hcju/stacksqa huggingface dataset:
Assume (1). Choose a diagram $i \mapsto X_i$ of finite discrete spaces such that $X = \lim X_i$. As each $X_i$ is Hausdorff and quasi-compact we find that $X$ is quasi-compact by Lemma \ref{lemma-inverse-limit-quasi-compact}. If $x, x' \in X$ are distinct points, then $x$ and $x'$ map to distinct points in some $X_i$. Hence $x$ and $x'$ have disjoint open neighbourhoods, i.e., $X$ is Hausdorff. In exactly the same way we see that $X$ is totally disconnected. \medskip\noindent Assume (2). Let $\mathcal{I}$ be the set of finite disjoint union decompositions $X = \coprod_{i \in I} U_i$ with $U_i$ nonempty open (and closed) for all $i \in I$. For each $I \in \mathcal{I}$ there is a continuous map $X \to I$ sending a point of $U_i$ to $i$. We define a partial ordering: $I \leq I'$ for $I, I' \in \mathcal{I}$ if and only if the covering corresponding to $I'$ refines the covering corresponding to $I$. In this case we obtain a canonical map $I' \to I$. In other words we obtain an inverse system of finite discrete spaces over $\mathcal{I}$. The maps $X \to I$ fit together and we obtain a continuous map $$ X \longrightarrow \lim_{I \in \mathcal{I}} I $$ We claim this map is a homeomorphism, which finishes the proof. (The final assertion follows too as the partially ordered set $\mathcal{I}$ is directed: given two disjoint union decompositions of $X$ we can find a third refining both.) Namely, the map is injective as $X$ is totally disconnected and hence $\{x\}$ is the intersection of all open and closed subsets of $X$ containing $x$ (Lemma \ref{lemma-connected-component-intersection-compact-Hausdorff}) and  the map is surjective by Lemma \ref{lemma-intersection-closed-in-quasi-compact}. By Lemma \ref{lemma-bijective-map} the map is a homeomorphism.
Let $i \mapsto X_i$ be a diagram of profinite spaces over the index category $\mathcal{I}$. Let us use the characterization of profinite spaces in Lemma \ref{lemma-profinite}. In particular each $X_i$ is Hausdorff, quasi-compact, and totally disconnected. By Lemma \ref{lemma-limits} the limit $X = \lim X_i$ exists. By Lemma \ref{lemma-inverse-limit-quasi-compact} the limit $X$ is quasi-compact. Let $x, x' \in X$ be distinct points. Then there exists an $i$ such that $x$ and $x'$ have distinct images $x_i$ and $x'_i$ in $X_i$ under the projection $X \to X_i$. Then $x_i$ and $x'_i$ have disjoint open neighbourhoods in $X_i$. Taking the inverse images of these opens we conclude that $X$ is Hausdorff. Similarly, $x_i$ and $x'_i$ are in distinct connected components of $X_i$ whence necessarily $x$ and $x'$ must be in distinct connected components of $X$. Hence $X$ is totally disconnected. This finishes the proof.
Write $X = \lim X_i$ as a limit of an inverse system of finite discrete spaces over a directed set $I$ (Lemma \ref{lemma-profinite}). Denote $f_i : X \to X_i$ the projection. For every point $x = (x_i) \in X$ a fundamental system of open neighbourhoods is the collection $f_i^{-1}(\{x_i\})$. Thus, as $X$ is quasi-compact, we may assume we have an open covering $$ X = f_{i_1}^{-1}(\{x_{i_1}\}) \cup \ldots \cup f_{i_n}^{-1}(\{x_{i_n}\}) $$ Choose $i \in I$ with $i \geq i_j$ for $j = 1, \ldots, n$ (this is possible as $I$ is a directed set). Then we see that the covering $$ X = \coprod\nolimits_{t \in X_i} f_i^{-1}(\{t\}) $$ refines the given covering and is of the desired form.
