Preserving Predimension Functions under Functional Convergences

Question

Definition 1. If ‎$‎‎‎\mathcal{L}‎$ ‎is a‎ ‎countable relational ‎language, ‎a ‎predimension ‎class ‎‎‎‎‎$‎C‎$ is a class ‎of $‎‎\mathcal{L}$-structures with ‎the ‎following ‎properties:‎

‎C1: ‎‎$‎‎‎\forall ‎M\in C~~~|M|<\aleph_{0}‎$‎

‎C2: $|‎\frac{C}{\cong}‎|\leq \aleph_{0}$

‎C3: ‎$C$ is closed under substructure.

C4: $C$ is closed under isomorphism.

‎Definition 2. A ‎‎predimension function ‎‎$‎‎\delta:C‎\longrightarrow ‎\mathbb{R}^{\geq 0}‎$ on the predimension class ‎$‎C‎$‎ ‎is a‎ ‎function ‎with ‎the ‎following ‎properties:‎

P1: ‎‎$‎‎\delta (‎\emptyset‎)=0$‎‎ ‎

P2: ‎‎$‎‎\forall M,N\in C~~M\cong N\Longrightarrow \delta (M)=\delta (N)$‎ ‎

P3: ‎‎$‎‎\forall M,N,P,Q\in C$

$Dom(M)=Dom(P)\cup Dom(Q)~,~Dom(N)=Dom(P)\cap Dom(Q)$

$\Longrightarrow\delta (M)+\delta (N)\leq \delta (P)+\delta (Q)$‎

P4: ‎$‎‎\neg \exists ‎\{ ‎M_{i}‎\}_{i\in \omega}\subseteq C~~~;~~~\forall i\in \omega~~~(M_{i}\subseteq M_{i+1}\Longrightarrow ‎\delta ‎(M_i)>‎\delta ‎(M_{i+1})‎‎)‎‎$‎

Remark 1. Finding predimension preserving transformations is important for producing various (non-linear) predimension functions and geometries with special properties. An example of such a special geometry is Hrushovski's amalgamation construction for refuting Zilber's Trichotomy Conjecture. The following lemma is about a predimension preserving transformation.

Lemma. Convex functions preserve predimensions. Precisely ‎if ‎$\delta:C‎\longrightarrow ‎\mathbb{R}^{\geq 0}‎$‎ ‎is a‎ ‎predimension ‎function ‎an‎d ‎$‎‎f:‎\mathbb{R}^{\geq 0}‎\longrightarrow‎ ‎\mathbb{R}^{\geq 0}‎$ is a‎ ‎function ‎with ‎the ‎following ‎properties‎:‎

F1: ‎‎$‎‎f(0)=0$‎

F2: ‎‎$‎‎‎\forall x~~~f'(x)\geq 0‎‎‎$‎

F3: ‎‎‎‎$‎‎‎\forall x~~~f''(x)\leq 0‎‎‎$‎‎‎

Then ‎$‎‎fo‎\delta‎:C‎\longrightarrow ‎‎\mathbb{R}^{\geq 0}‎$‎ ‎is a‎ ‎predimension ‎function.‎

Proof. Easy.

Remark 2. ‎The above lemma ‎provides a‎ ‎simple ‎tool ‎for ‎producing a‎ ‎wide ‎range ‎of ‎non-linear ‎predimension ‎functions ‎from a‎ ‎given ‎predimension. ‎For ‎example ‎one ‎can ‎produce a‎ ‎complicated ‎predimension ‎function ‎‎$‎‎‎‎M\mapsto Ln(|M|+1)‎$ using ‎the ‎trivial ‎predimension ‎‎$‎‎M\mapsto |M|$ ‎and ‎the ‎convex ‎transformation ‎‎$‎‎x‎\mapsto ‎Ln(x+1)‎$ ‎on ‎each ‎predimension ‎class ‎‎$‎C‎$‎.

Main Question. Which type of functional convergences can preserve predimension functions? (i.e. The limit of the predimension functions is a predimension itself.)

‎By a simple observation stated in my answer, the pointwise convergence cannot preserve the p4 property of predimensions. ‎

It seems the same "weakness" exists in uniform convergence too. In the below diagram one can see an imaginary situation that each function ‎$‎\delta_n‎$‎ is a predimension on ‎$‎C‎$ ‎(note that each $‎\delta_n‎$ is descending just on finite steps which doesn't violate p4) but ‎$‎\delta‎$ ‎which is ‎the unform limit of the sequence ‎$‎‎\{\delta_n\}_{n\geq 1}$ ‎is not a predimension on ‎$‎C‎$ ‎because it is strictly descending on an infinite increasing chain of structures in ‎$‎C‎$‎‎‎.‎

Question 1. Does the uniform convergence preserve predimensions? ‎

The positive side of the main question is much more important than finding counterexamples for "weak" convergences. Thus if the answer of the question 1 is negative, it is interesting to ask:‎ ‎

Question 2. ‎Is there a "strong" version of functional convergences which can preserve predimensions?

