Observation 2.2. For $A,B$ finite subsets of an $\mathcal {L}$ -structure C,

$$ \begin{align*}\delta(A\cup B)\leq \delta(A)+\delta(B)-\delta(A\cap B)\end{align*} $$

with equality if and only if there are no relations holding between A and B other than those inside A or those inside B. In this case, we say A and B are freely joined over $A\cap B$ and we write $A\cup B = A\oplus _{A\cap B}B$ .

Proof This is just inclusion–exclusion on the number of relations holding in $A\cup B$ .⊣

Proof Suppose $A\leq B$ . Then $\delta (A)\leq \delta (X\cup A)\leq \delta (X)+\delta (A)-\delta (X\cap A)$ . Thus, $\delta (X\cap A)\leq \delta (X)$ .

Now suppose $A\leq B\leq C$ . Let $A\subseteq X\subseteq C$ . Then $\delta (X\cap B)\leq \delta (X)$ since $B\leq C$ , and $\delta (A)=\delta (X\cap B\cap A)\leq \delta (X\cap B)$ since $A\leq B$ . So, $\delta (A)\leq \delta (X)$ .⊣

Proof Since $\delta (X\cup Y \cup Z)\leq \delta (X\cup Y)+\delta (X\cup Z)-\delta (X\cup (Y\cap Z))$ , we can subtract $\delta (X)$ from both sides and see that $\delta (Y\cup Z/X)\leq \delta (Y/X)+\delta (Z/X)-\delta (Y\cap Z/X)=-\delta (Y\cap Z/X)$ . Since Y and Z are simply algebraic over X, either $Y\cap Z=\emptyset $ or $\delta (Y\cap Z/X)>0$ . But since $X\leq C$ , $0\leq \delta (Y\cup Z/X)=-\delta (Y\cap Z/X)$ . So Y and Z are disjoint.⊣

Proof Take $X\supseteq A$ with minimal $\delta (X)$ and take X minimal as such (i.e., it has no proper subset containing A with the same value of $\delta $ ). The minimality of $\delta (X)$ implies that $X\leq C$ . Suppose X were not unique, then there would be another such set $Y\supseteq A$ with $\delta (Y)=\delta (X)$ . Then $\delta (X\cup Y)\leq \delta (X)+\delta (Y)-\delta (X\cap Y)<\delta (Y)=\delta (X)$ by minimality of the set X. But this would contradict the minimality of $\delta (X)$ .⊣

Observation 2.8. If $A\subseteq B\leq C$ , then $g_B(A)=g_C(A)$ .

Proof That $g_C(A)\leq g_B(A)$ is immediate since any $X\subseteq B$ so that $A\subseteq X$ and $\delta (X)<\delta (A)$ is also a subset of C. Suppose $X\subseteq C$ is so that $A\subseteq X$ and $\mathcal {L}_0\subseteq \mathcal {L}$ is a sublanguage so that restricting to this sublanguage, $\delta (X\vert \mathcal {L}_0)<\delta (A\vert \mathcal {L}_0)$ . Then $\delta (X\cap B)\leq \delta (X)< \delta (A)$ , so in the same sublanguage, B contains a set no larger than X witnessing that A is not strong. So $g_B(A)\leq g_C(A)$ .⊣

Observation 2.11. Fix $Y/X$ a minimally simply algebraic extension. There is a first-order formula which is true in any $C\in \mathcal {C}^\zeta _\mu $ and implies that C does not contain disjoint subsets $A,B^1,\ldots ,B^r$ of the form of $Y/X$ with $r>\mu (X,Y,g_C(A))$ .

Proof Let $\mathcal {L}'$ be the finite sublanguage of $\mathcal {L}$ guaranteed in Definition 2.7, and let m be an integer so that $\mu (X,Y,k)=\mu (X,Y,\infty )$ for any $k\geq m$ .

