Regula Krapf, Ph.D. 2017, University of Bonn

Regula Krapf successfully defended her PhD dissertation January 12, 2017 at the University of Bonn, with a dissertation entitled, “Class forcing and second-order arithmetic.”  I was a member of the dissertation examining committee. Peter Koepke was the dissertation supervisor.

Regula Krapf

Regula Krapf, Class forcing and second-order arithmetic, dissertation 2017, University of Bonn. (Slides)

Abstract. We provide a framework in a generalization of Gödel-Bernays set theory for performing class forcing. The forcing theorem states that the forcing relation is a (definable) class in the ground model (definability lemma) and that every statement that holds in a class-generic extension is forced by a condition in the generic filter (truth lemma). We prove both positive and negative results concerning the forcing theorem. On the one hand, we show that the definability lemma for one atomic formula implies the forcing theorem for all formulae in the language of set theory to hold. Furthermore, we introduce several properties which entail the forcing theorem. On the other hand, we give both counterexamples to the definability lemma and the truth lemma. In set forcing, the forcing theorem can be proved for all forcing notions by constructing a unique Boolean completion. We show that in class forcing the existence of a Boolean completion is essentially equivalent to the forcing theorem and, moreover, Boolean completions need not be unique.

The notion of pretameness was introduced to characterize those forcing notions which preserve the axiom scheme of replacement. We present several new characterizations of pretameness in terms of the forcing theorem, the preservation of separation, the existence of nice names for sets of ordinals and several other properties. Moreover, for each of the aforementioned properties we provide a corresponding characterization of the Ord-chain condition.

Finally, we prove two equiconsistency results which compare models of ZFC (with large cardinal properties) and models of second-order arithmetic with topological regularity properties (and determinacy hypotheses). We apply our previous results on class forcing to show that many important arboreal forcing notions preserve the $\Pi^1_1$-perfect set property over models of second-order arithmetic and also give an example of a forcing notion which implies the $\Pi^1_1$-perfect set property to fail in the generic extension.

Regula has now taken up a faculty position at the University of Koblenz.

Set-theoretic geology and the downward directed grounds hypothesis, Bonn, January 2017

This will be a talk for the University of Bonn Logic Seminar, Friday, January 13, 2017, at the Hausdorff Center for Mathematics.


Abstract. Set-theoretic geology is the study of the set-theoretic universe $V$ in the context of all its ground models and those of its forcing extensions. For example, a bedrock of the universe is a minimal ground model of it and the mantle is the intersection of all grounds. In this talk, I shall explain some recent advances, including especially the breakthrough result of Toshimichi Usuba, who proved the strong downward directed grounds hypothesis: for any set-indexed family of grounds, there is a deeper common ground below them all. This settles a large number of formerly open questions in set-theoretic geology, while also leading to new questions. It follows, for example, that the mantle is a model of ZFC and provably the largest forcing-invariant definable class. Strong downward directedness has also led to an unexpected connection between large cardinals and forcing: if there is a hyper-huge cardinal $\kappa$, then the universe indeed has a bedrock and all grounds use only $\kappa$-small forcing.


Transfinite game values in infinite chess, including new progress, Bonn, January 2017

This will be a talk January 10, 2017 for the Basic Notions Seminar, aimed at students, post-docs, faculty and guests of the Mathematics Institute, University of Bonn.

Bishop gateway terminals

Bishop cannon









Abstract. I shall give a general introduction to the theory of infinite games, using infinite chess — chess played on an infinite edgeless chessboard — as a central example. Since chess, when won, is won at a finite stage of play, infinite chess is an example of what is known technically as an open game, and such games admit the theory of transfinite ordinal game values. I shall exhibit several interesting positions in infinite chess with very high transfinite game values. The precise value of the omega one of chess is an open mathematical question.  This talk will include some of the latest progress, which includes a position with game value $\omega^4$.

It happens that I shall be in Bonn also for the dissertation defense of Regula Krapf, who will defend the same week, and who is one of the organizers of the seminar.

