hcju/stacksqa huggingface dataset:
Since $Z$ is quasi-compact, there is a reduction to the case where $I$ is finite (details omitted). Because $X$ is locally quasi-compact and $Z$ is quasi-compact, we can find a neighbourhood $Z \subset E$ which is quasi-compact, i.e., $E$ is quasi-compact and contains an open neighbourhood of $Z$ in $X$. If we prove the result after replacing $X$ by $E$, then the result follows. Hence we may assume $X$ is quasi-compact. \medskip\noindent We prove the result in case $I$ is finite and $X$ is quasi-compact by induction on $p$. The base case is $p = 0$. In this case we have $X = (X \setminus Z) \cup \bigcup W_i$. By Lemma \ref{lemma-relatively-compact-refinement} we can find a covering $X = V \cup \bigcup V_i$ by opens $V_i \subset W_i$ and $V \subset X \setminus Z$ with $\overline{V_i} \subset W_i$ for all $i$. Then we see that we obtain a solution of the problem posed by the lemma. \medskip\noindent Induction step. Assume the lemma proven for $p - 1$. Set $W_{j_0 \ldots j_{p - 1}}$ equal to the intersection of all $W_{i_0 \ldots i_p}$ with $\{j_0, \ldots, j_{p - 1}\} = \{i_0, \ldots, i_p\}$ (equality of sets). By induction there exists a solution for these opens, say $V_i \subset U_i$. It follows from our choice of $W_{j_0 \ldots j_{p - 1}}$ that we have $V_{i_0} \cap \ldots \cap V_{i_p} \subset W_{i_0 \ldots i_p}$ for all $(p + 1)$-tuples $i_0, \ldots, i_p$ where $i_a = i_b$ for some $0 \leq a < b \leq p$. Thus we only need to modify our choice of $V_i$ if $V_{i_0} \cap \ldots \cap V_{i_p} \not \subset W_{i_0 \ldots i_p}$ for some $(p + 1)$-tuple $i_0, \ldots, i_p$ with pairwise distinct elements. In this case we have $$ T = \overline{V_{i_0} \cap \ldots \cap V_{i_p} \setminus W_{i_0 \ldots i_p}} \subset  \overline{V_{i_0}} \cap \ldots \cap \overline{V_{i_p}} \setminus W_{i_0 \ldots i_p} $$ is a closed subset of $X$ contained in $U_{i_0} \cap \ldots \cap U_{i_p}$ not meeting $Z$. Hence we can replace $V_{i_0}$ by $V_{i_0} \setminus T$ to ``fix'' the problem. After repeating this finitely many times for each of the problem tuples, the lemma is proven.
Since $Z$ is quasi-compact, there is a reduction to the case where $I$ is finite (details omitted). We prove the result in case $I$ is finite by induction on $p$. \medskip\noindent The base case is $p = 0$. For $z \in Z \cap U_i$ and $z' \in Z \setminus U_i$ there exist disjoint opens $z \in V_{z, z'}$ and $z' \in W_{z, z'}$ of $X$. Since $Z \setminus U_i$ is quasi-compact (Lemma \ref{lemma-closed-in-quasi-compact}), we can choose a finite nunber $z'_1, \ldots, z'_r$ such that $Z \setminus U_i \subset W_{z, z'_1} \cup \ldots \cup W_{z, z'_r}$. Then we see that $V_z = V_{z, z'_1} \cap \ldots \cap V_{z, z'_r} \cap U_i$ is an open neighbourhood of $z$ contained in $U_i$ with the property that $\overline{V_z} \cap Z \subset U_i$. Since $z$ and $i$ where arbitrary and since $Z$ is quasi-compact we can find a finite list $z_1, i_1, \ldots, z_t, i_t$ and opens $V_{z_j} \subset U_{i_j}$ with $\overline{V_{z_j}} \cap Z \subset U_{i_j}$ and $Z \subset \bigcup V_{z_j}$. Then we can set $V_i = W_i \cap (\bigcup_{j : i = i_j} V_{z_j})$ to solve the problem for $p = 0$. \medskip\noindent Induction step. Assume the lemma proven for $p - 1$. Set $W_{j_0 \ldots j_{p - 1}}$ equal to the intersection of all $W_{i_0 \ldots i_p}$ with $\{j_0, \ldots, j_{p - 1}\} = \{i_0, \ldots, i_p\}$ (equality of sets). By induction there exists a solution for these opens, say $V_i \subset U_i$. It follows from our choice of $W_{j_0 \ldots j_{p - 