We will use Lemma \ref{lemma-profinite} to prove this. Since $\pi_0(X)$ is the image of a quasi-compact space it is quasi-compact (Lemma \ref{lemma-image-quasi-compact}). It is totally disconnected by construction (Lemma \ref{lemma-space-connected-components}). Let $C, D \subset X$ be distinct connected components of $X$. Write $C = \bigcap U_\alpha$ as the intersection of the open and closed subsets of $X$ containing $C$. Any finite intersection of $U_\alpha$'s is another. Since $\bigcap U_\alpha \cap D = \emptyset$ we conclude that $U_\alpha \cap D = \emptyset$ for some $\alpha$ (use Lemmas \ref{lemma-connected-components}, \ref{lemma-closed-in-quasi-compact} and \ref{lemma-intersection-closed-in-quasi-compact}) Since $U_\alpha$ is open and closed, it is the union of the connected components it contains, i.e., $U_\alpha$ is the inverse image of some open and closed subset $V_\alpha \subset \pi_0(X)$. This proves that the points corresponding to $C$ and $D$ are contained in disjoint open subsets, i.e., $\pi_0(X)$ is Hausdorff.
Let $X'$ and $Y'$ denote $X$ and $Y$ endowed with the constructible topology which are quasi-compact Hausdorff spaces by Lemma \ref{lemma-constructible-hausdorff-quasi-compact}. Part (1) says $X' \to Y'$ is continuous and follows immediately from the definitions. Part (3) follows as $f(X')$ is a quasi-compact subset of the Hausdorff space $Y'$, see Lemma \ref{lemma-quasi-compact-in-Hausdorff}. We have a commutative diagram $$ \xymatrix{ X' \ar[r] \ar[d] & X \ar[d] \\ Y' \ar[r] & Y } $$ of continuous maps of topological spaces. Since $Y'$ is Hausdorff we see that the fibres $X'_y$ are closed in $X'$. As $X'$ is quasi-compact we see that $X'_y$ is quasi-compact (Lemma \ref{lemma-closed-in-quasi-compact}). As $X'_y \to X_y$ is a surjective continuous map we conclude that $X_y$ is quasi-compact (Lemma \ref{lemma-image-quasi-compact}).
The only if part of this is Lemma \ref{lemma-fibres-spectral-map-quasi-compact}. Assume $f$ is continuous in the constructible topology. Let $V \subset Y$ be quasi-compact open. Then $V$ is open and closed in the constructible topology. Hence $f^{-1}(V)$ is open and closed in the constructible topology. Hence $f^{-1}(V)$ is quasi-compact in the constructible topology as $X$ is quasi-compact in the constructible topology by Lemma \ref{lemma-constructible-hausdorff-quasi-compact}. Since the identity $f^{-1}(V) \to f^{-1}(V)$ is surjective and continuous from the constructible topology to the usual topology, we conclude that $f^{-1}(V)$ is quasi-compact in the topology of $X$ by Lemma \ref{lemma-image-quasi-compact}. This finishes the proof.
Let $Z \subset E$ be a closed irreducible subset. Let $\eta$ be the generic point of the closure $\overline{Z}$ of $Z$ in $X$. To prove that $E$ is sober, we show that $\eta \in E$. If not, then since $E$ is closed in the constructible topology, there exists a constructible subset $F \subset X$ such that $\eta \in F$ and $F \cap E = \emptyset$. By Lemma \ref{lemma-generic-point-in-constructible} this implies $F \cap \overline{Z}$ contains a nonempty open subset of $\overline{Z}$. But this is impossible as $\overline{Z}$ is the closure of $Z$ and $Z \cap F = \emptyset$. \medskip\noindent Since $E$ is closed in the constructible topology, it is quasi-compact in the constructible topology (Lemmas \ref{lemma-closed-in-quasi-compact} and \ref{lemma-constructible-hausdorff-quasi-compact}). Hence a fortiori it is quasi-compact in the topology coming from $X$. If $U \subset X$ is a quasi-compact open, then $E \cap U$ is closed in the constructible topology, hence quasi-compact (as seen above). It follows that the quasi-compact open subsets of $E$ are the intersections $E \cap U$ with $U$ quasi-compact open in $X$. These form a basis for the topology. Finally, given two $U, U' \subset X$ quasi-compact opens, the intersection $(E \cap U) \cap (E \cap U') = E \cap (U \cap U')$ and $U \cap U'$ is quasi-compact as $X$ is spectral. This finishes the proof.
