Let $B$ be a Boolean algebra and $(X,\mathcal{T})$ be the corresponding Stone topological space to $B$ (the space $(\operatorname{Ult}(B),\mathcal{T}_B)$ where the elements of $\operatorname{Ult}(B) $ are the \operatorname{Ult}rafilters on $B$ and $\mathcal{T}_B$ is the topology, generated by the base $s_B(B)$, where $s_B(b)=\{u\in \operatorname{Ult}(B) \mid b\in u\}$. Let $\operatorname{Idl}(B)$ be the set of all ideals on $B$. We have that $I\subseteq B$ is ideal in $B$ iff $\overline{I}:=\{\overline{b} \mid b\in I\}$ is a filter in $B$.

Moreover, $\operatorname{Idl}(B)$ becomes a poset by the set-theoretical inclusion. My question is, are $(\mathcal{T},\subseteq)$ and $(\operatorname{Idl}(B),\subseteq)$ isomorphic (as posets)?

Also, $(\operatorname{Idl}(B),\subseteq)$ is a complete lattice (according to Introduction to Boolean algebras by S. Givant and P. Halmos). If the ideals are isomorphic to the topology as a poset, can the isomorphism be extended to lattice isomorphism?

My idea for a function $f:\mathcal{T}\rightarrow \operatorname{Idl}(B), u=\bigcup\{p_\alpha \mid \alpha\in A\}\mapsto \overline{u}=\bigcap\{\overline{p_\alpha} \mid \alpha\in A\}$ wouldn't work, because it would reverse the inclusions.

  • $\begingroup$ $B=\{0,a,\neg a, 1\}$ has $5$ ideals if I'm not mistaken, and $2$ ultrafilters; hence its Stone space has at most $4$ opens: so there can't be an isomorphism. $\endgroup$ – Max Jul 4 '18 at 16:00
  • $\begingroup$ There are 3 ultrafilers I think - $\{1\}, \{a,1\}, \{\neg a, 1\}$ and therefore 5 open sets in the Stone space - the empty set, all the ultrafilters on $B$ and the 3 elements of the base $\endgroup$ – user3701033 Jul 4 '18 at 16:43
  • $\begingroup$ $\{1\}$ is not an ultrafilter, it's only a filter $\endgroup$ – Max Jul 4 '18 at 16:43
  • $\begingroup$ oh, yes, my mistake, but aren't there only 4 ideals - $\{0\},\{0,a\},\{0,\neg a\}, \{0,a,\neg a, 1\}$ $\endgroup$ – user3701033 Jul 4 '18 at 16:50
  • $\begingroup$ Oh yes, right, my bad. $\endgroup$ – Max Jul 4 '18 at 16:54

It turns out I was wrong. I'll first deal with the case of a finite Boolean algebra to give some idea of what will happen.

For a finite boolean algebra $B$, $b\mapsto V(b)=\{u\mid b\in u\}$ is an isomorphism between $B$ and the open algebra of its Stone space (indeed, any open set is of the form $\displaystyle\bigcup_{i\in I}V(b_i)$ by definition, but $B$ is finite so we may choose $I$ finite, and so this becomes $V(\displaystyle\bigvee_{i\in I}b_i$)). This is actually just an instance of Stone duality, where it so happens that the clopen algebra is isomorphic to the open algebra.

Hence $\mathcal{T}_B$ is in this case isomorphic to $B$.

Now we also have a map $f:B\to \mathrm{Idl}(B)$ defined by $x\mapsto \{y\in B\mid y\leq x\}$. This map is order preserving : if $x\leq y$, $f(x)\subset f(y)$; and if $f(x)\subset f(y)$ then $x\in f(y)$ so $x\leq y$.

Hence $f$ is an order-isomorphism. Finally, to show that $f$ is surjective, pick $I$ an ideal and let $x= \bigvee I$ (which is well-defined as $I$ is finite- actually :) $x\in I$ because $I$ is finite. Hence clearly $f(x) \subset I$ because $I$ is an ideal, and $I\subset f(x)$ by definition.

Thus $\mathcal{T}_B\simeq B \simeq \mathrm{Idl}(B)$.

So indeed my finite "counterexample" had no way of working (and the same is true of any "counterexample"). We'll use what we just did for the general case.

For a general $B$, Stone duality yields an embedding $B\to \mathcal{T}_B$ (just as above actually), and our previous example also yields an embedding $B\to \mathrm{Idl}(B)$.

