I have just discovered that the set of ideals on an **infinite** join-semilattice is a boolean algebra (moreover it is a complete atomistic boolean algebra).

For me, it is a very counter-intuitive theorem (after all the set of ideals “should” not be boolean algebra, except of the finite case). Is it worth to call it a paradox?

See this link for details:

http://math.stackexchange.com/questions/1475328/when-ideals-are-a-boolean-algebra

Beyond being just a classroom curriculum requirement, Farley knows firsthand the importance of applying advanced mathematics to make the world a safer place. In fact, he actively applies his interests in lattice theory and the theory of ordered sets in the fight against terrorism. His work has been utilized in a number of ways and recognized both nationally and internationally.

Farley earned a degree in mathematics from Harvard University where he graduated summa cum laude. He then went on to Oxford University, United Kingdom where he obtained a Doctorate of Philosophy in mathematics. He also won the Senior Mathematical Prize and Johnson University Prize, Oxford’s highest mathematical awards. Farley has countless other awards and recognitions attributed to his many achievements in addition to these.

**What inspired you to pursue working with mathematical methods in counterterrorism?**

“I became interested in developing mathematical methods in counterterrorism around the time I was forced to flee the state of Tennessee, leaving many of my possessions behind, after receiving a few dozen death threats from supporters of the founder of the Ku Klux Klan. [Later on] I arrived at Massachusetts Institute of Technology as a visiting associate professor of applied mathematics and saw a flier for a talk entitled ‘Modeling the Al Qaeda Threat.’ The speaker was a Gordon Woo of Risk Management Solutions. It was, like all his talks, stimulating, and it led me to draft my first paper on this topic.”

**What are the responsibilities of your current role? **

“I am supervising a student from Iran, who is trying to discern the structure of the ‘perfect’ terrorist cell. [Recently] I met with a student about applying lattice theory to the control of teams of robots, with an eye toward applying lattice theory to military drones.”

**What is your favorite part about your daily duties?**

“My favorite part is that my work has actually been applied by the Ministry of National Security in Jamaica.”

**How has your education and training prepared you for your current role? **

“I am an expert in the theory of ordered sets. I model terrorist cells using ordered sets.”

**What do you do to continue your education and training?**

“I continue to learn algebra. [Recently] I read part of the book, *A Term of Commutative Algebra*, by Altman and MIT professor Kleiman.”

*Laura Catherine Hermoza has a lifelong love for writing. In addition to serving as a contributor to various media publications, she is also a published novelist of several books and works as a proofreader/editor. LC resides in Baltimore County.*

The 319th La Trobe University podcast interview is with Brian Davey, Emeritus Professor from the Department of Mathematics and Statistics at La Trobe University. He also writes mathematical songs and has some pretty mad skills with the devilsticks.

He joins the La Trobe University podcast to discuss order and lattice theory, a topic which he’s regularly researched and taught. He’s also the author of *Introduction to Lattices and Order* (with Hilary Priestley), which is published by the Cambridge University Press. You can also listen to him talk about three-valued logic. Have a question for Brian? Leave a comment below.

There’s lots more content like this in our Podcast Interview series. For more on this topic, check out the Mathematics collection in iTunes U or the RSS feed, and there’s lots more at La Trobe University on iTunes U.

]]>Recall that a partially ordered set (shortened as *poset*) is a pair where is a set and is a reflexive, transitive, and antisymmetric binary relation on . A directed acyclic graph, where if and only if there is a directed path from to , is a prime example of a poset.

If is a poset and if is defined as if and only if , then is a poset too: just think of the original poset, turned upside down—or, equivalently, to the order graph with all the edges reversed.

Given two posets , , a function is said to be *monotone* if implies . If is monotone as a function from to , then it is also monotone as a function from to : this simple observation often allows to make proofs in only one direction.

Let be a poset and let . An *upper bound* for is an element such that for every . A subset of may have a *least upper bound*, that is, an upper bound such that, if is any upper bound for , then : in this case, such least upper bound is usually indicated as . Dually, a *lower bound* for is an such that for every , and a *greatest lower bound* is an element such that, if is a lower bound for , then .