Proof of (1). Let $x \in \overline{E}$. Let $\{U_i\}$ be the set of quasi-compact open neighbourhoods of $x$. A finite intersection of the $U_i$ is another one. The intersection $U_i \cap E$ is nonempty for all $i$. Since the subsets $U_i \cap E$ are closed in the constructible topology we see that $\bigcap (U_i \cap E)$ is nonempty by Lemma \ref{lemma-constructible-hausdorff-quasi-compact} and Lemma \ref{lemma-intersection-closed-in-quasi-compact}. Since $X$ is a sober space and $\{U_i\}$ is a fundamental system of open neighbourhoods of $x$, we see that $\bigcap U_i$ is the set of generalizations of $x$. Thus $x$ is a specialization of a point of $E$. \medskip\noindent Part (2) is immediate from (1). \medskip\noindent Proof of (3). Assume $E'$ is as in (3). The complement of $E'$ is closed in the constructible topology (Lemma \ref{lemma-constructible}) and closed under specialization (Lemma \ref{lemma-open-closed-specialization}). Hence the complement is closed by (2), i.e., $E'$ is open.
Let $\{U_i\}$ be the set of quasi-compact open neighbourhoods of $x$. A finite intersection of the $U_i$ is another one. Let $\{V_j\}$ be the set of quasi-compact open neighbourhoods of $y$. A finite intersection of the $V_j$ is another one. If $U_i \cap V_j$ is empty for some $i, j$ we are done. If not, then the intersection $U_i \cap V_j$ is nonempty for all $i$ and $j$. The sets $U_i \cap V_j$ are closed in the constructible topology on $X$. By Lemma \ref{lemma-constructible-hausdorff-quasi-compact} we see that $\bigcap (U_i \cap V_j)$ is nonempty (Lemma \ref{lemma-intersection-closed-in-quasi-compact}). Since $X$ is a sober space and $\{U_i\}$ is a fundamental system of open neighbourhoods of $x$, we see that $\bigcap U_i$ is the set of generalizations of $x$. Similarly, $\bigcap V_j$ is the set of generalizations of $y$. Thus any element of $\bigcap (U_i \cap V_j)$ specializes to both $x$ and $y$.
Lemma \ref{lemma-profinite} shows the implication (1) $\Rightarrow$ (3). Irreducible components are closed, so if $X$ is totally disconnected, then every point is closed. So (3) implies (6). The equivalence of (6) and (5) is immediate, and (6) $\Leftrightarrow$ (7) holds because $X$ is sober. Assume (5). Then all constructible subsets of $X$ are closed (Lemma \ref{lemma-constructible-stable-specialization-closed}), in particular all quasi-compact opens are closed. So (5) implies (4). Since $X$ is sober, for any two points there is a quasi-compact open containing exactly one of them, hence (4) implies (2). Parts (4) and (8) are equivalent by the definition of the constructible topology. It remains to prove (2) implies (1). Suppose $X$ is Hausdorff. Every quasi-compact open is also closed (Lemma \ref{lemma-quasi-compact-in-Hausdorff}). This implies $X$ is totally disconnected. Hence it is profinite, by Lemma \ref{lemma-profinite}.