Both $\mathcal{T}_B$ and $\mathrm{Idl}(B)$ are complete lattices, so there is a chance we may extend this embedding to an isomorphism between the two lattices. Indeed, the image of $B$ is clearly "dense" in both of them (an open subset of the Stone space $O$ is $\displaystyle\bigcup_{x\in O} V(b_x)$ for some $b_x$; and an ideal $I$ is $\displaystyle\bigcup_{x\in I}f(x)$ for the same $f$ as before). We want to extend these embeddings to an isomorphism.

Hence let's proceed as follows:

For a given open set $O$, look at all decompositions $O=\displaystyle\bigcup_{i\in I}V(b_i)$, and compare the corresponding $\displaystyle\bigcup_{i\in I}f(b_i)$. I'll add a technical condition on those decompositions (to ensure that the second bit here is indeed an ideal): I require that $\{b_i, i\in I\}$ be closed under finite joins. Of course taking the closure under finite joins of such a decomposition doesn't change the corresponding open set, so I can make that requirement. This ensures that $\displaystyle\bigcup_{i\in I}f(b_i)$ is indeed an ideal.

Suppose $(b_i)_{i\in I}, (c_j)_{j\in J}$ are two such decompositions.

Let $y\in \displaystyle\bigcup_{i\in I}f(b_i)$. Then for some $i\in I$, $y\leq b_i$. Hence $V(y) \subset V(b_i)$. Hence $V(y) \subset O$ so $V(y) \subset \displaystyle\bigcup_{j\in J}V(c_j)$. Because $V(y)$ is closed, and the Stone space is compact, there are $j_1,...,j_n \in J$ such that $V(y)\subset \displaystyle\bigcup_{k=1}^nV(c_{j_k})$. Hence $V(y) \subset V(\bigvee_k c_{j_k})$ hence $y\leq \bigvee_k c_{j_k}$. But each $c_{j_k}$ is in $\displaystyle\bigcup_{j\in J}f(c_j)$ so their join must be in it too (it's a finite join; and by the technical condition I imposed earlier, this big union is an ideal), hence $y$ is in it as well. Therefore $\displaystyle\bigcup_{i\in I}f(b_i) \subset \displaystyle\bigcup_{j\in J}f(c_j)$. The converse inclusion obviously holds by symmetry.

Hence we have a well-defined map $L:\mathcal{T}_B\to \mathrm{Idl}(B)$.

It's surjective: if $I$ is an ideal, write $I=\displaystyle\bigcup_{b\in I}f(b)$, then $I=L(\displaystyle\bigcup_{b\in I}V(b))$ (because $(b)_{b\in I}$ is a correct decomposition of this open set, as it is closed under finite joins).

It is order-preserving : note that in the above proof of well-definition, what I actually proved was that if $(b_i)$ was an appropriate decomposition of $O$, $(c_j)$ one for $O'$ and $O\subset O'$, then $L(O)\subset L(O')$.

Now assume $L(O)\subset L(O')$, and let $(b_i), (c_j)$ be respective appropriate decompositions of $O,O'$. Then $b_i \in L(O)$, hence $b_i\in L(O')$, hence there is $j$ such that $b_i\leq c_j$. Hence $V(b_i)\subset V(c_j)$. Hence $V(b_i)\subset O'$. Taking a union on $i$ yields $O\subset O'$.

Hence this is an isomorphism.

Is this a complete lattice isomorphism ? Well yes, an order-isomorphism between complete lattices is automatically a lattice-isomorphism.

Let $(L,\leq), (T,\leq)$ two lattices, and $f: L\to T$ an order isomorphism. I'll for instance prove that $f(x)\land f(y) = f(x\land y)$: $f(x\land y)\leq f(x)$ and $f(x\land y) \leq f(y)$ because $f$ is order-preserving. Hence $f(x\land y)\leq f(x)\land f(y)$. Now let $z$ be such that $f(z) = f(x)\land f(y)$ ($f$ is surjective). Then $f(z)\leq f(x)$ so $z\leq x$. Similarly, $z\leq y$. Hence $z\leq x\land y$. Thus $f(z)\leq f(x\land y)$. Conclusion: $f(x\land y) = f(x)\land f(y)$.

Order-isomorphisms preserve everything that can be phrased in terms of the order (where you are right, though, is that an order morphism need not preserve sups and infs, and for those in general it is reasonable to ask the question; but for an isomorphism, there's no need.

EDIT: here's an easier proof of the isomorphism, because the converse of what I called $L$ is easier to make explicit.