Remark 3. For the answer of the question 2 I am looking for those convergences which work for an arbitrary predimension class and an arbitrary sequence of predimensions on it.

Answer 1

Here is a simple observation which shows the pointwise convergence doesn't preserve predimentions.

Theorem. ‎Pointwise convergence doesn't preserve predimentions. Precisely there is a countable relational first order language ‎$‎\mathcal{L}‎$ ‎and a‎ ‎predimension ‎class ‎‎$‎‎C$ ‎of ‎‎‎$‎‎‎\mathcal{L}‎$-structures ‎and a‎ ‎sequence ‎‎‎$‎‎\{\delta_{n}\}_{n\in \omega}$ ‎of ‎predimension functions ‎on ‎‎$‎C‎$ ‎and a ‎non-predimension ‎‎function ‎‎$‎‎\delta :C‎\longrightarrow ‎‎\mathbb{R}^{\geq 0}‎$‎ ‎such ‎that ‎‎$\forall M\in C~~~‎‎\delta (M)=lim_{n‎\rightarrow ‎‎\infty‎‎}\delta_{n}(M)$‎‎.

Proof. It is easy to see that pointwise convergence preserves the properties P1, P2 and P3. Thus for producing a counterexample we should show that pointwise convergence destroys P4 in some special situations. In order to do this consider $\mathcal{L}=\{(R_j,2)~|~j\geq 1\}$ and a sequence ‎$‎‎\langle G_i~|~i\geq 1\rangle$ of ‎$‎‎\mathcal{L}$-structures ‎(undirected ‎graphs with colored edges) as follows:‎ ‎ ‎$‎‎‎\forall i\geq 1 ~~~Dom(G_i):=\{0,1,\cdots, i\}‎$‎

‎$‎‎‎\forall i,j\geq 1~~~‎‎‎‎‎‎R_j^{G_i}:=\{(m,n)\in (i+1)\times (i+1)~|~m=j~\vee~n=j\}$‎‎‎

Trivially we have ‎$‎‎G_1\subseteq G_2\subseteq G_3\subseteq \cdots$. Now let ‎$‎‎C:=‎\bigcup‎_{i\geq 1}age(G_i)$ ‎and for each ‎$‎‎n\geq 1$‎ ‎define ‎‎$‎‎\delta_n:C‎\longrightarrow ‎\mathbb{R}‎$ ‎to ‎be ‎‎$‎‎\forall M\in C~~~\delta_{n}(M):=‎\frac{n+1}{n+2}|M|-\sum_{j=1}^{n}\frac{|R_{j}^{M}|‎}{2j}$.‎‎

Fact 1. ‎‎$‎C‎$ ‎is a‎ ‎predimension ‎class.‎

Proof. ‎Straightforward.

Fact 2. ‎$‎\forall ‎‎‎n\geq 1~~~\forall M\in C~~~\delta_{n}(M)‎\geq 0‎$‎

Proof. Note that the definition of ‎$‎‎\delta_n$ ‎depends ‎on ‎interpretations ‎of ‎‎$‎‎R_1,\cdots R_n$ ‎so‎ it suffices to prove ‎$\forall n\geq i\geq 1~~~‎\forall ‎M\subseteq G_{i}~~~\delta_{n}(M)‎\geq 0‎$‎ ‎which ‎follows ‎from ‎the ‎following argument:‎ ‎ ‎‎Only ‎the ‎colors ‎‎$‎‎R_1,\cdots , R_i$ ‎are ‎used in ‎‎‎‎$‎G_i‎$ ‎and ‎‎$‎‎M$ ‎is produced ‎by ‎removing ‎‎$‎‎i+1-|M|$ ‎vertexes ‎from ‎$‎G_i‎$‎. ‎By the definition of ‎$‎G_i‎$ ‎i‎t ‎is ‎easy ‎to ‎see that ‎‎$\sum_{j=1}^{n}\frac{|R_{j}^{M}|‎}{2j}\leq |M|-1‎‎$ ‎and ‎so ‎‎$‎‎\delta_{n}(M)=‎\frac{n+1}{n+2}|M|-\sum_{j=1}^{n}\frac{|R_{j}^{M}|‎}{2j}\geq 1-‎\frac{|M|}{n+2}\geq 0$ because $‎‎|M|\leq |G_i|=i+1\leq n+1<n+2$‎.