Since $f$ is spectral it defines a continuous map between $X$ and $Y$ in the constructible topology. By Lemmas \ref{lemma-constructible-hausdorff-quasi-compact} and \ref{lemma-bijective-map} it follows that $X \to Y$ is a homeomorphism in the constructible topology. Let $U \subset X$ be quasi-compact open. Then $f(U)$ is constructible in $Y$. Let $y \in Y$ specialize to a point in $f(U)$. By the last assumption we see that $f^{-1}(y)$ specializes to a point of $U$. Hence $f^{-1}(y) \in U$. Thus $y \in f(U)$. It follows that $f(U)$ is open, see Lemma \ref{lemma-constructible-stable-specialization-closed}. Whence $f$ is a homeomorphism. To prove the lemma in case specializations lift along $f$ one shows instead that $f(Z)$ is closed if $X \setminus Z$ is a quasi-compact open of $X$.
Let $I$ be a directed set. Let $X_i$ be an inverse system of finite sober spaces over $I$. Let $X = \lim X_i$ which exists by Lemma \ref{lemma-limits}. As a set $X = \lim X_i$. Denote $p_i : X \to X_i$ the projection. Because $I$ is directed we may apply Lemma \ref{lemma-describe-limits}. A basis for the topology is given by the opens $p_i^{-1}(U_i)$ for $U_i \subset X_i$ open. Since an open covering of $p_i^{-1}(U_i)$ is in particular an open covering in the profinite topology, we conclude that $p_i^{-1}(U_i)$ is quasi-compact. Given $U_i \subset X_i$ and $U_j \subset X_j$, then $p_i^{-1}(U_i) \cap p_j^{-1}(U_j) = p_k^{-1}(U_k)$ for some $k \geq i, j$ and open $U_k \subset X_k$. Finally, if $Z \subset X$ is irreducible and closed, then $p_i(Z) \subset X_i$ is irreducible and therefore has a unique generic point $\xi_i$ (because $X_i$ is a finite sober topological space). Then $\xi = \lim \xi_i$ is a generic point of $Z$ (it is a point of $Z$ as $Z$ is closed). This finishes the proof.
Write $W = \{0, 1\}$ where $0$ is a specialization of $1$ but not vice versa. Let $I$ be a set. The space $\prod_{i \in I} W$ is spectral by Lemma \ref{lemma-directed-inverse-limit-finite-sober-spectral-spaces}. Thus we see that a subspace of $\prod_{i \in I} W$ closed in the constructible topology is a spectral space by Lemma \ref{lemma-spectral-sub}. \medskip\noindent For the converse, let $X$ be a spectral space. Let $U \subset X$ be a quasi-compact open. Consider the continuous map $$ f_U : X \longrightarrow W $$ which maps every point in $U$ to $1$ and every point in $X \setminus U$ to $0$. Taking the product of these maps we obtain a continuous map $$ f = \prod f_U : X \longrightarrow \prod\nolimits_U W $$ By construction the map $f : X \to Y$ is spectral. By Lemma \ref{lemma-fibres-spectral-map-quasi-compact} the image of $f$ is closed in the constructible topology. If $x', x \in X$ are distinct, then since $X$ is sober either $x'$ is not a specialization of $x$ or conversely. In either case (as the quasi-compact opens form a basis for the topology of $X$) there exists a quasi-compact open $U \subset X$ such that $f_U(x') \not = f_U(x)$. Thus $f$ is injective. Let $Y = f(X)$ endowed with the induced topology. Let $y' \leadsto y$ be a specialization in $Y$ and say $f(x') = y'$ and $f(x) = y$. Arguing as above we see that $x' \leadsto x$, since otherwise there is a $U$ such that $x \in U$ and $x' \not \in U$, which would imply $f_U(x') \not \leadsto f_U(x)$. We conclude that $f : X \to Y$ is a homeomorphism by Lemma \ref{lemma-spectral-bijective}.
