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In any lattice $L$, an ideal is a subset $I\subseteq L$ that is downward closed, and moreover $a,b\in I$ implies $a\vee b\in I$. We denote by ${\cal I}(L)$ the set of ideals of $L$, ordered by set inclusion.

Let $F(\mathbb{N})$ denote the lattice of finite subsets of $\mathbb{N}$, ordered by set inclusion. Is ${\cal I}(F(\mathbb{N}))$ isomorphic to some "familiar" lattice?

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There is quite an easy isomorphism that you may just have overlooked: ${\cal I}(F(\mathbb{N})) \cong {\cal P}(\mathbb{N})$, where ${\cal P}(\mathbb{N})$ denotes the powerset of $\mathbb{N}$. The isomorphism is very natural (not in a strict mathematical sense):

$$ I\in {\cal I}(F(\mathbb{N})) \mapsto \bigcup I.$$

Let's call this isomorphism $\Phi$ and we set $L:= F(\mathbb{N})$.

It's easy to see that $\Phi$ is a lattice homomorphism, so let's concentrate on injectivity and surjectivity.

Injectivity of $\Phi$. Take $J_1, J_2 \in {\cal I}(L)$ and suppose $\Phi(J_1)=\Phi(J_2)$. Let $e=\{n_1,\ldots,n_r\}\in J_1$. Since $\bigcup J_1=\bigcup J_2$ we get $n_i\in \bigcup J_2$ for all $i=1,\ldots,r$. Now since $J_2$ is an ideal, this implies $\{n_i\}\in J_2$ for all $i$, which in turn implies

$$\{n_1\}\cup\ldots\cup\{n_r\} = \{n_1,\ldots,n_r\} = e \in J_2.$$ So we proved that $J_1\subseteq J_2$, and a similar argument shows $J_2\subseteq J_1$. So $J_1=J_2$ and $\Phi$ is injective.

Surjectivity of $\Phi$. For $A\in{\cal P}(\mathbb{N})$ let $J:=\{e\in L: e\subseteq A\}$. Clearly $J$ is an ideal and $\Phi(J)=A$, proving that $\Phi$ is surjective.

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    $\begingroup$ Good answer. A more high-level way to see this isomorphism is through the duality between join-semilattices and algebraic lattices. Algebraic lattices are described in detail in the book A Course in Universal Algebra by Burris and Sankappanavar. Of course, the duality between join-semilattices and algebraic lattices is a part of a larger framework of dualities between ordered sets and complete lattices. $\endgroup$
    – Joseph Van Name
    Commented Jun 1, 2015 at 17:11
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    $\begingroup$ If $X$ is a join-semilattice, then $\mathcal{I}(X)$ is an algebraic lattice, and if $L$ is an algebraic lattice, then the set $K(L)$ of compact elements in $L$ is a join-semilattice. If $L$ is an algebraic lattice, then $L\simeq\mathcal{I}(K(L))$, and if $X$ is a join-semilattice, then $X\simeq K(\mathcal{I}(X))$. The lattice $P(\mathbb{N})$ is an algebraic lattice and $K(P(\mathbb{N}))=F(\mathbb{N})$. Therefore, $\mathcal{I}(F(\mathbb{N}))=I(K(P(\mathbb{N})))\simeq P(\mathbb{N})$. See newton.case.edu/papers/algfca.pdf for a proof of this duality. $\endgroup$
    – Joseph Van Name
    Commented Jun 1, 2015 at 17:12
  • $\begingroup$ Very nice @JosephVanName - thanks for pointing out this approach! $\endgroup$
    – Dominic van der Zypen
    Commented Jun 2, 2015 at 6:58

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