A partially ordered set such that every subset of have both a greatest lower bound and a least upper bound , is called a *complete lattice*.

It follows from the definition that a complete lattice is nonempty. In fact, it must contain at least an element and an element , although these two may coincide.

Power sets, ordered by inclusion, are complete lattices: the greatest lower bound of a family of subsets is the intersection of the subsets, and the least upper bound is the union. The real line is not a complete lattice: hovewer, any closed and bounded interval of the real line is a complete lattice. Moreover, if is a complete lattice, and with then the *interval* is a complete lattice: if , then for every , thus too.

Finite boolean algebras are complete lattices. Infinite boolean algebra are only guaranteed to have finite meets and joins: indeed, the boolean algebra of the recursive subsets of is not a complete lattice. To see this, let be a recursively enumerable subset of which is not recursive: then for a suitable total recursive function we have .

The following statement was proved by Knaster in 1928 for power sets, then by Tarski in 1955 for arbitrary complete lattices.

**Knaster-Tarski fixed point theorem.** *Let be a complete lattice with top element and bottom element ; let be a monotone function, and let . Then:*

*is a complete lattice.**The smallest fixed point of satisfies*

*and the largest fixed point of satisfies*

*If is finite, then the sequence defined by and for converges to in finitely many iterations, and the sequence defined by and**for*converges to in finitely many iterations.

Observe that is not, in general, a full sublattice of : the least upper bound of in may not coincide with the least upper bound of in . To understand why, let where the natural ordering of is extended by putting and for every . Let be defined by and if : then the greatest upper bound of is in , but in .

*Proof.* We first prove point 2. Call the set of *post-fixed points*, and its least upper bound. Then for every we have , thus also , i.e., is an upper bound for : then , so that the greatest lower bound of the set of post-fixed point is itself a post-fixed point! But if is post-fixed, then is post-fixed as well as is monotone: as is by construction the least upper bound of the set of the post-fixed points, . By antisymmetry of the order relation, , i.e., is actually a fixed point: but if the least upper bound of the overfixed points is a fixed point, then it cannot help but be the greatest fixed point as well! Dually, has a least fixed point, which is the greatest lower bound of the set of *pre-fixed* points.

We now use point 2 to prove point 1. Let and let . By monotonicity of , for every and we have , so that too: as clearly for , is also a monotone function from to itself. As the latter is a complete lattice, has a least fixed point in it: if satisfies for every , then it also satisfies , thus also . Hence, this is the least upper bound of in , although it might happen that . Dually, has a greatest lower bound in , which might be strictly greater than the greatest lower bound of in .

Finally, suppose is finite. Then the sequence where and also has finitely many elements, so there exist and such that . But by construction, so and . If too, then : then is a fixed point and is greater than every fixed point, so ti must be the greatest fixed point . Dually, if and , then , the least fixed point, for some .

The Knaster-Tarski theorem can be used to define specific constructions as least or greatest fixed points of monotone functions over complete lattices: if the latter are finite, then it also provides an algorithm for their construction. An example of this is *bisimilarity*. Recall that a *bisimulation* on a family of processes is a binary relation such that, for every :

- if then there exists such that and ; and
- if then there exists such that and .

Two processes are *bisimilar* if there is a bisimulation that contains them. As any union of bisimulations is a bisimulation, whatever is, there exists a largest bisimulation on , called bisimilarity, and usually indicated by .

If is finite, then bisimilarity can be constructed via the Knaster-Tarski fixed point theorem. In fact, consider the powerset as a complete lattice according to inclusion, and define a function by saying that if and only if conditions 1 and 2 of bisimulation are satisfied for . Then not only is monotone, but bisimulations are precisely those such that : hence, bisimilarity, which is the least upper bound of the family of bisimulations, is precisely the greatest fixed point of !

Coming back to the Schröder-Bernstein theorem: let and be injective functions. Then

is a monotone function on the complete lattice . Surely is a bijection. By the Knaster-Tarski theorem, there exists such that : but for such we have , so that is a bijection too! Define then a bijection as follows:

- if , let ;
- otherwise, let be the unique such that .

Observe that the standard proof is obtained by explicitly constructing a fixed point for .