Let $\rho (A)$ say that $A\vert \mathcal {L}'\cong X\vert \mathcal {L}'$ . For each $k\in \omega $ , let $\psi _k(A,C^1,\ldots C^k)$ say that the sets are disjoint, and that $ \operatorname {\mathrm {tp_{r.q.f.}}}(C^j/A)\supseteq \operatorname {\mathrm {tp_{r.q.f.}}}(Y/X)$ for each j. For each $l\leq m$ , let $\phi _l$ say that $g_C(A)=l$ . Finally, let $\theta $ be the formula which says $\forall A$ , if $\rho (A)$ holds, then

$$\begin{align*}\bigvee_{l<m}\left( \phi_l(A)\wedge \neg \exists Z \psi_{\mu(X,Y,l)+1}(A,Z)\right)\vee \left(\bigwedge_{l<m}\neg \phi_l(A)\wedge \neg \exists Z \psi_{\mu(X,Y,\infty)+1}(A,Z)\right). \end{align*}$$

Let $\Omega $ be the collection of all the extensions $Y'/X'$ so that $ \operatorname {\mathrm {tp_{r.q.f.}}}(Y'/X')= \operatorname {\mathrm {tp_{r.q.f.}}}(Y/X)$ and $X'\vert \mathcal {L}'\cong X\vert \mathcal {L}'$ . Then $\theta $ says that the $\mu $ -bound is obeyed for each extension in $\Omega $ . Thus $\theta $ is true in every $C\in \mathcal {C}^\zeta _{\mu }$ and implies that C respects the $\mu $ -bound for $Y/X$ .⊣

Observation 2.12. Let $A\subseteq B$ be finite $\mathcal {L}$ -structures. Let $A'\subseteq B'$ be formed by removing some occurrence $R(\bar {b})$ of some relation R from B. If $B'/A'$ is of the form of $Y/X$ , then either $B/A$ is of the form of $Y/X$ or $\bar {b}\subseteq A$ .

Proof A careful reading of Lemma 3 of [Reference Hrushovski13] will show that this is what is proved there. The full proof appears as Lemma 42 in [Reference Andrews4].⊣

Proof It is immediate that $E\models \zeta $ because $\zeta $ is preserved under free joins. Let $X\subseteq E$ . Then $\delta (X)=\delta (X\cap B_1)+\delta (X\cap B_2)-\delta (X\cap A)\geq \delta (X\cap B_2)\geq 0$ , since X is the free-join of $X\cap B_1$ with $X\cap B_2$ over $X\cap A$ and $A\leq B_1$ . Thus $E\in \mathcal {C}_0$ .

Suppose that $Y/X$ is a minimally simply algebraic extension, and $F,C^1,\ldots ,C^r$ are disjoint subsets of E so that each $C^i/F$ is of the form of $Y/X$ . We restrict E to $\mathcal {L}_{Y/X}$ for all tuples outside of F.

By Lemma 2.13, there are four cases to consider:

• • One of the $C^i$ is contained in $B_1\smallsetminus A$ and $F\subseteq A$ . Since $B_1\smallsetminus A$ is simply algebraic over A, we have that $C^i=B_1\smallsetminus A$ . In this case, we have that r is at most one more than the number of disjoint extensions over F of the form of $Y/X$ in $B_2$ . Since $B_2\leq E$ , Observation 2.8 shows that $g_{B_2}(F)=g_{E}(F)$ . So, if $r>\mu (X, Y,g_E(F))$ , then in $B_2$ we already have $g_{B_2}(F)$ disjoint extensions over F of the form of $Y/X$ . Since $B_1\smallsetminus A/F$ is minimally simply algebraic and of the form of $Y/X$ , each of these extensions is of the form of $B_1\smallsetminus A/F$ . Thus we have (2.14) above.

• • $F\cup \bigcup _{i\leq r}C^i$ is entirely contained in $B_1$ or $B_2$ . If it’s contained in $B_2$ , then since $B_2\in \mathcal {C}_\mu $ and $B_2\leq E$ , we see that $r\leq \mu (X,Y,g_{B_2}(F))=\mu (X,Y,g_{E}(F))$ . If it’s contained in $B_1$ , then since $B_1\in \mathcal {C}_\mu $ , we have that $r\leq \mu (X,Y,g_{B_1}(F))$ . Either $r\leq \mu (X,Y,g_{B_1}(F))\leq \mu (X,Y,g_{E}(F))$ or we are in case (2.14) above.