1}}$ that we have $V_{i_0} \cap \ldots \cap V_{i_p} \subset W_{i_0 \ldots i_p}$ for all $(p + 1)$-tuples $i_0, \ldots, i_p$ where $i_a = i_b$ for some $0 \leq a < b \leq p$. Thus we only need to modify our choice of $V_i$ if $V_{i_0} \cap \ldots \cap V_{i_p} \not \subset W_{i_0 \ldots i_p}$ for some $(p + 1)$-tuple $i_0, \ldots, i_p$ with pairwise distinct elements. In this case we have $$ T = \overline{V_{i_0} \cap \ldots \cap V_{i_p} \setminus W_{i_0 \ldots i_p}} \subset \overline{V_{i_0}} \cap \ldots \cap \overline{V_{i_p}} \setminus W_{i_0 \ldots i_p} $$ is a closed subset of $X$ not meeting $Z$ by our property (3) of the opens $V_i$. Hence we can replace $V_{i_0}$ by $V_{i_0} \setminus T$ to ``fix'' the problem. After repeating this finitely many times for each of the problem tuples, the lemma is proven.
This follows from the discussion above and Categories, Lemma \ref{categories-lemma-limits-products-equalizers}. It follows from the description above that the forgetful functor commutes with limits. Another way to see this is to use Categories, Lemma \ref{categories-lemma-adjoint-exact} and use that the forgetful functor has a left adjoint, namely the functor which assigns to a set the corresponding discrete topological space.
Recall that $\lim X_i$ is a subspace of $\prod X_i$. By Theorem \ref{theorem-tychonov} this product is quasi-compact. Hence it suffices to show that $\lim X_i$ is a closed subspace of $\prod X_i$ (Lemma \ref{lemma-closed-in-quasi-compact}). If $\varphi : j \to k$ is a morphism of $\mathcal{I}$, then let $\Gamma_\varphi \subset X_j \times X_k$ denote the graph of the corresponding continuous map $X_j \to X_k$. By Lemma \ref{lemma-graph-closed} this graph is closed. It is clear that $\lim X_i$ is the intersection of the closed subsets $$ \Gamma_\varphi \times \prod\nolimits_{l \not = j, k} X_l \subset \prod X_i $$ Thus the result follows.
In the proof of Lemma \ref{lemma-inverse-limit-quasi-compact} we have seen that $X = \lim X_i$ is the intersection of the closed subsets $$ Z_\varphi = \Gamma_\varphi \times \prod\nolimits_{l \not = j, k} X_l $$ inside the quasi-compact space $\prod X_i$ where $\varphi : j \to k$ is a morphism of $\mathcal{I}$ and $\Gamma_\varphi \subset X_j \times X_k$ is the graph of the corresponding morphism $X_j \to X_k$. Hence by Lemma \ref{lemma-intersection-closed-in-quasi-compact} it suffices to show any finite intersection of these subsets is nonempty. Assume $\varphi_t : j_t \to k_t$, $t = 1, \ldots, n$ is a finite collection of morphisms of $\mathcal{I}$. Since $\mathcal{I}$ is cofiltered, we can pick an object $j$ and a morphism $\psi_t : j \to j_t$ for each $t$. For each pair $t, t'$ such that either (a) $j_t = j_{t'}$, or (b) $j_t = k_{t'}$, or (c) $k_t = k_{t'}$ we obtain two morphisms $j \to l$ with $l = j_t$ in case (a), (b) or $l = k_t$ in case (c). Because $\mathcal{I}$ is cofiltered and since there are finitely many pairs $(t, t')$ we may choose a map $j' \to j$ which equalizes these two morphisms for all such pairs $(t, t')$. Pick an element $x \in X_{j'}$ and for each $t$ let $x_{j_t}$, resp.\ $x_{k_t}$ be the image of $x$ under the morphism $X_{j'} \to X_j \to X_{j_t}$, resp.\ $X_{j'} \to X_j \to X_{j_t} \to X_{k_t}$. For any index $l \in \Ob(\mathcal{I})$ which is not equal to $j_t$ or $k_t$ for some $t$ we pick an arbitrary element $x_l \in X_l$ (using the axiom of choice). Then $(x_i)_{i \in \Ob(\mathcal{I})}$ is in the intersection  $$ Z_{\varphi_1} \cap \ldots \cap Z_{\varphi_n} $$ by construction and the proof is complete.