Transfinite game values in infinite chess | The mate-in-$n$ problem of infinite chess is decidable | A position in infinite chess with game value $\omega^4$ | more on infinite chess | Slides

The rearrangement number

  • A. Blass, J. Brendle, W. Brian, J. D. Hamkins, M. Hardy, and P. B. Larson, “The rearrangement number.” (manuscript under review)  
    author = {Andreas Blass and J\"org Brendle and Will Brian and Joel David Hamkins and Michael Hardy and Paul B. Larson},
    title = {The rearrangement number},
    journal = {},
    year = {},
    volume = {},
    number = {},
    pages = {},
    month = {},
    note = {manuscript under review},
    url = {},
    eprint = {1612.07830},
    archivePrefix = {arXiv},
    primaryClass = {math.LO},
    abstract = {},
    keywords = {},
    source = {},

Abstract.  How many permutations of the natural numbers are needed so that every conditionally convergent series of real numbers can be rearranged to no longer converge to the same sum? We show that the minimum number of permutations needed for this purpose, which we call the rearrangement number, is uncountable, but whether it equals the cardinal of the continuum is independent of the usual axioms of
set theory. We compare the rearrangement number with several natural variants, for example one obtained by requiring the rearranged series to still converge but to a new, finite limit. We also compare the rearrangement number with several well-studied
cardinal characteristics of the continuum. We present some new forcing constructions designed to add permutations that rearrange series from the ground model in particular ways, thereby obtaining consistency results going beyond those that follow from comparisons with familiar cardinal characteristics. Finally we deal briefly with some variants concerning rearrangements by a special sort of permutations and with rearranging some divergent series to become (conditionally) convergent.

This project started with Michael Hardy’s question on MathOverflow, How many rearrangements must fail to alter the value of a sum before you conclude that none do? I had proposed in my answer that we should think of the cardinal in question as a cardinal characteristic of the continuum, the rearrangement number, since we could prove that it was uncountable and that it was the continuum under MA, and had begun to separate it from other familiar cardinal characteristics. Eventually, the research effort grew into the collaboration of this paper. What a lot of fun!

Colloquium talk at Vassar | Lecture notes talk at CUNY | the original MathOverflow question

Here are the lecture notes for an introductory talk on the topic I had given at the Vassar College Mathematics Colloquium

There are no regular polygons in the hexagonal lattice, except triangles and hexagons

I should like to follow up my post last week about the impossibility of finding regular polygons in the integer lattice. This time, however, we shall consider the hexagonal and triangular lattices.  It is easy to find lattice points that form the vertices of an equilateral triangle or a regular hexagon.


And similarly, such figures can be found also in the triangular lattice.

hex-tri-gridIndeed, the triangular and hexagonal lattices are each refinements of the other, so they will allow exactly the same shapes arising from lattice points.

But can you find a square? How about an octagon?

Question.  Which regular polygons can be formed by vertices of the hexagonal or triangular lattices?

The answer is that in fact there are no other types of regular polygons to be found! I’d like to prove this by means of some elementary classical reasoning.

Theorem. The only non-degenerate regular polygons that arise from vertices in the hexagonal or triangular lattices are the equilateral triangles and regular hexagons.

This theorem extends the analysis I gave in my post last week for the integer lattice, showing that there are no regular polygons in the integer lattice, except squares.

Proof.  The argument for the hexagonal and triangular lattices uses a similar idea as with the integer lattice, but there is a little issue with the square and pentagon case. We can handle both the hexagonal and triangular lattices at the same time. The crucial fact is that both of these lattices are invariant by $120^\circ$ rotation about any lattice vertex.

To get started, suppose that we can find a square in the hexagonal lattice. square We may rotate this square by $120^\circ$ about the vertex $a$, and the square vertices will all land on lattice-vertex points. Next, we may rotate the resulting square about the point $b$, and again the vertices will land on lattice points. So we have described how to transform any square vertex point $a$ to another lattice point $c$ which is strictly inside the square. By applying that operation to each of the four vertices of the square, we thereby arrive by symmetry at a strictly smaller square. Thus, for any square in the hexagonal or triangular lattice, there is a strictly smaller square. But if there were any square in the lattice at all, there would have to be a square with smallest side-length, since there are only finitely many lattice distances less than any given length. So there can be no square in the hexagonal or triangular lattice.