• • $r\leq \delta (F)$ . Then it’s automatic that $r\leq \mu (X,Y,g_E(F))$ as $\mu (X,Y,m)$ is always $\geq \delta (X)=\delta (F)$ .

• • For one $C^i$ , setting $Z=(F\cap A)\cup (C^i\cap B_2)$ , we see $\delta (Z/Z\cap A)<0$ . Let $Y=Z\smallsetminus A$ . Then $\delta (Y/A)<0$ . Further, one of the $C^j$ is contained in $B_1\smallsetminus A$ , so $\lvert Y\rvert \leq \lvert C^i\cap B_2\rvert \leq \lvert B_1\smallsetminus A\rvert $ . Since one of the $C^j$ is contained in $B_1\smallsetminus A$ , all of the relations between the set Y and A which were retained when we took the reduct above are in $\mathcal {L}_{B_1/A}$ , so $\delta (Y\vert \mathcal {L}_{B_1/A}/A\vert \mathcal {L}_{B_1/A})<0$ . Thus, case (2.14) holds.⊣

Proof We may assume that there is no $B'$ so that $A\leq B'\leq B_1$ as otherwise we can first amalgamate this $B'$ with $B_2$ over A. Thus, either $B_1$ is $A\cup \{x\}$ where x is unrelated to any element in A or $B_1$ is simply algebraic over A, say minimally simply algebraic over $F\subseteq A$ . In the first case, the free-join suffices. In the second case, the free-join works unless one of the three conditions enumerated in the Algebraic Amalgamation Lemma holds. The second and third cannot hold, because $A\leq B_2$ . Thus we can assume that $B_1\smallsetminus A$ is minimally simply algebraic over $F\subseteq A$ and that in $B_2$ there are $C^1,\ldots C^{r}$ which are $\mu (F,B_1\smallsetminus A,g_{B_2}(F))$ disjoint extensions of the form of $B_1\smallsetminus A/F$ . Since $A\leq B_1$ and $A\leq B_2$ , we have $g_A(F)=g_{B_1}(F)=g_{B_2}(F)$ . Thus, it cannot be that all of these $C^j$ are contained in A, since then $B_1$ would have violated the $\mu $ -bound. Without loss of generality, $C^1\not \subseteq A$ . $C^1$ cannot be partially in A since $A\leq B_2$ . Thus $C^1\subseteq B_2\smallsetminus A$ . Since $A\leq B_2$ , there are no extra relations in $ \operatorname {\mathrm {tp_{r.q.f.}}}(C^1/A)$ other than those in $ \operatorname {\mathrm {tp_{r.q.f.}}}(B_1\smallsetminus A/F)$ , and $A\cup C^1\leq B_2$ . Thus, we can form g by sending $B_1\smallsetminus A$ to $C^1$ over A.⊣

Proof Suppose (1)–(3) are true. To see (2’)(a) holds, since $\zeta $ is universal it suffices to check that $\zeta $ holds on every finite substructure of $\mathcal {M}$ . Every finite $A\subseteq \mathcal {M}$ is contained in its finite self-sufficient closure C, which is strong in $\mathcal {M}$ . As $C\in \mathcal {C}_\mu ^\zeta $ by (2), we have $C\models \zeta $ and thus $A\models \zeta $ . Similarly by $C\in \mathcal {C}_\mu ^\zeta $ , (2’)(b) holds on A. Finally, to see (2’)(c), let $A=F\cup \bigcup _{i\leq r}C^i$ and again consider the self-sufficient closure C of A. Then since $C\in \mathcal {C}_\mu ^\zeta $ by (2), we have that $r\leq \mu (X,Y,g_{C}(F))$ , but $g_C(F)=g_{\mathcal {M}}(F)$ because $C\leq \mathcal {M}$ . This shows that (2’) holds. (3’) and (3”) hold similarly by applying the algebraic amalgamation lemma, i.e., the algebraic amalgamation lemma shows that it would keep us in $\mathcal {C}^\zeta _\mu $ to have these sets, and property (3) then gives us the needed sets inside $\mathcal {M}$ .