Suppose that $V \subset X$ is retrocompact open in $X$. It suffices to show that $V \cap U$ is retrocompact in $U$ by Lemma \ref{lemma-inverse-images-constructibles}. To show this let $W \subset U$ be open and quasi-compact. Then $W$ is open and quasi-compact in $X$. Hence $V \cap W = V \cap U \cap W$ is quasi-compact as $V$ is retrocompact in $X$.
Suppose that $V, W \subset U$ are retrocompact open in $U$. Then $V, W$ are retrocompact open in $X$ (Lemma \ref{lemma-composition-quasi-compact}). Hence $V \cap (U \setminus W) = V \cap (X \setminus W)$ is constructible in $X$. We conclude since every constructible subset of $U$ is a finite union of subsets of the form $V \cap (U \setminus W)$.
If $E$ is constructible in $X$, then by Lemma \ref{lemma-open-immersion-constructible-inverse-image} we see that $E \cap V_j$ is constructible in $V_j$ for all $j$. Conversely, suppose that $E \cap V_j$ is constructible in $V_j$ for each $j = 1, \ldots, m$. Then $E = \bigcup E \cap V_j$ is a finite union of constructible sets by Lemma \ref{lemma-quasi-compact-open-immersion-constructible-image} and hence constructible.
Suppose that $V \subset X$ is retrocompact open in $X$. It suffices to show that $V \cap Z$ is retrocompact in $Z$ by Lemma \ref{lemma-inverse-images-constructibles}. To show this let $W \subset Z$ be open and quasi-compact. The subset $W' = W \cup (X \setminus Z)$ is quasi-compact, open, and $W = Z \cap W'$. Hence $V \cap Z \cap W = V \cap Z \cap W'$ is a closed subset of the quasi-compact open $V \cap W'$ as $V$ is retrocompact in $X$. Thus $V \cap Z \cap W$ is quasi-compact by Lemma \ref{lemma-closed-in-quasi-compact}.
Suppose that $V \subset X$ is retrocompact open in $X$. It suffices to show that $V \cap T$ is retrocompact in $T$ by Lemma \ref{lemma-inverse-images-constructibles}. To show this let $W \subset T$ be open and quasi-compact. By assumption (2) we can find a quasi-compact open $W' \subset X$ such that $W = T \cap W'$ (details omitted). Hence $V \cap T \cap W = V \cap T \cap W'$ is the intersection of $T$ with  the quasi-compact open $V \cap W'$ as $V$ is retrocompact in $X$. Thus $V \cap T \cap W$ is quasi-compact.
Suppose that $V \subset Z$ is retrocompact open in $Z$. Consider the open subset $\tilde V = V \cup (X \setminus Z)$ of $X$. Let $W \subset X$ be quasi-compact open. Then $$ W \cap \tilde V = \left(V \cap W\right) \cup \left((X \setminus Z) \cap W\right). $$ The first part is quasi-compact as $V \cap W = V \cap (Z \cap W)$ and $(Z \cap W)$ is quasi-compact open in $Z$ (Lemma \ref{lemma-closed-in-quasi-compact}) and $V$ is retrocompact in $Z$. The second part is quasi-compact as $(X \setminus Z)$ is retrocompact in $X$. In this way we see that $\tilde V$ is retrocompact in $X$. Thus if $V_1, V_2 \subset Z$ are retrocompact open, then $$ V_1 \cap (Z \setminus V_2) = \tilde V_1 \cap (X \setminus \tilde V_2) $$ is constructible in $X$. We conclude since every constructible subset of $Z$ is a finite union of subsets of the form $V_1 \cap (Z \setminus V_2)$.