The same construction works with pentagons. pentagon

If there is a pentagon in the lattice, then we may rotate it about point $a$, and then again about point $b$, resulting in point $c$ strictly inside the pentagon, which leads to an infinite sequence of strictly smaller pentagon, whose sizes (by similarity) go to zero. So there can be no pentagon in the hexagonal or triangular lattices.

If we attempt to apply this argument with triangles or hexagons, then the process simply leads back again to points in the original figure. But of course, since triangles and hexagons do arise in these grids, we didn’t expect the construction to work with them.

triangle hexagon





But also, this particular two-step rotation construction does not succeed with the heptagon (7-sides) or larger polygons, since the resulting point $c$ ends up outside the original heptagon, which means that the new heptagon we construct ends up being larger, rather than smaller than the original.








Fortunately, for the 7-gon and higher, we may fall back on the essential idea used in the square lattice case. Namely, because the interior angles of these polygons are now larger than $120^\circ$, we may simply rotate each side of the polygon by $120^\circ$ and thereby land at a lattice point. In this way, we construct a proportionally smaller instance of the same regular $n$-gon, and so there can be no smallest instance of such a polygon.









In summary, in every case of a regular polygon except the equilateral triangle and the regular hexagon, we found that by means of $120^\circ$ rotations we were able to find a strictly smaller instance of the polygon. Therefore, there can be no instances of such polygons arising from lattice points in the hexagonal or triangular tilings. QED

See my earlier post: there are no regular polygons in the integer lattice, except squares.

There are no nondegenerate regular polygons in the integer lattice, except for squares

Consider a regular square lattice, formed by a grid of intersection points of uniformly spaced parallel horizontal and vertical lines in the plane.  One may easily find points that form the vertices of a square in this lattice.
But can one find points that form the vertices of regular hexagon? Or a regular pentagon? How about an equilateral triangle? And so on.


The answer is that one cannot find these shapes using vertices in the square lattice.

Theorem. There is no regular pentagon in the square lattice, and no regular hexagon, no regular heptagon, and so on. Indeed, the only nondegenerate regular polygons to be found using vertices in a square lattice are squares themselves.

I think this must be a classical result. I was inspired by Vaughn Climenhaga’s beautiful image in his Proof without words answer on MathOverflow, which handled the case of a hexagon. Reflecting upon it, I realized that the same idea works with other regular polygons, and I endeavored to produce the corresponding images, below.

Proof. Suppose that we could find five vertices in the square lattice that formed a regular pentagon.  Construct on each side a perpendicular of the same length, as pictured in brown below. Since the lattice is invariant under $90^\circ$ rotations centered at a lattice point, each of these new points is still a lattice point. And by symmetry, they form the vertices of a (proportionally smaller) regular pentagon. Therefore, there can be no regular pentagon at all in the square lattice, since if there is one, it is clear that there would have to be a smallest instance.   pentagon

An alternative way to argue is: by similarity, the size of the smaller pentagon shrinks by the same factor with each step, and so in the limit the size approaches zero; but clearly, we cannot have a lattice-point regular pentagon whose size is smaller than the square lattice spacing itself. So there is no regular pentagon in the square lattice.

The same argument works with larger regular polygons.  The main point to realize is that for all regular $n$-gons, where $n>4$, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular $n$-gon is strictly smaller than the original. This completes the proof for all $n$-gons for $n>4$.
















The case of an equilateral triangle requires special care. If one attempts the same construction idea as above, building the perpendicular on the edges of a triangle, then the resulting triangle becomes larger, rather than smaller, and so the proof method breaks down.