One direction is given by Lemma \ref{lemma-directed-inverse-limit-finite-sober-spectral-spaces}. For the converse, assume $X$ is spectral. Then we may assume $X \subset \prod_{i \in I} W$ is a subset closed in the constructible topology where $W = \{0, 1\}$ as in Lemma \ref{lemma-spectral-closed-in-product-two-point-space}. We can write $$ \prod\nolimits_{i \in I} W = \lim_{J \subset I\text{ finite }} \prod\nolimits_{j \in J} W $$ as a cofiltered limit. For each $J$, let $X_J \subset \prod_{j \in J} W$ be the image of $X$. Then we see that $X = \lim X_J$ as sets because $X$ is closed in the product with the constructible topology (detail omitted). A formal argument (omitted) on limits shows that $X = \lim X_J$ as topological spaces.
The limit $Z = \lim Z_i$ exists by Lemma \ref{lemma-limits}. Denote $X'_i$ the space $X_i$ endowed with the constructible topology and $Z'_i$ the corresponding subspace of $X'_i$. Let $a : j \to i$ in $\mathcal{I}$ be a morphism. As $f_a$ is spectral it defines a continuous map $f_a : X'_j \to X'_i$. Thus $f_a|_{Z_j} : Z'_j \to Z'_i$ is a continuous map of quasi-compact Hausdorff spaces (by Lemmas \ref{lemma-constructible-hausdorff-quasi-compact} and \ref{lemma-closed-in-quasi-compact}). Thus $Z' = \lim Z_i$ is quasi-compact by Lemma \ref{lemma-inverse-limit-quasi-compact}. The maps $Z'_i \to Z_i$ are continuous, hence $Z' \to Z$ is continuous and a bijection on underlying sets. Hence $Z$ is quasi-compact as the image of the surjective continuous map $Z' \to Z$ (Lemma \ref{lemma-image-quasi-compact}).
Denote $X'_i$ the space $X_i$ endowed with the constructible topology and $Z'_i$ the corresponding subspace of $X'_i$. Let $a : j \to i$ in $\mathcal{I}$ be a morphism. As $f_a$ is spectral it defines a continuous map $f_a : X'_j \to X'_i$. Thus $f_a|_{Z_j} : Z'_j \to Z'_i$ is a continuous map of quasi-compact Hausdorff spaces (by Lemmas \ref{lemma-constructible-hausdorff-quasi-compact} and \ref{lemma-closed-in-quasi-compact}). By Lemma \ref{lemma-nonempty-limit} the space $\lim Z'_i$ is nonempty. Since $\lim Z'_i = \lim Z_i$ as sets we conclude.
Observe that $$ p_i^{-1}(E) \setminus p_i^{-1}(F) = \lim_{a : j \to i} f_a^{-1}(E) \setminus f_a^{-1}(F) $$ Since $f_a$ is a spectral map, it is continuous in the constructible topology hence the set $f_a^{-1}(E) \setminus f_a^{-1}(F)$ is closed in the constructible topology. Hence Lemma \ref{lemma-inverse-limit-spectral-spaces-nonempty} applies to show that the LHS is nonempty if and only if each of the spaces of the RHS is nonempty.