Now we suppose (1),(2’),(3’),(3”). Suppose $C\leq \mathcal {M}$ . Then for any set $A\subseteq C$ , we have $g_C(A)=g_{\mathcal {M}}(A)$ . Thus condition (2’) ensures that $C\in \mathcal {C}^\zeta _\mu $ , so (2) holds. Property (3) follows from (3’) and (3”) exactly as the strong amalgamation lemma follows from the algebraic amalgamation lemma.⊣

Proof The first two conditions in (2’) are easy to see are elementary. For the third, we use the formulas from Observation 2.11. Carefully examining the formula $\theta $ produced in Observation 2.11, one can see they are equivalent to $\forall \exists $ formulas. Alternatively, to see that (2’) is equivalent to a collection of $\forall \exists $ -sentences, it suffices to see that it is preserved in unions of chains. Suppose $M_0\subseteq M_1\subseteq \cdots \subseteq \bigcup _i M_i=N$ are so that each $M_i$ satisfies (2’). Note that $\lim _i g_{M_i}(A)=g_N(A)$ for every $A\subseteq N$ . Suppose there are $F,C^1,\ldots, C^r$ disjoint subsets of N and each extension $C^j/F$ is of the form of $Y/X$ . Then take i large enough that $M_i$ contains $F\cup \bigcup _i C^i$ and $g_{M_i}(A)=g_{N}(A)$ . Then since $M_i$ satisfies (2’), we see that $r\leq \mu (X,Y,g_{M_i}(A))=\mu (X,Y,g_N(A))$ .

For (3’), it suffices to note that a finite set being strong in the structure is defined by an infinite schema of universal sentences.

For (3”), the fact that it is first-order to determine the value of $\mu (X,Y,m)$ (as in Observation 2.11) and that it is first-order to be strong enough so that $\mu (G,H,g_C(G))=\mu (G,H,g_{M\oplus _B C}(G))$ suffices to make this first order. Let us see that it is preserved in unions of chains. Again suppose that $M_0\subseteq M_1\subseteq \cdots \subseteq \bigcup _i M_i=N$ and each $M_i$ satisfies (3”). Let $B\subseteq N$ , $B,C\in \mathcal {C}^\zeta _\mu $ be as in the hypothesis of (3”). If B is strong enough in N that there is no Y with $\lvert Y\rvert <\lvert C\smallsetminus B\rvert $ and $\delta (Y\vert \mathcal {L}_{C/B}/B\vert \mathcal {L}_{C/B})<0$ , then this is true in every $M_i$ which contains B as the non-existence of this Y is a universal condition. Similarly, since for every $A\subseteq N$ , we have that $\lim _i g_{M_i}(A)=g_N(A)$ , and $M_i\leq M_i\oplus _B C$ and $N\leq N\oplus _B C$ , we also have that $\lim _i g_{M_i\oplus _B C}(A)=g_{N\oplus _B C}(A)$ . Thus if there is no $H/G$ an extension in C so that $\mu (G,H,g_C(G))\neq \mu (G,H,g_{N\oplus _B C}(G))$ , then the same is true in $M_i$ for any large enough $M_i$ . Thus, in a large enough $M_i$ , we must have $\mu (F,C\smallsetminus B,g_{M_i}(F))=\mu (F,C\smallsetminus B,g_{N}(F))$ disjoint extensions over F of the form of $C\smallsetminus B$ over F. Thus, we have these extensions in N as well.⊣

Proof Since any countable elementary extension of $\mathcal {M}$ is isomorphic with $\mathcal {M}$ , we see that there are only countably many m-types in $\text {Th} (\mathcal {M})$ for each m. Thus, there is some countable saturated model of the theory of $\mathcal {M}$ . But this, being an elementary extension of $\mathcal {M}$ , is isomorphic with $\mathcal {M}$ .⊣

Proof Let B be the self-sufficient closure of A, so $\delta (B)=d(A)$ . Then $B\leq \mathcal {M}$ . Thus $\{x\}\cup B\leq \mathcal {M}$ and $\{y\}\cup B\leq \mathcal {M}$ . Using property (3) and a back-and-forth along strong substructures, we see that $(\mathcal {M},Bx)\cong _B (\mathcal {M},By)$ , which implies the needed isomorphism.⊣

Proof Suppose $d(\{x\}\cup A)=d(A)$ . Let B be the self-sufficient closure of A, so $\delta (B)=d(A)$ . Lemma 2.6 shows that B is algebraic over A.