Nevertheless, one can reduce the equilateral triangle case to a hexagon: if you could find an equilateral triangle in the square lattice, then since the lattice is invariant under translation via a lattice-point line segment, it follows that we could build a regular hexagon. But we have already showed that we cannot find a regular hexagon in the square lattice, and so we cannot find an equilateral triangle.


So we’ve completed the proof for all nondegenerate regular polygons, except the square. QED

For those who might be interested, here is the tex code I used to generate the nested polygons. The code accepts input $n$ to produce a regular $n$-gon with the perpendiculars constructed.




\foreach \k in {0,...,\d} {
\foreach \i in {0,...,\m} {
\coordinate (x) at (360*\i/\n+\f:\R);
\coordinate (y) at (360*\i/\n+360/\n+\f:\R);
% the perp indicator
\ifthenelse{\k=0\AND\i=\m\AND\n<20}{\draw[very thin,black!\c!white] ($(x)!.85!(y)$) node (p) {} --
($(p)!1!90:(y)$) node (q) {} -- ($(q)!1!90:(p)$);}{}
% connecting to lower level
\draw [line width=.5*exp(-\k/2),GonColorC!\c!white] (y) -- ($(y)!1!-90:(x)$);
% edge and vertices of current gon
\draw [line width=1.5*exp(-\k/2),GonColor!\c!white] (x) node
[circle,scale=.3*\R,black!\c!white,fill=black!\c!white] {} -- (y);

\draw (\f:\R) node [circle,scale=.3*\R,black!\c!white,fill=black!\c!white] {};


\foreach\n in {5,6,7,8,9,10,12,16,20} {


See my follow-up post on the regular polygons arising in the hexagonal and triangular latticeshex-grid

Mathematical logic, GC Math 712, Spring 2017

I shall be teaching Mathematical Logic, Math 712 at the CUNY Graduate Center in Spring 2017. The course is 4.5 credits, and it will meet Tuesdays and Thursdays, 2:00 pm – 3:30 pm.  There are no official pre-requisites, other than a willingness to engage with graduate-level mathematics. Students will benefit from having taken the first semester logic course, Math 711.


Course Description. This course is a graduate introduction to mathematical logic, in which we shall cover a variety of topics united by the theme of Gödel’s incompleteness theorem. In particular, during the semester we shall give several independent proofs of that theorem and its generalizations from different perspectives. We shall do so in the context of our studies of computability theory, decidability, strongly undecidable structures, the arithmetic hierarchy, arithmetization, definability and selected topics in proof theory, model theory and set theory, including the hierarchy of consistency strength. Students will complete weekly problem sets and write a term paper.

Check back here later for further information about the course.

A lifetime of hot air

We’ve been making some Fermi estimations in the Math for Liberal Arts class I am teaching, and today we discuss the following question:

Question. If you collect all the hot-air that you have breathed in your life, what would the volume be? If you made a hot-air balloon, would it be able to lift you and all your possessions?

To answer, let’s start with the first part. How much do I breathe? If I imagine inhaling and then exhaling a deep, big breath, I figure I could inflate a small paper bag, perhaps well over one liter, but probably not as much as two liters. But my passive resting breathing is probably much less than a big deep breath. So let’s figure a half liter per ordinary passive breath. How often do I breathe? Well, in the swimming pool, I can hold my breath under water for a minute or even two minutes (in my younger swim-team days); but if I hold my breath right now, I can say that it does start to feel a little unnatural, like I should take my next breath, even after just about five or ten seconds, even though this impulse could be resisted longer. It seems to me that my body wants to take another breath in about that time. If we breathe every five seconds, that would mean 12 breaths per minute, so let’s say ten breaths per minute, which would mean a volume of 5 liters per minute. That makes $5\times 60=300$ liters per hour, or $300*24=7200$ liters per day. In a year, this would be $7200\times 365$, which is less than 7000 times 400, which is 2,800,000 liters per year. Let’s say 2.5 million liters per year of hot air. Times 50 years would make $125$ million liters of hot air in all.