Let $E = \bigcup_{i = 1, \ldots, n} U_i \cap V_i^c$ with $U_i, V_i$ retrocompact open in $X$. Let $W \subset X$ be quasi-compact open. Then $E \cap W = \bigcup_{i = 1, \ldots, n} U_i \cap V_i^c \cap W$. Thus it suffices to show that $U \cap V^c \cap W$ is quasi-compact if $U, V$ are retrocompact open and $W$ is quasi-compact open. This is true because $U \cap V^c \cap W$ is a closed subset of the quasi-compact $U \cap W$ so Lemma \ref{lemma-closed-in-quasi-compact} applies.
Observe that any retrocompact subset $T$ of $X$ has a basis for the induced topology consisting of quasi-compact opens. In particular this holds for any constructible subset (Lemma \ref{lemma-constructible-is-retrocompact}). Write $E = E_1 \cup \ldots \cup E_n$ with $E_i = U_i \cap V_i^c$ where $U_i, V_i \subset X$ are retrocompact open. Note that $E_i = E \cap E_i$ is constructible in $E$ by Lemma \ref{lemma-intersect-constructible-with-constructible}. Hence $F \cap E_i$ is constructible in $E_i$ by Lemma \ref{lemma-intersect-constructible-with-constructible}. Thus it suffices to prove the lemma in case $E = U \cap V^c$ where $U, V \subset X$ are retrocompact open. In this case the inclusion $E \subset X$ is a composition $$ E = U \cap V^c \to U \to X $$ Then we can apply Lemma \ref{lemma-closed-constructible-image} to the first inclusion and Lemma \ref{lemma-quasi-compact-open-immersion-constructible-image} to the second.
Note that $T$ is quasi-compact and open in $\overline{T}$. Using our basis of quasi-compact opens we can write $T = U \cap \overline{T}$ where $U$ is quasi-compact open in $X$. Then $U \setminus T = U \cap T^c$ is retrocompact in $U$ as $T^c$ is retrocompact in $X$. Hence the inclusion $T \subset X$ can be written as the composition of the inclusion $T \subset U$ of a closed subset with retrocompact complement and the inclusion $U \subset X$ which is retrocompact by our assumption on intersections of quasi-compact opens. Thus the lemma is a consequence of Lemmas \ref{lemma-quasi-compact-open-immersion-constructible-image} and \ref{lemma-closed-constructible-image}.
Assume $E$ is constructible and $Z \subset X$ irreducible closed. Then $E \cap Z$ is constructible in $Z$ by Lemma \ref{lemma-constructible-map-Noetherian}. Hence $E \cap Z$ is a finite union of nonempty locally closed subsets $T_i$ of $Z$. Clearly if none of the $T_i$ is open in $Z$, then $E \cap Z$ is not dense in $Z$. In this way we see that (1) implies (2). \medskip\noindent Conversely, assume (2) holds. Consider the set $\mathcal{S}$ of closed subsets $Y$ of $X$ such that $E \cap Y$ is not constructible in $Y$. If $\mathcal{S} \not = \emptyset$, then it has a smallest element $Y$ as $X$ is Noetherian. Let $Y = Y_1 \cup \ldots \cup Y_r$ be the decomposition of $Y$ into its irreducible components, see Lemma \ref{lemma-Noetherian}. If $r > 1$, then each $Y_i \cap E$ is constructible in $Y_i$ and hence a finite union of locally closed subsets of $Y_i$. Thus $E \cap Y$ is a finite union of locally closed subsets of $Y$ too and we conclude that $E \cap Y$ is constructible in $Y$ by Lemma \ref{lemma-constructible-Noetherian-space}. This is a contradiction and so $r = 1$. If $r = 1$, then $Y$ is irreducible, and by assumption (2) we see that $E \cap Y$ either (a) contains an open $V$ of $Y$ or (b) is not dense in $Y$. In case (a) we see, by minimality of $Y$, that $E \cap (Y \setminus V)$ is a finite union of locally closed subsets of $Y \setminus V$. Thus $E \cap Y$ is a finite union of locally closed subsets of $Y$ and is constructible by Lemma \ref{lemma-constructible-Noetherian-space}. This is a contradiction and so we must be in case (b). In case (b) we see that $E \cap Y = E \cap Y'$ for some proper closed subset $Y' \subset Y$. By minimality of $Y$ we see that $E \cap Y'$ is a finite union of locally closed subsets of $Y'$ and we see that $E \cap Y' = E \cap Y$ is a finite union of locally closed subsets of $Y$ and is constructible by Lemma \ref{lemma-constructible-Noetherian-space}. This contradiction finishes the proof of the lemma.
