alexandrov topology on $\Omega$ whenever $U$ is open, and this is where I struggle. The Alexandrov topology has a basis consisting of the principal open sets $\omega\upset = \{\omega' \in \Omega \mid \omega \le \omega'\}$ for $\omega \in \Omega$, so it is enough to show that $\chi_U^{-1}(\omega\upset)$ is open in $T$ for all $\omega \in \Omega$.

Thinking of $\omega$ as a subset of $\{*\}$, it is easy to see that $\chi_U^{-1}(\omega\upset) = \bigcap_{x \in \omega} U$. This is an intersection of copies of $U$ over a subsingleton, but unfortunately the axioms of a topological space only tell us that finite intersections are open, so I'm not sure how to continue from here.

Answer 1

Much of the following is from Tom de Jong helping me think it through.

Consider the space $\mathbb S$ whose point set is $\Omega$ the propositions, and the open sets are of the form $$\{\varphi \mid p \lor (q \land \varphi)\}$$ where $p$ and $q$ range over all propositions such that $p$ implies $q$. Denote this as $O_{p\le q}$. We first prove this is a topology. (Not a topological base, but an entire topology.)

For finite intersections we have $$\bigcap_i O_{p_i\le q_i} = \left\{ \varphi ~\middle|~ \bigwedge_i p_i \lor (q_i \land \varphi) \right\}.$$ We use distributivity to expand the finite conjunction. This gives you the disjunction of $\bigwedge_i p_i$ with a bunch of things that contain $\varphi$. We can use distributivity again to pull $\varphi$ out. This puts the intersection in the desired form.

For arbitrary unions, we similarly have $\exists i. p_i \lor (q_i\land \varphi)$ on the right. We can push the existential inside the disjunction, and use infinite distributivity to pull $\land \varphi$ out. So we indeed have a topology.

Let $\mathbb S'$ be the topological space over $\Omega$ whose opens are upwards closed subsets. Before we show $\mathbb S$ is the true Sierpinski space, let's first prove $\mathbb S'$ is the false prophet. Consider the open set $O_{\bot\le\top} = \{\top\}$ in $\mathbb S$. If $\mathbb S'$ had the universal property, then there will be a map $f : \mathbb S\to\mathbb S'$. It must be the identity map on the underlying set, because by construction $f^{-1} \{\top\} = \{\top\}$ and so $f(\varphi) = \top \iff \varphi = \top$. Since $f$ is continuous, we consider the open $\{\varphi \mid \neg\neg\varphi\}$ in $\mathbb S'$ (obviously upwards closed). The preimage, which is the same set, must also be open in $\mathbb S$, i.e. it can be expressed as some $O_{p\le q}$. Since $\top$ is in this subset and $\bot$ is not, we are forced to set $p = \top$ and $q =\top$, which means this subset is the entire set, i.e. double negation elimination is true.

Now for the proof that $\mathbb S$ satisfies the universal property. Suppose $T$ is a topological space and $U$ its open set, consider the mapping $f(t) = (t \in U)$ from $T$ to $\mathbb S$. We just need to show it's continuous. Now the preimage of $O_{p\le q}$ is $$\{ t \mid p \lor (q \land t \in U) \}.$$ We know $\{ t \mid p\}$ is open, since it's the union $\bigcup_p T$. Similarly $\{t \mid q\}$ is also open. Simply taking unions and intersections proves our claim.

By the way, this is kind of similar to Phoa's principle, which says for some particular distributive lattice $L$, every endofunction is monotonous and of the form $f(t) = p \lor (q \land t)$.

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There is yet another space $\mathbb{S}''$ given by topologizing $\Omega$ by the Scott topology. A subset $U$ is open iff it is upwards closed and if a directed disjunction lands in $U$, then one of the elements lands in $U$ already. We shall prove that $\mathbb{S}'' = \mathbb{S}$. An open set of $\mathbb{S}$ is obviously open in $\mathbb{S}''$. So we take an open $U \subseteq \mathbb{S}''$ and show that it must be of the form $O_{p \le q}$.

Consider the directed set given by $J = \{n \in \{0, 1\} \mid (n = 0) \lor \varphi\}$. Let the family $p_{i \in J}$ be given by $p_0 = \bot$ and $p_1 = \top$. Then $\bigvee_{i \in J} p_i = \varphi$. Therefore $$ \varphi \in U \iff (\bot \in U) \lor (\varphi \land (\top \in U)) $$ Hence $U = O_{(\bot \in U) \le (\top \in U)}$. This shows that $\mathbb S'' = \mathbb S$.