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Let us call a lattice $(L,\leq)$ interval-isomorphic if for all $a<b \in L$ we have $L \cong [a,b]$, where $[a,b]=\{x\in L:a\leq x\leq b\}$.

Are there $2^{\aleph_0}$ pairwise non-isomorphic interval-isomorphic lattices on the ground set $\omega$?

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    $\begingroup$ Are there $3$? Maybe I'm just being an idiot here, but the only ones I can think of are the countable atomless Boolean algebra, and the linearly ordered set $\mathbb Q \cap [0,1]$. $\endgroup$ – Will Brian Sep 2 '19 at 16:16
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    $\begingroup$ Is this property preserved under direct products? $\endgroup$ – Sam Hopkins Sep 2 '19 at 23:29
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    $\begingroup$ @SamHopkins Usually not. Since $L$ embeds as an interval into $L\times L$ (say, by $x\mapsto(x,a)$ for a fixed element $a$), $L\times L$ has the property only if and only if (1) $L$ has the property and (2) $L\simeq L\times L$. $\endgroup$ – Emil Jeřábek supports Monica Sep 4 '19 at 8:28
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    $\begingroup$ The generic (or "typical") $\{0,1\}$-lattice (the Fraisse limit of all finite $\{0,1\}$-lattices) is another example. $\endgroup$ – Goldstern Sep 5 '19 at 17:56
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    $\begingroup$ Gabor Czedli has several papers about such lattices, which he calls "fractal": math.u-szeged.hu/~czedli/listak/publist.html $\endgroup$ – Goldstern Sep 5 '19 at 17:57
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This is not a solution, but a description of a potential solution. I conjecture that ``Yes, there are $2^{\aleph_0}$-many self-similar countable lattices.''

I am pretty sure that any irreducible continuous geometry has the property that any two proper nontrivial intervals are isomorphic to each other. I am less confident that the whole lattice is isomorphic to any of its intervals, but it seems plausible.

But continuous geometries are uncountable, so they do not answer the question. Nevertheless, they have countable dense sublattices. One way to construct such a countable dense sublattice was described by von Neumann:

Let $\mathbb D$ be a countable division ring and let $PG(\mathbb D, 2^n-1)$ be $(2^n-1)$-dimensional projective space over $\mathbb D$ equipped with a dimension function normalized so that the dimension of the whole space is $1$. There are dimension-preserving embeddings

$PG(\mathbb D, 1)\leq PG(\mathbb D, 3)\leq PG(\mathbb D, 7)\leq \cdots .$

Let $PG(\mathbb D, \omega)$ be the direct limit of these embeddings. This is a countable, complemented, modular lattice with normalized dimension function taking values in the dyadic rationals. It is also a metric space with metric given by $d(a,b) = \dim(a\vee b)-\dim(a\wedge b)$.

The continuous geometry over $\mathbb D$, $CG(\mathbb D)$, is the (uncountable) metric completion of $PG(\mathbb D, \omega)$. But let's hold off and not complete this lattice. Instead, let's stay with the countable and very homogeneous modular lattice $PG(\mathbb D, \omega)$. I think it is a good candidate for a countable lattice isomorphic to each of its proper intervals.

$PG(\mathbb D, \omega)$ encodes information about $\mathbb D$. I do not know the circumstances when $PG(\mathbb D, \omega)\cong PG(\mathbb D', \omega)$ implies $\mathbb D\cong \mathbb D'$, but in Birkhoff's paper Von Neumann and Lattice Theory the author write ``Curiously, the real and quaternion continuous geometries are isomorphic.'' This indicates that the related implication $CG(\mathbb D)\cong CG(\mathbb D')$ implies $\mathbb D\cong \mathbb D'$ can fail, but that Birkhoff found it curious when it happened, even for closely related division algebras. I'm not sure what this says about the implication for $PG(\mathbb D, \omega)$ in place of $CG(\mathbb D)$.

But let's suppose that it is often the case that $\mathbb D\not\cong \mathbb D'$ implies $PG(\mathbb D, \omega)\not\cong PG(\mathbb D', \omega)$ There surely ARE $2^{\aleph_0}$-many choices for $\mathbb D$. For example, $\mathbb D = \mathbb Q[\sqrt{p_1}, \sqrt{p_2},\ldots]$, where we adjoin some set of square roots of primes to $\mathbb Q$, is a countable field, and fields constructed this way are not isomorphic if they are contructed from different sets of primes.

Thus, I propose the lattices $PG(\mathbb D, \omega)$ for countable fields $\mathbb D$ as likely candidates to solve this problem.

[In the comments to the question it is noted that the countable atomless Boolean algebra is an example of a countable self-similar lattice. This example may be thought of as arising from the above construction starting with $\mathbb D$ equal to the ``field of one element''. By this I mean replace $PG(\mathbb D, 2^n-1)$ in von Neumann's construction with the power set lattice with $2^n$ atoms.]

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