Assume $E$ is a quasi-compact open of $X$. By Lemma \ref{lemma-describe-limits} we can write $E = p_i^{-1}(U_i)$ for some $i$ and some open $U_i \subset X_i$. Write $U_i = \bigcup U_{i, \alpha}$ as a union of quasi-compact opens. As $E$ is quasi-compact we can find $\alpha_1, \ldots, \alpha_n$ such that $E = p_i^{-1}(U_{i, \alpha_1} \cup \ldots \cup U_{i, \alpha_n})$. Hence $E_i = U_{i, \alpha_1} \cup \ldots \cup U_{i, \alpha_n}$ works. \medskip\noindent Assume $E$ is a constructible closed subset. Then $E^c$ is quasi-compact open. So $E^c = p_i^{-1}(F_i)$ for some $i$ and quasi-compact open $F_i \subset X_i$ by the result of the previous paragraph. Then $E = p_i^{-1}(F_i^c)$ as desired. \medskip\noindent If $E$ is general we can write $E = \bigcup_{l = 1, \ldots, n} U_l \cap Z_l$ with $U_l$ constructible open and $Z_l$ constructible closed. By the result of the previous paragraphs we may write $U_l = p_{i_l}^{-1}(U_{l, i_l})$ and $Z_l = p_{j_l}^{-1}(Z_{l, j_l})$ with $U_{l, i_l} \subset X_{i_l}$ constructible open and $Z_{l, j_l} \subset X_{j_l}$ constructible closed. As $\mathcal{I}$ is cofiltered we may choose an object $k$ of $\mathcal{I}$ and morphism $a_l : k \to i_l$ and $b_l : k \to j_l$. Then taking $E_k = \bigcup_{l = 1, \ldots, n} f_{a_l}^{-1}(U_{l, i_l}) \cap f_{b_l}^{-1}(Z_{l, j_l})$ we obtain a constructible subset of $X_k$ whose inverse image in $X$ is $E$.
The limit $X = \lim X_i$ exists (Lemma \ref{lemma-limits}) and is quasi-compact by Lemma \ref{lemma-inverse-limit-spectral-spaces-quasi-compact}. \medskip\noindent Denote $p_i : X \to X_i$ the projection. Because $\mathcal{I}$ is cofiltered we can apply Lemma \ref{lemma-describe-limits}. Hence a basis for the topology on $X$ is given by the opens $p_i^{-1}(U_i)$ for $U_i \subset X_i$ open. Since a basis for the topology of $X_i$ is given by the quasi-compact open, we conclude that a basis for the topology on $X$ is given by $p_i^{-1}(U_i)$ with $U_i \subset X_i$ quasi-compact open. A formal argument shows that $$ p_i^{-1}(U_i) = \lim_{a : j \to i} f_a^{-1}(U_i) $$ as topological spaces. Since each $f_a$ is spectral the sets $f_a^{-1}(U_i)$ are closed in the constructible topology of $X_j$ and hence $p_i^{-1}(U_i)$ is quasi-compact by Lemma \ref{lemma-inverse-limit-spectral-spaces-quasi-compact}. Thus $X$ has a basis for the topology consisting of quasi-compact opens. \medskip\noindent Any quasi-compact open $U$ of $X$ is of the form $U = p_i^{-1}(U_i)$ for some $i$ and some quasi-compact open $U_i \subset X_i$ (see Lemma \ref{lemma-inverse-limit-spectral-spaces-constructible}). Given $U_i \subset X_i$ and $U_j \subset X_j$ quasi-compact open, then $p_i^{-1}(U_i) \cap p_j^{-1}(U_j) = p_k^{-1}(U_k)$ for some $k$ and quasi-compact open $U_k \subset X_k$. Namely, choose $k$ and morphisms $k \to i$ and $k \to j$ and let $U_k$ be the intersection of the pullbacks of $U_i$ and $U_j$ to $X_k$. Thus we see that the intersection of two quasi-compact opens of $X$ is quasi-compact open. \medskip\noindent Finally, let $Z \subset X$ be irreducible and closed. Then $p_i(Z) \subset X_i$ is irreducible and therefore $Z_i = \overline{p_i(Z)}$ has a unique generic point $\xi_i$ (because $X_i$ is a spectral space). Then $f_a(\xi_j) = \xi_i$ for $a : j \to i$ in $\mathcal{I}$ because $\overline{f_a(Z_j)} = Z_i$. Hence $\xi = \lim \xi_i$ is a point of $X$. Claim: $\xi \in Z$. Namely, if not we can find a quasi-compact open containing $\xi$ disjoint from $Z$. This would be of the form $p_i^{-1}(U_i)$ for some $i$ and quasi-compact open $U_i \subset X_i$. Then $\xi_i \in U_i$ but $p_i(Z) \cap U_i = \emptyset$ which contradicts $\xi_i \in \overline{p_i(Z)}$. So $\xi \in Z$ and hence $\overline{\{\xi\}} \subset Z$. Conversely, every $z \in Z$ is in the closure of $\xi$. Namely, given a quasi-compact open neighbourhood $U$ of $z$ we write $U = p_i^{-1}(U_i)$ for some $i$ and quasi-compact open $U_i \subset X_i$. We see that $p_i(z) \in U_i$ hence $\xi_i \in U_i$ hence $\xi \in U$. Thus $\xi$ is a generic point of $Z$. We omit the proof that $\xi$ is the unique generic point of $Z$ (hint: show that a second generic point has to be equal to $\xi$ by showing that it has to map to $\xi_i$ in $X_i$ since by spectrality of $X_i$ the irreducible $Z_i$ has a unique generic point). This finishes the proof.