It is clear that (1) implies (2). Assume (2). Consider the set $\mathcal{S}$ of closed subsets $Y$ of $X$ containing $x$ such that $E \cap Y$ is not a neighbourhood of $x$ in $Y$. If $\mathcal{S} \not = \emptyset$, then it has a minimal element $Y$ as $X$ is Noetherian. Suppose $Y = Y_1 \cup Y_2$ with two smaller nonempty closed subsets $Y_1$, $Y_2$. If $x \in Y_i$ for $i = 1, 2$, then $Y_i \cap E$ is a neighbourhood of $x$ in $Y_i$ and we conclude $Y \cap E$ is a neighbourhood of $x$ in $Y$ which is a contradiction. If $x \in Y_1$ but $x \not\in Y_2$ (say), then $Y_1 \cap E$ is a neighbourhood of $x$ in $Y_1$ and hence also in $Y$, which is a contradiction as well. We conclude that $Y$ is irreducible closed. By assumption (2) we see that $E \cap Y$ is dense in $Y$. Thus $E \cap Y$ contains an open $V$ of $Y$, see Lemma \ref{lemma-characterize-constructible-Noetherian}. If $x \in V$ then $E \cap Y$ is a neighbourhood of $x$ in $Y$ which is a contradiction. If $x \not \in V$, then $Y' = Y \setminus V$ is a proper closed subset of $Y$ containing $x$. By minimality of $Y$ we see that $E \cap Y'$ contains an open neighbourhood $V' \subset Y'$ of $x$ in $Y'$. But then $V' \cup V$ is an open neighbourhood of $x$ in $Y$ contained in $E$, a contradiction. This contradiction finishes the proof of the lemma.
(See also remark below.) If $X$ is not quasi-compact, there exists an open covering $X = \bigcup_{i \in I} U_i$ such that no finite number of $U_i$ cover $X$. Let $Z$ be the subset of the power set $\mathcal{P}(I)$ of $I$ consisting of $I$ and all nonempty finite subsets of $I$. Define a topology on $Z$ with as a basis for the topology the following sets: \begin{enumerate} \item All subsets of $Z\setminus\{I\}$. \item For every finite subset $K$ of $I$ the set $U_K := \{J\subset I \mid J \in Z, \ K\subset J \})$. \end{enumerate} It is left to the reader to verify this is the basis for a topology. Consider the subset of $Z \times X$ defined by the formula $$ M = \{(J, x) \mid J \in Z, \ x \in \bigcap\nolimits_{i \in J} U_i^c)\} $$ If $(J, x) \not \in M$, then $x \in U_i$ for some $i \in J$. Hence $U_{\{i\}} \times U_i \subset Z \times X$ is an open subset containing $(J, x)$ and not intersecting $M$. Hence $M$ is closed. The projection of $M$ to $Z$ is $Z-\{I\}$ which is not closed. Hence $Z \times X \to Z$ is not closed. \medskip\noindent Assume $X$ is quasi-compact. Let $Z$ be a topological space. Let $M \subset  Z \times X$ be closed. Let $z \in Z$ be a point which is not in $\text{pr}_1(M)$. By the Tube Lemma \ref{lemma-tube} there exists an open $U \subset Z$ such that $U \times X$ is contained in the complement of $M$. Hence $\text{pr}_1(M)$ is closed.