Now we show that $x\in \operatorname {\mathrm {acl}}(B)$ , which suffices since $B\subseteq \operatorname {\mathrm {acl}}(A)$ . Fix E to be a set so that $\delta (E)=d(\lbrace x \rbrace \cup A)=d(A)$ and $\lbrace x \rbrace \cup A\subseteq E$ . Then $\delta (E\cup B)\leq \delta (E)+\delta (B)-\delta (E\cap B)$ . If E does not contain B, then $\delta (E\cap B)>\delta (B)$ by minimality of B. Then $\delta (E\cup B)<\delta (E)=d(A)$ , a contradiction. So E contains B.

Take a sequence of extensions $B=B_0\subset B_1\subset \cdots B_n=E$ with each $B_{i+1}$ chosen to be minimal containing $B_i$ contained in E with $\delta (B_{i+1})=d(A)$ . It follows from $\delta (B_{i})=d(A)$ that each $B_i\leq \mathcal {M}$ . Then $B_{i+1}$ is a simply algebraic extension over $B_i$ . Thus, the $\mu $ -bound ensures that there are not infinitely many extensions over $B_i$ which are disjoint and of the form of $B_{i+1}/B_i$ , and Lemma 2.4 shows that any two extensions of the form of $B_{i+1}/B_i$ must be disjoint. Thus $B_{i+1}$ is algebraic over $B_i$ . Conclude that E is algebraic over B.⊣

Proof In the previous two lemmas, we saw that over any set there is a unique non-algebraic type realized in $\mathcal {M}$ . Since $\mathcal {M}$ is saturated, this implies that $\text {Th} (\mathcal {M})$ is strongly minimal.⊣

Proof Observe that Lemma 2.21 holds for any model satisfying (2’). Let N be infinite and satisfy (2’) and (3”). Let $N'\succeq N$ contain a countable indiscernible sequence $I=(x_i)_{i\in \omega }$ . We observe that $d(\{x_i\mid i<k\})=k$ . Were this not the case, then we would have some $x_i\in \operatorname {\mathrm {acl}}(\{x_j\mid j<i\})$ by Lemma 2.21, which cannot happen by indiscernibility. Thus every finite initial subsequence of I, and thus every finite subsequence of I is strong in $N'$ . Let $M\preceq N'$ be countable and contain I. Then M satisfies (1), (2’), (3’), and (3”), with (3’) witnessed by I. Thus $M\cong \mathcal {M}_\mu ^\zeta $ by Lemma 2.16. So $N\models \text {Th} (\mathcal {M}_\mu ^\zeta )$ .

By Lindström’s test [Reference Hodges12, 8.3.4] and Lemmas 2.17 and 2.22, $\text {Th} (\mathcal {M}^\zeta _\mu )$ is model complete.⊣

Proof As in [Reference Hrushovski13, Lemma 15], the geometry associated to any hypergraph via the d function is flat. Lemmas 2.20 and 2.21 show that d is the dimension function of the acl-geometry in $\mathcal {M}$ .⊣

Proof Since we are trying to show that $B_0$ is large (i.e., contains either B, $B_1$ or $B_2$ , depending on the case), we can assume that there is no proper subset $B_0^{\prime }\subset B_0$ which has $\delta (B_0^{\prime }/A_0)\leq 0$ . For every $b_i\in B$ clearly $\delta (b_i/A_0) = 1$ , so $|B_0|>1$ . Then for every $b_i\in B_0$ , the element $b_i$ must appear in at least two relations in $A_0\cup B_0$ or else $B_0\smallsetminus \lbrace b_i \rbrace $ contradicts the minimality of $B_0$ . Now, for every $i\notin \lbrace 1,t+1 \rbrace $ , if $b_i\in B_0$ , then $(b_{i-1},a_{i-1},b_i)$ , $(b_i,a_i,b_{i+1})$ are edges in $A_0\cup B_0$ (if $i=2t$ , then take $a_i = g$ and $b_{i+1} = b_1$ ). For $i\in \lbrace 1,t+1 \rbrace $ , if $b_i\in B_0$ , then at least one of $b_{i-1}$ , $b_{i+1}$ (for $i=1$ , take $b_{i-1}$ to be $b_{2t}$ ) must be an element of $B_0$ . Hence, $R[A_0\cup B_0]$ , the restriction of R to $A_0\cup B_0$ , must contain at least one of $R_1$ , $R_2$ . In particular, $b_1,b_{2t}\in B_0$ . Since $t\geq k+1$ , also in any case $A\smallsetminus \lbrace g,h \rbrace \subseteq A_0$ . Furthermore, if $R_2\subseteq R[A_0\cup B_0]$ , then also $g\in A_0$ . I.e., if $g\notin A_0$ , then $R_1\subseteq R[A_0\cup B_0]$ .