Now, each liter of air fills a cube 10 cm on a side, and one thousand of these fit into a cubic meter. So we’ve got 125,000 cubic meters of hot air. This is the same as a cube 50 meters by 50 meters by 50 meters. That is my hot-air balloon! Filled with a lifetime of hot air. (This is considerably larger, more than one hundred times as large, as a typical recreational hot-air balloon, which I understand are usually under 1000 cubic meters. From this point of view, it would seem likely that it could lift me and all my possessions, although body temperature may be much less than is achieved in those balloons.)

If the air was at my body temperature ($98.6^\circ$ F), then would it be able to lift me and all my possessions? Well, let’s see how much it would lift. Hot air expands in proportion to temperature (from absolute zero). If it is a day like today, about $50^\circ$ degrees F, which is about a 50 degree F difference, and absolute zero is minus 460 F. So this is about a 10 per cent increase in temperature. (In metric: we have body temperature of about 37 degrees, and it is about 10 degrees Celsius today, so a difference of 27 degrees, and absolute zero is minus 270, so about ten per cent increase, as I had said.)

So the heat of the hot air caused it to expand in volume by ten percent. The buoyant force of the hot-air balloon is exactly the weight of this displaced air, by the Archimedean principle. Thus, the lifting force of my hot-air balloon will be equal to the mass of air filling ten percent of the volume we computed. How much does air weigh? I happen to remember from my high school science class that one mole of air at one atmosphere of pressure is about 22 liters (my teacher had a cube of exactly that size sitting on his desk, to help us to visualize it). And I also know that air is mainly nitrogen, which forms the molecule $N_2$, and since nitrogen in the periodic table has an atomic number of 14, the molecule $N_2$ has a mass of 28 grams per mole. So air weighs about 28 grams per 22 liters, which is about 1.3 grams per liter. Each cubic meter is one thousand liters, and so 1.3 kilograms per cubic meter (this is much larger than I had expected—air weighs more than one kilogram per cubic meter!). My hot air in total was 125,000 cubic meters, and we said that because of the temperature difference, the volume expanded by ten percent, or 12,500 cubic meters. This expansion would displace an equal volume of air, which weighs 1.3 kg per cubic meter. Thus, the displaced air weighs $12,500\times 1.3\approx 16,000$ kilograms, or about 16 metric tons. So all my hot air, at body temperature in a giant hot air balloon on a chilly day, would have a lifting force able to lift 16 metric tons.

Would this lift me and all my possessions? Do I own 16 tons of stuff? Well, thankfully, I don’t own a car, which would be a ton or more by itself. But I do own a lot of books, a piano, an oven, a dishwasher, some heavy furniture, paintings, and various other items, as well as a collection of large potted plants on my terrace. It seems likely to me that I could fit most if not all my possessions within 16 tons. To gain a little confidence in this, let’s estimate the mass of my books. My wife and I have about twelve large shelves filled with books, each about 2 meters, and then I have about 3 more such shelves filled with books in my offices at the university. If we count half of the home books as mine, plus my office books, that makes 9 shelves times two meters, for about 20 meters of books. If the books are about 25 cm tall on average, and 15 cm deep, that makes $20\times .25\times .15=.75$ cubic meters of books. Let’s round up to one cubic meter of books. How much does a cubic meter of paper weigh? Well, one ream of copy paper weighs about 2 kilograms, and that is a volume of 8.5 by 11 by 2 inches. One meter is about 33 inches, and so we could fill one cubic meter with a pile of reams of copy paper 3 by 3 by 17, which would be 163 reams, or about 300 kilograms. So not even a half ton of books! So I can definitely lift all my most important possessions within the 16 tons.

Final answer: Yes, if we filled a giant balloon with all the hot-air I have breathed in my life, at body temperature, then it would lift me and all my possessions.

On number sense — Would one day of NYC coffee fill the Statue of Liberty? And other fun questions…

Today I gave a lecture on what I call number sense, using a process of estimation and approximation in order to calculate various unknown quantities, including a few fantastical ones. How much coffee is made per day in New York City? Would it fill up the Statue of Liberty? Approximately how many babies are born in New York City each day? If you made a stack of quarters to reach the distance to the moon, what would the dollar-value be? If you piled those quarters in a heap, would it fit in Central Park? How much does the Empire State Building weigh?