Since every point of $Y$ is closed, we see from Lemma \ref{lemma-closed-in-quasi-compact} that the closed subset $f^{-1}(y)$ of $X$ is quasi-compact for all $y \in Y$. Thus, by Theorem \ref{theorem-characterize-proper} it suffices to show that $f$ is closed. If $E \subset X$ is closed, then it is quasi-compact (Lemma \ref{lemma-closed-in-quasi-compact}), hence $f(E) \subset Y$ is quasi-compact (Lemma \ref{lemma-image-quasi-compact}), hence $f(E)$ is closed in $Y$ (Lemma \ref{lemma-quasi-compact-in-Hausdorff}).
As $X$ is not Jacobson there exists a closed set $Z$ and an open set $U$ in $X$ such that $Z \cap U$ is nonempty and does not contain points closed in $X$. As $X$ has a basis of quasi-compact open sets we may replace $U$ by an open quasi-compact neighborhood of a point in $Z\cap U$ and so we may assume that $U$ is quasi-compact open. By Lemma \ref{lemma-quasi-compact-closed-point}, there exists a point $x \in Z \cap U$ closed in $Z \cap U$, and so $\{x\}$ is locally closed but not closed in $X$.
From  Lemma \ref{lemma-jacobson-equivalent-locally-closed} above, we just have to see that if $U$ is open in $X$ then $U\cap X_0$ is retrocompact in $X_0$ if and only if $U$ is retrocompact in $X$. This follows if we prove that for $U$ open in $X$ then $U\cap X_0$ is quasi-compact if and only if $U$ is quasi-compact. From  Lemma \ref{lemma-jacobson-inherited} it follows that we may replace $X$ by $U$ and assume that $U = X$. Finally notice that any collection of opens $\mathcal{U}$ of $X$ cover $X$ if and only if they cover $X_0$, using the Jacobson property of $X$ in the closed $X\setminus \bigcup \mathcal{U}$ to find a point in $X_0$ if it were nonempty.
Assume $f$ is closed. Let $y' \leadsto y$ in $Y$ and any $x'\in X$ with $f(x') = y'$ be given. Consider the closed subset $T = \overline{\{x'\}}$ of $X$. Then $f(T) \subset Y$ is a closed subset, and $y' \in f(T)$. Hence also $y \in f(T)$. Hence $y = f(x)$ with $x \in T$, i.e., $x' \leadsto x$. \medskip\noindent Assume $f$ is open, $X$ Noetherian, every irreducible closed subset of $X$ has a generic point, and $Y$ is Kolmogorov. Let $y' \leadsto y$ in $Y$ and any $x \in X$ with $f(x) = y$ be given. Consider $T = f^{-1}(\{y'\}) \subset X$. Take an open neighbourhood $x \in U \subset X$ of $x$. Then $f(U) \subset Y$ is open and $y \in f(U)$. Hence also $y' \in f(U)$. In other words, $T \cap U \not = \emptyset$. This proves that $x \in \overline{T}$. Since $X$ is Noetherian, $T$ is Noetherian (Lemma \ref{lemma-Noetherian}). Hence it has a decomposition $T = T_1 \cup \ldots \cup T_n$ into irreducible components. Then correspondingly $\overline{T} = \overline{T_1} \cup \ldots \cup \overline{T_n}$. By the above $x \in \overline{T_i}$ for some $i$. By assumption there exists a generic point $x' \in \overline{T_i}$, and we see that $x' \leadsto x$. As $x' \in \overline{T}$ we see that $f(x') \in \overline{\{y'\}}$. Note that $f(\overline{T_i}) = f(\overline{\{x'\}}) \subset \overline{\{f(x')\}}$. If $f(x') \not = y'$, then since $Y$ is Kolmogorov $f(x')$ is not a generic point of the irreducible closed subset $\overline{\{y'\}}$ and the inclusion $\overline{\{f(x')\}} \subset \overline{\{y'\}}$ is strict, i.e., $y' \not \in f(\overline{T_i})$. This contradicts the fact that $f(T_i) = \{y'\}$. Hence $f(x') = y'$ and we win.
