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Do there exist connected proper smooth $\mathbb{C}$-schemes $X_i$ ($\forall i\in \mathbb{Z}_{>0}$) with $\mathrm{dim}_{\mathbb{C}}X_i=i$ such that $X_i$ admits an immersion into $X_{i+1}$ and any connected proper smooth $\mathbb{C}$-scheme admits an immersion into $X_j$ for some $j\in \mathbb{Z}_{>0}$? If this is not possible for $\mathbb{Z}_{>0}$, is this possible for $\mathbb{Z}_{>n}$ for some $n$?

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This is too long for a comment. You can arrange this by blowing up.

Let $(Y_i,y_i)_{i\in \mathbb{Z}_{>0}}$ be a countable collection of connected, proper, smooth $\mathbb{C}$-schemes with specified closed point $y_i$ such that every connected, proper, smooth $\mathbb{C}$-scheme admits a closed immersion in the open $Y_i\setminus\{y_i\}$ For some choice of $y_i$. In my previous answer, each $Y_i$ is the total space of a fibration where every connected, proper, smooth $\mathbb{C}$-scheme is isomorphic to one of the fibers. Thus, you can repeat each of those schemes twice, and on each of the two copies, choose specified closed points that are in two distinct fibers. Denote by $n_i$ the dimension of $Y_i$.

Now define $Z_{i,n_i}$ to be the blowing up of $Y_i$ at $y_i$. The exceptional divisor is isomorphic to $\mathbb{P}^{n_i -1}$. Choose a complete flag of projective subspaces in the exceptional divisor, and denote by $Z_{i,m}$ the term with dimension $m$ for every $m=0,\dots,n_i-1$. Define the sequence $r_i$ inductively by $r_1=n_1$ and $r_{i+1}=r_i + n_{i+1}$.

For each $i$, define $X_{r_i}$ to equal the product $Z_{1,n_1}\times_{\text{Spec}\ \mathbb{C}} \dots \times_{\text{Spec}\ \mathbb{C}} Z_{i,n_i}$. For every integer $m=0,\dots,n_i$, define $X_{r_{i-1}+m}$ to equal the inverse image of $Z_{i,m}$ under the projection, $$X_{r_i}\to Z_{i,n_i}.$$

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  • $\begingroup$ can we also require that for each $i$ there exists a $\mathbb{C}$-immersion $X_i\rightarrow X_{i+1}$ inducing an isomorphism on $\mathbb{Z}$-singular cohomology of the associated complex analytic spaces (in cohomological degrees$\leq 2i$)? $\endgroup$
    – user74900
    Mar 30 '19 at 10:43
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    $\begingroup$ @AknazarKazhymurat. Certainly the construction above does not give that. The problem here, and with the previous MathOverflow question, is that the stack in groupoids of proper, smooth schemes is not an algebraic stack (I explain why in my article on "Artin's axioms, composition and moduli spaces"). Thus, each of these constructions of "countable families" that represent each geometric point are ad hoc. There is no reason for these ad hoc constructions to have any good properties. $\endgroup$ Apr 1 '19 at 11:26

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