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Let $(P,\leq)$ be a countably infinite poset with the property that whenever $a<b\in P$ then $P\cong [a,b]$.

Question. If $P$ does contain elements $a,b$ with $a<b$, does this imply that $P \cong [0,1]\cap\mathbb{Q}$, or that $P$ is isomorphic to the nonzero countable atomless Boolean algebra?

Note. Thanks to user @YCor for suggestions to make this a better question.

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  • $\begingroup$ Thanks - I will edit the question to avoid this special case! $\endgroup$
    – Dominic van der Zypen
    Nov 9, 2017 at 9:50
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    $\begingroup$ The atomless Boolean algebra should also work. $\endgroup$
    – Asaf Karagila ♦
    Nov 9, 2017 at 10:06
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    $\begingroup$ I guess this is the poset of clopen subset in a Cantor set (which is indeed the unique atomless nonzero countable Boolean algebra). $\endgroup$
    – YCor
    Nov 9, 2017 at 10:38
  • $\begingroup$ Thanks for your answers, Asaf & YCor. Do you want to post it as an answer? Or should I perhaps delete the post? $\endgroup$
    – Dominic van der Zypen
    Nov 9, 2017 at 11:03
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    $\begingroup$ @YCor Your example of the computable sets modulo finite is the same as the (unique) countable atomless Boolean algebra. $\endgroup$
    – Joel David Hamkins
    Nov 9, 2017 at 11:19

3 Answers 3

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How about this. Let $P$ have "levels" indexed by $\mathbb{Q} \cap [0,1]$. Levels $0$ and $1$ each contain a single element, all other levels contain a countably infinite set of elements. An element at level $a$ lies below one at level $b$ if and only if $a <b$. It's clearly not a Boolean algebra, but also clearly has the desired property.

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    $\begingroup$ Indeed, $P$ is isomorphic to any interval in the lexicographic product $\mathbb{Q}\times A$ where $A$ is a countably infinite discrete poset (=antichain). A similar construction will work in $B\times A$ where $B$ is the countable atomless Boolean algebra, and also when $A$ is replaced with a finite antichain of any given cardinal. $\endgroup$
    – YCor
    Nov 9, 2017 at 14:32
  • $\begingroup$ (Sequel of my previous comment) Maybe some more general construction works when $A$ is chosen of variable cardinal. This should encompass Joel's construction as well. $\endgroup$
    – YCor
    Nov 9, 2017 at 14:37
  • $\begingroup$ Namely, choose a subset $J$ of $\{1,2,\dots,\infty\}$. Choose a function $f_J:\mathbf{Q}\to J$ such that every fiber of $f$ is dense ($f$ is unique modulo the action of $\mathrm{Aut}(\mathbf{Q},\le)$ if I'm correct). Define $X_J=\{(x,n):x\in\mathbf{Q},1\le n< f_J(x)+1\}$ with the lexicographic order where the second factor (integers with infinity) is viewed as an antichain. Then all intervals in $X_J$ are isomorphic; choose one, say $Y_J=[(0,1)(1,1)]$. Nik's example should be isomorphic to $Y_{\{\infty\}}$ while Joel's example is maybe $Y_{\{1,3\}}$. $\endgroup$
    – YCor
    Nov 9, 2017 at 14:46
  • $\begingroup$ If $X$ is a poset, we can define an equivalence relation on $X$ by $x\simeq y$ if for all $z$, $x\le z\Leftrightarrow y\le z$ and $z\le x\Leftrightarrow z\le x$. Clearly equivalences classes form antichains, and we can say that $X$ is reduced if these are only singleton. The quotient $\bar{X}$ of any poset by any this equivalence relation is reduced. Then we can ask about the original question assuming $P$ reduced. $\endgroup$
    – YCor
    Nov 9, 2017 at 14:56
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The generic (or random) bounded partial order $R$, i.e., the Fraïssé limit of all finite bounded orders $(P,<,0,1)$ (with $0=\min P$, $1=\max P$), is fractal. It is saturated, so no two incomparable elements have a least upper bound.

Similarly, the generic (locally finite) bounded lattice $L$ is isomorphic to each of its proper intervals. It contains every finite lattice as a sublattice, and hence does not satisfy any laws (other than the laws satisfied by all lattices). In particular, $L$ is not distributive and not even modular.

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  • $\begingroup$ Does it have the property I call "reduced" in my comment to Nik's answer? (I call a poset $X$ non-reduced if there exist $x\neq y$ such that for all $z$, $z\le x\Leftrightarrow z\le y$ and $x\le z\Leftrightarrow y\le z$). $\endgroup$
    – YCor
    Nov 9, 2017 at 21:30
  • $\begingroup$ For any two incomparable $x,y$ there will be infinitely many $z\le x$ which are not $\le y$. This is true both for the lattice and for the partial order. $\endgroup$
    – Goldstern
    Nov 9, 2017 at 21:35
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Here is another construction. Start with $0<1$ and add an antichain of three points $a$, $b$, $c$ between them, making a copy of $M_3$. Next, we iteratively ensure the interval-isomorphism property by adding new points to each new interval we created. That is, whenever we add a new point to the overall order, then we also add a copy of that point into all the resulting intervals that we have created. In countably many steps, this will make a (non-linear) countably infinite partial order with the interval-isomorphism property, but it is not a Boolean algebra, since $a$ has no complement, and indeed, it has a copy of $M_3$ and hence is not a distributive lattice.

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  • $\begingroup$ The resulting order has a tri-fold fractal nature. Probably it would make a nice picture, if someone is handy with that, since you add the three points between $0$ and $1$, and then three points in all those six intervals, and then three points in all the resulting smaller intervals, and so on. $\endgroup$
    – Joel David Hamkins
    Nov 9, 2017 at 14:47
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    $\begingroup$ It seems to me that this poset is actually a lattice, and in fact a modular lattice. That suggests an analogous construction that begins with the minimal non-modular lattice (a chain of 4 points and a chain of 3, with their top points identified and their bottom points identified); I would guess that the result of your construction would still be a lattice, but of course not modular. $\endgroup$
    – Andreas Blass
    Nov 9, 2017 at 14:53
  • $\begingroup$ Yes, that seems right to me. It seems you can start with any finite order between $0$ and $1$ and then do the same iterative thing, and you will never add new points except inside intervals of the original order. $\endgroup$
    – Joel David Hamkins
    Nov 9, 2017 at 15:23

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