Consider, for each ideal $I$, the open set $M(I) = \displaystyle\bigcup_{i\in I}V(i)$. Then $M$ is clearly increasing. Moreover, if $M(I)\subset M(J)$, then for $i\in I$, $V(i)\subset M(J)$ so again by a compactness argument we get finitely many $j_1,...,j_n \in J$ with $V(i)\subset \bigcup_k V(j_k)$ and so $V(i)\subset V(\bigvee_k j_k)$, and hence $i\leq \bigvee_k j_k$, implying $i\in J$.

Thus $M$ is an embedding. Moreover, for $O=\displaystyle\bigcup_{i\in I}V(b_i)$, denoting $P=$ the ideal generated by $\{b_i, i\in I\}$ we get $O= M(P)$, as we may describe $P$ in terms of finite meets of the $b_i$'s.

This proof is quicker, but I found the other one because of the direction I set things up.

  • $\begingroup$ Why does the technical condition guarantee that the union is an ideal? Particularly, how does it guarantee the joint closure for 2 elements of $\cup_{i\in I} f(b_i)$. Also, why doesn't it change the representation of an open set? $\endgroup$ – user3701033 Jul 4 '18 at 19:14
  • $\begingroup$ Well if $x\leq b_i, y\leq b_j$, then $x\lor y \leq b_i \lor b_j$, but $b_i\lor b_j = b_k$ for some $k$ (that's the technical condition), hence $x\lor y \in f(b_k)$: we indeed get an ideal ( I assume it's clear how the union is downwards closed ? ). $\endgroup$ – Max Jul 4 '18 at 19:27
  • $\begingroup$ Why it doesn't change the open set to add all the finite meets: Let $O=\cup_{i\in I}V(b_i)$ and let $(c_j)$ be the new family (that is, with finite meets added). Then clearly $O\subset \cup_{j\in J}V(c_j)$, because each $b_i$ is a $c_j$. Conversely, let $u\in V(c_j)$. $c_j$ is the meet of $b_{i_1},...,b_{i_n}$. Thus $b_{i_1}\lor...\lor b_{i_n} \in u$. But $u$ is an ultrafilter: if a finite meet is in an ultrafilter, then one of the elements of the meet is in the ultrafilter (that's a classical property). Hence $u\in b_{i_k}$ for some $k$, thus $u\in O$ $\endgroup$ – Max Jul 4 '18 at 19:30
  • $\begingroup$ Thank you, took me a while to digest the proof, I think you swapped the notions of meet and joint (or I did), but it's all ok. $\endgroup$ – user3701033 Jul 4 '18 at 20:11
  • $\begingroup$ I probably did, I always get them confused (in my mind "meet" should be the supremum because if people meet they are more numerous; but when they are joined they get closer and so "smaller" - I guess it's a bad idea to have images in your head about this sort of stuff). I'll try to correct it $\endgroup$ – Max Jul 4 '18 at 20:17

In the notes by KP Hart here, it's exercise 3 on page 6:

If $I$ is an ideal in $B$ then $f(I):= \bigcup \{s_B(b) : b \in I\}$ is open in the Stone space, and vice versa if $O$ is open in the Stone space $g(O) = \{b \in B: s_B(b) \subseteq O\}$ is an ideal in $B$. The first is trivial, as open sets are closed under unions and the second is almost immediate and follows from the definitions quite simply. The only minor thing is whether we allow for the whole Stone space and the empty set (which depends on your exact definition of ideal). $f(B) = \operatorname{Stone}(X)$, but $B$ is a "trivial ideal", $F(\emptyset) = \emptyset$ (and some authors require ideals to be non-empty, althoug the empty set does voidly satisfy the two defining axioms for ideals $I$ of $B$:

  1. $\forall x\in I: \forall y\in B : y \le x \to y \in I$.
  2. $\forall x,y \in I: x \lor y \in I$.

It's clear that $f$ and $g$ are monotonous and each other's inverse between the set of ideals under inclusion and the topology on the Stone space under inclusion. So they are order isomorphisms hence lattice isomorphisms too (@Max' answer goes a little into that).

The correspondence $F \to \bigcap \{s_B(b): b \in F\}$ is one between the filters on $B$ and the closed sets of the Stone space. This one reverses inclusion: a larger filter gives a smaller closed set. This is exercise 4 in the mentioned notes.

This is all just Stone duality theory: points (in Stone spaces) correspond to ultrafilters (in BA), closed sets to filters, open sets to ideals, subspaces to quotients, continuous images to homomorphisms. etc. etc.


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