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?



Set of ideals of the set of finite subsets of $\mathbb{N}$
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.