We conclude with a very interesting converse to the Knaster-Tarski theorem, proved by Anne Davis in 1955, in an article that appeared on the same issue as Tarski’s—actually, immediately after it! Recall that a *lattice* is a poset where every *finite* subset has a least upper bound and a greatest lower bound.

**Davis’ characterization of complete lattices.** *Let be a lattice. Suppose that every monotone function has a fixed point. Then is a complete lattice.*

An interesting exercise, is to construct a proof of Davis’ theorem by showing that meets and joins of arbitrary sets can be seen as fixed points of suitable monotone functions.

]]>As an ex-lattice gauge theorist myself I think there are some aspects of it that people working on more sexy subjects such as quantum gravity would benefit from understanding better. In particular they should understand how the phase diagram of QCD at high temperature and density is being charted using these non-perturbative methods. The reason they need to know this is that a similar phase structure should exist in quantum gravity and there is likely to be a strong (but approximate) correspondence through AdS/CFT duality that relates quantum gravity to a QCD-like theory.

In the QCD theory of the strong interactions there is believed to be a temperature known as the Hadgdorn temperature above which nuclear matter breaks down into a quark gluon plasma. This happens at around 10 billion degrees Kelvin. In quantum gravity according to string theory (if you don’t like string theory dont switch off, this is just a short diversion) there is another Hagdorn temperature at around the Planck scale. That’s about 10^{32} degrees Kelvin. What happens there?

According to string theory the length of strings becomes very large and effectively the concept of the string breaks down. Sometimes string theorists call this the topological phase of string theory because they think that spacetime loses its geometry in the hotter phase. The truth is that not much is known about what really happens because most of string theory is based on perturbative calculations and phase transitions are very non-perturbative. What might happen is that not only geometry of space-time is lost but topology too. In that case it should be called the non-topological phase, or pregeometric phase. To put it another way, spacetime evaporates. Even if you don’t believe in string theory you might still consider this possibility. Some non-string theorists talk about geometrogenesis which is the process of cooling from the high temperature pregeometric phase to the more familiar geometric phase at the start of the big bang.

For now we can get some feel for the phase structure of quantum gravity by looking at the phase structure of QCD which brings me to one of the ICHEP talks from yesterday. However I’ll do that in a separate post in case people get confused and think it was about quantum gravity.

]]>Let’s denote where is a strong partitioning an element of the complete lattice . Our conjecture is trivially equivalent to the statement that is closed under arbitrary meets and joins.

That is closed regarding any joins is obvious. To finish proving the conjecture we need to show that is closed under arbitrary meets. In this post I prove weaker result that is closed under finite meets.

I hope this finite case may serve as a model for the general infinite case. However it seems that generalizing it to infinite case is non-trivial.

**Theorem** Let is a distributive complete lattice and is a strong partitioning of some element of this lattice. Then is closed under finite meets.

**Proof** Let .

Then

Applying the formula twice we get

But for any exist such that and . So .

]]>**Conjecture** The complete lattice generated by a strong partitioning of an element of a complete lattice is equal to .

**Proposition** Provided that this conjecture is true, we can prove that the complete lattice generated by a strong partitioning of an element of a complete lattice is a complete atomic boolean lattice with the set of its atoms being (Note: So is completely distributive).

**Proof** Completeness of is obvious. Let . Then exists such that . Let . Then and . is the biggest element of . So we have proved that is a boolean lattice.

Now let prove that is atomic with the set of atoms being . Let and . If then either or where , and . Because is a strong partitioning, and . So .

Finally we will prove that elements of are not atoms. Let and . Then where and . If is an atom then what is impossible. **QED**

The above conjecture as a step to solution to the original conjecture may also be considered for the polymath research problem. Or maybe we should research both these two problems in a single polymath set, as the solution of one of them may inspire the solution of the other of these two problems.

]]>I will call *weak partitioning* of a set such that

.

I will call *strong partitioning* of a set such that

.

**Question:** Do exist complete lattices for which weak partitioning and strong partitioning are not the same?

If this conjecture (that it is the same for arbitrary complete lattices) is indeed false for arbitrary complete lattices, we must find the cases when it is true. (I strongly suspect that it is true for distributive complete lattices.)

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