These kinds of back-of-the-envelope calculations, in my view, have at their heart the idea that one can solve a difficult or seemingly impossible problem by breaking it into more manageable pieces. We don’t just pull a final answer out of the air, but rather make simplifying assumptions and informed estimates for related quantities that we are more familiar with or have knowledge about, and then use that information to derive a better estimate for the main quantity. For most of these questions, at the outset we may have little idea what would be a reasonable answer, but by the end, we gain some insight and find ourselves a little closer to the truth.

These kind of calculations are also known as Fermi estimations, in light of Fermi’s remarkable ability to make surprisingly accurate estimations on the basis of little or no hard data. The wikipedia page (thanks to Artie Prendergrast-Smith for mentioning this link in the comments) emphasizes that even in a case where the estimate is significantly off the true value, nevertheless we may still find value in the Fermi calculation, because it focusses our attention to the reasons for the divergence. In discovering which of the assumptions underlying our calculations was wrong, we come to a deeper understanding of the true situation.

In the lecture, I began with some very easy cases. For example, how many seats were in the auditorium? The students estimated that there were approximately 12 seats per row and about 10 rows, so 120 seats in all. How old was one of the students, in seconds? Well, he was 18 years old, and so we could simply multiply each year by 365 days, times 24 hours per day, times 60 minutes per hour, times 60 seconds per minute, to get
$$18\times 365\times 24\times 60\times 60\approx 600,000,000 \text{ seconds}.$$
One student objected about leap days, since 365 should be 365.25 or so. But I pointed out that this difference was not as important as it might seem, since already we had made far larger rounding assumptions. For example, the student was not exactly 18 years old, but 18 years old and some several months; by using 18 years only, we made a bigger difference in the answer than caused by the leap-day issue, which would be a difference of only five days or so over 18 years. For the same reason, we should feel free to round the numbers to make the calculation easier. We are aiming at a ballpark estimate rather than an exact answer.

Let’s now do some more interesting cases.

anthora-cupCoffee in New York. How much coffee is made each day in New York? Would it fill the Statue of Liberty? First, let me say that I really don’t have any definite information about how much coffee is made each day in New York, and I fear that my own coffee-obsessed perspective will lead me to over-estimate the amount, but let’s give it a try. New York City has a population of approximately 10 million people. Some of those people, like myself, drink a large amount of coffee each day, but many of the others do not drink coffee at all. I would think that a sizable percentage of the NYC population does drink coffee, perhaps as much as a third or even half consumes coffee daily. Many of those coffee-drinkers have more than one cup per day, and also surely more coffee is made than consumed. So it seems reasonable to me to estimate that we have approximately one medium cup of coffee per person on average per day in New York. Basically, we’re saying that the heavy coffee drinkers and the made-but-not-sold coffee approximately makes up for those who abstain, making the average about one cup per person. So we are talking about 10 million cups of coffee per day. A medium cup of brewed coffee at Starbucks is I think about 12-16 ounces, a little less than a pint, and so let’s say about 3 cups per liter. This amounts to roughly 3 million liters of coffee.

Would it fill the Statue of Liberty? The statue itself is, I estimate, about twenty stories tall, counting the base, and if each story is 15 ft, or 5 meters, that would mean 100 meters tall, counting the base. But I think that the base is about half the height, so let’s say 50 meters for the actual statue itself. I’ve never been inside the statue, but my students say that it is about 10 meters across inside, a little more at the bottom than near the top. If we approximated it as a rectangular solid, that would give a volume of $10\times 10\times 50$ cubic meters, or 5000 cubic meters. But since the statue tapers as you go up, particularly in the arm holding the torch, it really is more like a cone than a rectangular solid, and so we should divide by three. But let’s divide just by two, because she isn’t quite as tapered as a cone. So the Statue of Liberty has a volume of approximately 2500 cubic meters. One cubic meter can be thought of as a 10 by 10 by 10 array of little 10cm cubes, and each of those is exactly one liter. So a cubic meter is 1000 liters, and therefore the Statue of Liberty has a volume of $2500\times 1000=2.5$ million liters. But since we had 3 million liters of coffee, the answer our calculation arrives at is: Yes, one day’s worth of New York coffee would fill up the Statue of Liberty!