Fact 3. ‎$\forall M\in C~~~lim_{n‎\rightarrow ‎\infty‎}\delta_{n}(M)< \infty‎$‎

Proof. ‎Note that for each given ‎$‎M‎\in C$ ‎there is a natural number ‎$‎‎i\geq 1$ ‎and a‎ ‎structure ‎‎$‎N\in C‎$ ‎such ‎that ‎‎$‎‎M\cong N\subseteq G_i$. ‎Also‎ just‎ ‎finite ‎number ‎of ‎‎symbols ‎in ‎‎$‎‎\mathcal{L}$ ‎have ‎non-empty ‎interpretations in ‎$‎‎N$‎. ‎Thus ‎‎$‎\forall ‎‎‎‎‎n>i~~~‎\delta‎_{n}(M)=\delta_{n}(N)=‎\frac{n+1}{n+2}|M|-\sum_{j=1}^{i}\frac{|R_{j}^{M}|‎}{2j}‎$. Define $‎‎a:=\sum_{j=1}^{i}\frac{|R_{j}^{M}|‎}{2j}$ so ‎$lim_{n‎\rightarrow ‎\infty‎}\delta_{n}(M)=lim_{n‎\rightarrow ‎\infty‎} (‎\frac{n+1}{n+2}|M|-a)=|M|-a<\infty‎‎$‎‎

‎Fact 4. ‎$\forall n\geq 1~~~\delta_{n}:C‎‎\longrightarrow ‎‎\mathbb{R}^{\geq 0}‎‎‎‎$ ‎is a‎ ‎predimension ‎function.‎ ‎

Proof. ‎By the fact ‎2 ‎the ‎range ‎of ‎each ‎‎$‎‎‎\delta‎_n$ ‎lies ‎in ‎‎$‎‎\mathbb{R}^{\geq 0}‎$‎. ‎Also ‎it ‎is ‎straightforward ‎to ‎check ‎that ‎for ‎fixed ‎‎$‎n‎$ ‎the ‎function ‎‎$‎‎‎\delta‎_n$ satisfies all properties of a‎ ‎predimension.‎

Fact 5. ‎If we define ‎the function $\delta‎$ ‎by‎ $\forall M\in C~~~\delta (M):=‎‎‎‎lim_{n‎\rightarrow ‎\infty‎}\delta_{n}(M)‎‎‎‎$ then it is a well-defined function from ‎$‎C‎$ ‎to ‎‎‎$‎‎‎‎\mathbb{R}^{\geq 0}‎$‎‎ ‎which‎ is not a predimension‎‎.‎

Proof. ‎By the fact ‎3 ‎‎$‎‎\delta$ ‎is ‎well-defined ‎and ‎according to the ‎fact 2 ‎its ‎range ‎lies ‎in ‎‎$‎‎\mathbb{R}^{\geq 0}$. ‎It ‎is ‎not a‎ ‎predimension ‎because ‎it ‎doesn't ‎satisfy ‎the ‎property ‎P4 for ‎the ‎following ‎sequence ‎$‎‎‎\langle ‎H_i~|~i\geq 1‎\rangle‎‎$‎ ‎of ‎structures ‎in ‎‎$‎C‎$. ‎‎

‎$‎‎‎\forall i\geq 1 ~~~H_i:=G_{2i}\setminus\{2i-1\}‎$ ‎ ‎(i.e. The induced subgraph of ‎$‎G_{2i}‎$‎ by removing the vertex labled by ‎$‎‎2i-1$‎).‎ Trivially ‎we have ‎$‎‎H_1\subseteq H_2\subseteq H_3\subseteq \cdots$ ‎but:‎

$‎\forall ‎i\geq 1~~~\delta (H_i)=|H_i|-\sum_{j=1}^{2i}‎\frac{|R_{j}^{H_i}|}{2j}$

$=(2i+1-1)-(‎\frac{1}{1}+‎\frac{2}{2}+‎‎\cdots‎+‎\frac{2i-2}{2i-2}+‎\frac{0}{2i-1}+‎\frac{2i-1}{2i}‎‎‎‎‎‎)=1+‎\frac{1}{2i}‎‎‎$ thus $‎\delta‎(H_1)>‎\delta ‎(H_2)>‎\delta‎(H_3)>‎‎\cdots ‎\geq 1‎$ ‎which ‎violates ‎p4 ‎and ‎completes ‎the ‎proof ‎of ‎the ‎theorem ‎‎‎.

Remark. ‎In order to find a counterexample for the question 1 ‎it ‎seems ‎possible ‎to ‎modify ‎the ‎argument ‎of the ‎above theorem ‎ ‎by ‎transformation ‎of ‎the ‎predimensions $‎\langle ‎\delta_n~|~n\geq 1‎\rangle‎$ ‎using a sequence of suitable ‎convex functions which are uniformly convergent too. (e.g. Each function $f_m(x)=‎\frac{x}{1+mx}‎$ preserves predimensions and ‎also we have $‎‎f_{n}‎\rightrightarrows 0‎$).‎