Part (1) is a special case of Lemma \ref{lemma-inverse-limit-spectral-spaces-constructible}. Part (2) is a special case of Lemma \ref{lemma-inverse-limit-spectral-spaces-equal} as quasi-compact opens are both open and closed in the constructible topology. Parts (3) and (4) follow formally from (1) and (2) and the fact that taking inverse images of subsets commutes with taking unions and intersections.
Let $W \subset X$ satisfy (1). Then $W$ is closed in the constructible topology, hence quasi-compact in the constructible topology (by Lemmas \ref{lemma-constructible-hausdorff-quasi-compact} and \ref{lemma-closed-in-quasi-compact}), hence quasi-compact in the topology of $X$ (because opens in $X$ are open in the constructible topology). Thus (2) holds. \medskip\noindent It is clear that (2) implies (3) by taking $E = W$. \medskip\noindent Let $X$ be a spectral space and let $E \subset W$ be as in (3). Since every point of $W$ specializes to a point of $E$ we see that an open of $W$ which contains $E$ is equal to $W$. Hence since $E$ is quasi-compact, so is $W$. If $x \in X$, $x \not \in W$, then $Z = \overline{\{x\}}$ is disjoint from $W$. Since $W$ is quasi-compact we can find a quasi-compact open $U$ with $W \subset U$ and $U \cap Z = \emptyset$. We conclude that (4) holds. \medskip\noindent If $W = \bigcap_{j \in J} U_j$ then setting $I$ equal to the set of finite subsets of $J$ and $U_i = U_{j_1} \cap \ldots \cap U_{j_r}$ for $i = \{j_1, \ldots, j_r\}$ shows that (4) implies (5). It is immediate that (5) implies (1). \medskip\noindent Let $I$ and $U_i$ be as in (5). Since $W = \bigcap U_i$ we have $W = \lim U_i$ by the universal property of limits. Then $W$ is a spectral space by Lemma \ref{lemma-directed-inverse-limit-spectral-spaces}. Let $U \subset X$ be an open neighbourhood of $W$. Then $E_i = U_i \cap (X \setminus U)$ is a family of constructible subsets of the spectral space $Z = X \setminus U$ with empty intersection. Using that the spectral topology on $Z$ is quasi-compact (Lemma \ref{lemma-constructible-hausdorff-quasi-compact}) we conclude from Lemma \ref{lemma-intersection-closed-in-quasi-compact} that $E_i = \emptyset$ for some $i$.
Since $E$ is constructible, it is quasi-compact and hence Lemma \ref{lemma-make-spectral-space} applies to $W$. If $E$ is constructible, then $E$ is constructible in $U_i$ for all $i \in I$. Hence $U_i \setminus E$ is spectral by Lemma \ref{lemma-spectral-sub}. Since $W \setminus E = \bigcap (U_i \setminus E)$ we have $W \setminus E = \lim U_i \setminus E$ by the universal property of limits. Then $W \setminus E$ is a spectral space by Lemma \ref{lemma-directed-inverse-limit-spectral-spaces}.