If $h\notin A_0$ , then as $b_1\in B_0$ and $b_1$ must appear in two relation, we have $b_2,b_{2t}\in B_0$ , and consequently $R_1,R_2\subseteq R[A_0\cup B_0]$ . This is case (1), so we may assume $h\in A_0$ . Now, if $g\notin A_0$ , then as we’ve seen we must be in case (2). Finally, assume we are not in cases (1) or (2), in particular $A=A_0$ . Since $B_1\nsubseteq B_0$ , also $R_1\nsubseteq R[A_0\cup B_0]$ . Then $R_2\subseteq R[A_0\cup B_0]$ and we are in case (3).

Now for the additional part, assume instead that $B_0$ is minimal such that $\delta (B_0/A_0)< 0 $ . Again, if $b_i\in B_0$ then it must appear in at least two relations in $R[A_0\cup B_0]$ . By what we’ve shown, for some $i\in \lbrace 1,2 \rbrace $ , $B_i\subseteq B_0$ . Observe that $\delta (B_i/A_0)\geq \delta (B_i/A) = 0$ , so there exists $b_j\in B_0\smallsetminus B_i$ . As before, this implies that also $B_{3-i}\subseteq B_0$ , i.e., $B_0 = B$ . A short check yields that for any $A\smallsetminus \lbrace g,h \rbrace \subseteq X \subset A$ , we have $\delta (B/X)\geq 0$ . So $A_0=A$ and indeed $\delta (B/A) < 0$ .⊣

Proof First we show that no subset of $D_t$ has $\delta (X)<2$ unless $\lvert X\rvert \leq 1$ . Assume to the contrary that $|X|\geq 2$ , $\delta (X) < 2$ . Then there must be edges in X, hence $X\cap A$ and $X\cap B$ are non-empty. $\delta (D_t)> 1$ , so $X\neq D_t$ . Then by Lemma 3.2, $\delta (X\cap B/X\cap A) \geq 0$ . So, $1-\delta (X\cap A) \geq \delta (X)-\delta (X\cap A)\geq 0$ which means we must have $1\geq \delta (X\cap A) =|X\cap A|\geq 1$ since there are no relations holding in the set A. But then again by Lemma 3.2, $\delta (X)>\delta (X\cap A) = 1$ in contradiction to $\delta (X)<2$ .

We show $D_t$ embeds strongly into $\mathcal {M}_\mu $ . Embed $\lbrace a_1,\dots , a_k, b_1 \rbrace $ onto an independent subset of $\mathcal {M}_\mu $ . Because the isomorphism type of a ternary edge is in $\mathcal {C}_\mu $ , by genericity of $\mathcal {M}_\mu $ , any two independent points extend to an edge. Embed $b_2$ onto an extension of $b_{1}, a_1$ to an edge. Now embed $b_3$ onto an extension of $b_2, a_2$ to an edge. Continue in this manner until $b_{2t}$ has been embedded. Now embed g onto an extension of $b_{2t}, b_1$ to an edge and h onto an extension of $b_1, b_{t+1}$ to an edge. Observe that each embedded point must be new. Now, since $D_t\leq \mathcal {M}_{\mu }$ , we have $D_t\in \mathcal {C}_{\mu }$ by property (2) of $\mathcal {M}_{\mu }$ .⊣

Observation 3.4. Let C be minimally simply algebraic over F. From the definition of a minimally simply algebraic extension, it follows that every element of F appears in at least one edge in $F\cup C$ that is not an edge of F, and every element of C appears in at least two edges in $F\cup C$ (unless $\lvert C\rvert =1$ ).

