# What lattices are isomorphic to $R^{N}$ for some $N$, equipped with the product order?

What lattices are isomorphic to $\mathbb{R}^{N}$ for some $N\in \mathbb{N}$, equipped with the canonical order?

Remark:

When I say $\mathbb{R}^N$, I don’t mean it to be a vector space. Instead, I refer to the Cartesian product of $N$ totally ordered sets, ${(A_i, \geq_i)}_{i=1}^{N}$ each of which is isomorphic to $\mathbb{R}$ equipped with its canonical order. Therefore, the canonical order on the Cartesian product of $A_1 \times A_2 \times...A_N$ operates as the following:

For $x=(x_1,x_2,...,x_N)$ and $y=(y_1,...,y_N)$ both in $A_1 \times A_2 \times...A_N$:

$x \succeq y$ if for all $i$, $x_i \geq_i y_i$.

Also, $x\succ y$ if $x\succeq y$ and there is at least one $i$ such that $x_i >_i y_i$.

For example, we know that the lattice must be distributive, since $(A_1 \times A_2\times ... A_N, \succeq)$ is a distributive lattice. Also, we know that the partial order must be dense. Moreover, the lattice must be unbounded, Dedekind complete, and separable in it's order topology. But I'm looking for the simplest necessary and sufficient conditions. Thanks!

• I'm unsure what kind of answer you are seeking, since being isomorphic to $\mathbb{R}^N$ seems like an intrinsic necesssary and sufficient condition. Perhaps you are looking for unexpected examples? Or you want simple-to-check necessary and sufficient criterion? – Joel David Hamkins Sep 18 '16 at 21:20
• Thanks for your comment. I'm looking for rather subtle yet simple conditions, like the ones that I explained in my remark. For example, it's well known that for a totally ordered set to be isomorphic to R with the canonical order, it's necessary and sufficient for it to be unbounded, order-dense, separable in its order topology, and Dedekind complete. I'm looking for a result with this flavor. Again, to reemphasize, I'm not looking for a particular N, as long as some N exists so that the above holds. Hope this explains my intentions. – pedram Sep 18 '16 at 21:37
• OK. One obvious necessary condition is that if $x<y$, then there is $x'$ between, such that the interval $[x,x']$ is linear, dense and complete (that is, like an interval in $\mathbb{R}$). – Joel David Hamkins Sep 18 '16 at 21:52
• That's right. I also had that in mind, but I didn't want to assume it as a condition by itself. Do you think there are other conditions that imply that property? – pedram Sep 18 '16 at 21:55
• I was able to prove that if a totally ordered interval, $(x,y)$ exists, and every totally ordered interval $(x,y)$ is a subset of another totally ordered interval $(z,t)$ where $z \prec x$ and $y \prec t$, then assuming the ordered-dense property, distributivity, conditional completeness, and unboundedness, give us $R^{|I|}$ for some set $I$. However, I prefer not to assume the existence of a totally ordered interval. – pedram Sep 18 '16 at 22:03

It is possible for me to give a transparent order-theoretic characterization of the lattice $\mathbb R^N$ if you allow me to phrase it in terms of functionals.
The Priestley dual of $\mathbb R^N$ is the coproduct of $N$ copies of the Priestley dual of $\mathbb R$. The Priestley dual of $\mathbb R$ is a chain whose elements are the lattice homomorphisms $\chi\colon \mathbb R\to {\mathbf 2}$. These homomorphisms are constant $0$ or $1$, or else they map $(\infty,r)$ to $0$ and $[r,\infty)$ to $1$ or $(-\infty,r]$ to $0$ and $(r,\infty)$ to $1$. The order is the pointwise order on functionals. You can visualize this as the lexicographic product of $\mathbb R$ with the $2$-element chain (a copy of the reals with each element doubled), with a new top and bottom attached. The topology is defined by taking as a basic clopen set each $U_r = \{\chi\;|\;\chi(r)=1\}$ for $r\in \mathbb R$. It is not hard to express this topology on functionals entirely in terms of the order on functionals, and it is not hard to recognize this topological chain.
The coproduct of $N$ copies takes the disjoint union of the copies and identifies the top elements of the copies and identifies the bottom elements.
(So the characterization of $\mathbb R^N$ is: it is distributive and its dual looks like $\ldots$.)