Let $E$ be constructible and stable under generalization. Let $Y \subset X$ be an irreducible closed subset with generic point $\xi \in Y$. If $E \cap Y$ is nonempty, then it contains $\xi$ (by stability under generalization) and hence is dense in $Y$, hence it contains a nonempty open of $Y$, see Lemma \ref{lemma-characterize-constructible-Noetherian}. Thus $E$ is open by Lemma \ref{lemma-characterize-open-Noetherian}. This proves (2). To prove (1) apply (2) to the complement of $E$ in $X$.
Let $x \in X$ be a point. We will show that $\delta - \delta'$ is constant in a neighbourhood of $x$. We may replace $X$ by an open neighbourhood of $x$ in $X$ which is Noetherian. Hence we may assume $X$ is Noetherian and sober. Let $Z_1, \ldots, Z_r$ be the irreducible components of $X$ passing through $x$. (There are finitely many as $X$ is Noetherian, see Lemma \ref{lemma-Noetherian}.) Let $\xi_i \in Z_i$ be the generic point. Note $Z_1 \cup \ldots \cup Z_r$ is a neighbourhood of $x$ in $X$ (not necessarily closed). We claim that $\delta - \delta'$ is constant on $Z_1 \cup \ldots \cup Z_r$. Namely, if $y \in Z_i$, then $$ \delta(x) - \delta(y) = \delta(x) - \delta(\xi_i) + \delta(\xi_i) - \delta(y) = - \text{codim}(\overline{\{x\}}, Z_i) + \text{codim}(\overline{\{y\}}, Z_i) $$ by Lemma \ref{lemma-dimension-function-catenary}. Similarly for $\delta'$. Whence the result.
We will use repeatedly that an open subspace of a catenary space is catenary, see Lemma \ref{lemma-catenary} and that a Noetherian topological space has finitely many irreducible components, see Lemma \ref{lemma-Noetherian}. In the proof of Lemma \ref{lemma-dimension-function-unique} we saw how to construct such a function. Namely, we first replace $X$ by a Noetherian open neighbourhood of $x$. Next, we let $Z_1, \ldots, Z_r \subset X$ be the irreducible components of $X$. Let $$ Z_i \cap Z_j = \bigcup Z_{ijk} $$ be the decomposition into irreducible components. We replace $X$ by $$ X \setminus \left( \bigcup\nolimits_{x \not \in Z_i} Z_i \cup \bigcup\nolimits_{x \not \in Z_{ijk}} Z_{ijk} \right) $$ so that we may assume $x \in Z_i$ for all $i$ and $x \in Z_{ijk}$ for all $i, j, k$. For $y \in X$ choose any $i$ such that $y \in Z_i$ and set $$ \delta(y) = - \text{codim}(\overline{\{x\}}, Z_i) + \text{codim}(\overline{\{y\}}, Z_i). $$ We claim this is a dimension function. First we show that it is well defined, i.e., independent of the choice of $i$. Namely, suppose that $y \in Z_{ijk}$ for some $i, j, k$. Then we have (using Lemma \ref{lemma-catenary-in-codimension}) \begin{align*} \delta(y) & = - \text{codim}(\overline{\{x\}}, Z_i) + \text{codim}(\overline{\{y\}}, Z_i) \\ & = - \text{codim}(\overline{\{x\}}, Z_{ijk}) - \text{codim}(Z_{ijk}, Z_i) + \text{codim}(\overline{\{y\}}, Z_{ijk}) + \text{codim}(Z_{ijk}, Z_i) \\ & = - \text{codim}(\overline{\{x\}}, Z_{ijk}) + \text{codim}(\overline{\{y\}}, Z_{ijk}) \end{align*} which is symmetric in $i$ and $j$. We omit the proof that it is a dimension function.