Observation 3.5. Suppose $E=B_1\oplus _A B_2$ where every point in $B_1\smallsetminus A$ appears in at most k edges in $B_1$ . Let $F, C^1,\dots , C^r\subseteq E$ be disjoint such that each $C^i/F$ is a minimally simply algebraic extension. If $F\nsubseteq B_2$ , then it is immediate from Observation 3.4 that $r\leq k$ .

Proof We start with the following claim:

Choose $t>|M|$ greater than the index of any relation symbol appearing in M, in particular greater than $\max \{g_M(A) \mid A\not \leq M\}$ , and also large enough so that for every $A, B\subseteq M$ such that $B/A$ is of the form of an extension $Y/X$ , $\mu (X,Y,m) = \mu (X,Y,\infty )$ for every $m\geq t$ . Denote $E:=E_t$ . By the claim, this guarantees that for any $A\subseteq M$ , if $A\not \leq M$ , then $g_E(A) = g_M(A)$ . If $A\leq M$ , then $\mu (A,B,g_E(A)) = \mu (A,B,\infty )$ whenever $B\subseteq M$ is minimally simply algebraic over A. In any case, whenever $A,B\subseteq M$ are such that $A/B$ is of the form of an extension $Y/X$ , then $\mu (X,Y,g_E(A)) = \mu (X,Y, g_M(A))$ .

Now assume for a contradiction that $F, C^1,\dots , C^r\subseteq E$ are disjoint such that each $C^i/F$ is of the form of a minimally simply algebraic extension $Y/X$ , and $r>\mu (X, Y, g_E(F))$ . By construction, every point in $D_t\smallsetminus X$ is in at most three relations in $E_t$ , so 3-permissiveness and Observation 3.5 imply $F\subseteq M$ . By what we’ve shown, at most $\mu (X, Y, g_E(F)) = \mu (X, Y, g_M(F))$ of the $C^i$ are contained in M. Without loss of generality assume $C^1\nsubseteq M$ . Then by $\delta (C^1\cap D_t/F\cup (C^1\cap M)) \leq 0$ , Lemma 3.2 implies $|C^1|> t> |M|$ and so none of the $C^i$ can be fully contained in M. Using the same reasoning for $\delta (C^2/F\cup (C^2\cap M))\leq 0$ , again by Lemma 3.2, we see that $C_1$ and $C_2$ intersect inside $D_t$ , in contradiction.⊣

Proof Clearly $\delta (X, E_t) \leq \delta (X, M)$ . Let $X\subseteq Z\subseteq E_t$ , then

$$\begin{align*}\delta(Z) = \delta(Z\cap M) + \delta(Z\cap D_t/Z\cap \bar{c}). \end{align*}$$

The first summand is at least $\delta (X, M)$ and the second summand is at least $-1$ . Thus, to witness $\delta (X,E_t)< \delta (X,M)$ , we must have that $\delta (Z\cap M)=\delta (X, M)$ and $\delta (Z\cap D_t/Z\cap F) = {-}1$ . By Lemma 3.2, this means $D_t\subseteq Z$ , and in particular $\bar {c}\subseteq Z\cap M$ . Conversely, if Y such as in the statement exists then $Y\cup D_t$ witnesses $\delta (X,E_t) \leq \delta (X,M)-1$ .⊣

Proof If $|\bar {b}|\geq 4$ , then by labeling $\bar {b} = (a_1,\dots , a_{|\bar {b}|-2}, g,h)$ Lemma 3.7 and Lemma 3.8 give us the desired B.

Otherwise, construct $M'$ by adding new points $p_1,\dots , p_{4-|b|}$ to M and no new edges. Let $\bar {b}'$ be the concatenation of $\bar {b}$ with $(p_1,\dots, p_{4-|\bar {b}|})$ . Now apply the above to M and $\bar {b}'$ consecutively $5-|b|$ times. To use Lemma 3.8 note that after the penultimate application, every set containing $\bar {b}$ can be extended to a set containing $\bar {b}'$ without changing its $\delta $ value.⊣

Observation 3.11. A minimally simply algebraic extension $X\subseteq Y$ is k-unblockable if and only if for any k-permissive $\mu $ , if $Y\in \mathcal {C}_\mu $ and $M\equiv \mathcal {M}_\mu $ then for any $Z\cong X$ in M, M contains $\mu (X,Y,g_M(Z))$ disjoint extensions over Z each of the form of $Y/X$ .

Proof Let $\mu $ be k-permissive such that $B\in \mathcal {C}_\mu $ , let $Z\in \mathcal {C}_\mu $ contain A, and let $E=B\oplus _A Z$ . Suppose $F,C^1,\ldots, C^r$ are disjoint extensions of the form of a minimally simply algebraic extension $Y/X$ with $r>\mu (X,Y,g_{E}(F))$ . By k-permissiveness of $\mu $ , we have $r>k$ . Thus, by Observation 3.5 it must be that $F\subseteq Z$ . Now we observe that every $C^i$ is either contained in $B\smallsetminus A$ or is contained in Z, for if it were partially but not totally in $B\smallsetminus A$ , then we would have $\delta (C^i/Z)<0$ showing that $Z\not \leq E$ , which is a contradiction. So, either $F\cup \bigcup _{i\leq r}C^i\subseteq Z$ or one of the $C^i$ is contained in $B\smallsetminus A$ , which implies that $F\subseteq A$ . Since B is minimally simply algebraic over A, this implies $C^i=B\smallsetminus A$ and $F=A$ . So, we conclude that there are already $\mu (X,Y,g_{E}(F))=\mu (X,Y,g_M(F))$ disjoint extensions of the form $Y/X$ over F in M.⊣

Proof We define a sequence of 2-unblockables over $\hat {A} = \lbrace a_1,\dots , a_k,g \rbrace $ , where $k\geq 2$ and all the elements of $\hat {A}$ are distinct. For every $l>k$ , by $a_l$ we mean $a_i$ where $i\equiv _k l$ , $1\leq i\leq k$ . Let $B=\lbrace b_1,\dots , b_{2t} \rbrace $ be new elements, where $t>k+1$ is arbitrary. Define $\hat {D}_t$ to be the hypergraph whose set of edges is

$$\begin{align*}R = \{(b_i,a_i,b_{i+1}) \mid 1\leq i< 2t\}\cup\lbrace (b_{2t}, g, b_1 \rbrace). \end{align*}$$

Observe that $\hat {D}_t$ is in fact $D_t\smallsetminus \{h\}$ , so $\hat {D}_t\leq D_t$ . Lemma 3.3 says that $D_t\in \mathcal {C}_\mu $ , thus $\hat {D}_t\in \mathcal {C}_{\mu }$ as well.

There is a clear bijection between elements of B and edges in R, so $\delta (B/A) = 0$ . By Lemma 3.2, there are no $A_0\subseteq \hat {A}$ , $\emptyset \neq B_0\subset B$ such that $\delta (B_0/A_0) \leq 0$ , so B is minimally simply algebraic over A. Finally, 2-unblockability is immediate by Lemma 3.12.⊣

Proof By Lemma 3.12 we only need to show that B is minimally simply algebraic over A. To see this, let $\emptyset \neq X\subset B$ and observe that there are strictly more elements in $X\smallsetminus A$ than there are edges in X.⊣

Proof If $|C|=1$ , then $C\cup F$ is an edge, i.e., isomorphic to $P_3$ . Otherwise, each $a_i\in C$ must appear in at least two edges in $C\cup F$ , hence $a_{i-1},a_{i+1}\in C\cup F$ . Proceeding this way in both directions, since by assumption $F\cup C\neq A$ , we find $a_m, a_M\in F$ distinct such that $a_j\in C$ for every $m<j<M$ (where if $j>k$ we replace it with $j-k$ ). By definition of minimal simple algebraicity, these are exactly the points in $F\cup C$ , with $F=\lbrace a_m, a_M \rbrace $ .⊣