Well, we do not have perfect confidence in our estimates and assumptions — for example, perhaps there are many fewer coffee drinkers in New York than we estimated or perhaps we underestimated the volume of the Statue of Liberty. Since the estimated volumes were of basically similar magnitudes, we aren’t really entitled to say that definitely the coffee would fill up the Statue of Liberty. Rather, what we have come to know is that those two volumes are comparably similar in size; they are in the same ballpark.

Elevator trips. While riding downtown last weekend with my son on the subway, a crowded 4 train, we overheard the group standing next to us talking about elevators. One lady said, “My elevator company serves as many elevator trips in New York City in five days as the population of the entire world,” and the rest of her group, impressed, nodded affirmatively in reply. But my thoughts, upon hearing that, were to make a quick calculation. Suppose all 10 million NYC residents rode an elevator 10 times every day, which is way too high (probably one trip per person per day is more reasonable, since many people live and work in buildings without elevators). Even in this extreme case of ten trips per person per day, it would mean only 100 million trips total per day, or 500 million trips over 5 days. This is much less than the world population, and so no way is that person’s claim true, especially since there are also many elevator companies. I thought of mentioning my calculation to those people on the subway, but decided against it. Walking out of the subway in the East Village, however, I asked my son (14 years old) whether he heard those people talking about elevators, and he replied, “Oh, yes, and when they said that, I calculated it in my head: no way is that true.” He then proceeded to explain his calculation, similar to mine. Yay, Horatio!

The Chicago marathon. In the run-up to the Chicago marathon this year, on a route that would wind through the windy city streets, Newsweek magazine reported, “Chicago Marathon organizers expect 1.7 million fans to line the route this year.” (Thanks to the critical math commentary of Mark Iris for bringing this example to my attention.) The organizers had emphasized the economic impact of these spectators, many of whom would presumably be eating in Chicago restaurants and staying at Chicago hotels. But is this a reasonable number?

Let’s calculate. A marathon is approximately 26 miles, and the route has two sides for spectators, so let’s round it to 50 miles of spectator viewing spots. Each mile is about 1800 yards, so we have $50\times 1800=90000$ yards of viewing spots. Each spectator, standing shoulder-to-shoulder, with all their stuff, takes up about a yard of space. So if the marathon was lined on both sides for the entire route with spectators standing shoulder-to-shoulder, this would amount to about 90,000 spectators. In order to have 1.7 million spectators, therefore, they would have to be lined up behind each other. But even if the spectators were 10 people deep on each side for the entire route, which is a vast crowd, it would still amount only to 900,000 people. We would have basically to double this to get to 1.7 million. So, if the live event really had 1.7 million spectators lining the route, then it would mean that the race was lined 20 people deep on each side for the entire route. No way is this number correct! I have never had the chance to attend the Chicago marathon, but at the New York marathon, which I assume is comparable, I know that while there are thick crowds at the finish line in Central Park and at some of the other prominent or especially interesting race locations, most of the rest of the route is much thinner, and at the typical nothing-special location, one could expect easily to have a front-row spot.

College of Staten IslandHow many bricks on the college campus? Our campus, covered with some lovely woods and green meadows, has a number of brick Georgian buildings. Most of these are the same standard size as the mathematics department, but there are also some larger buildings such as the library, the performing arts center, the campus center and the gymnasium. At my lecture, the students and I agreed that altogether it amounted to about 30 buildings of the size of the math building. This building is approximately 30 meters by 70 meters, which would make a perimeter of 200 meters, but because the building has wings and is not purely rectangular, let’s add another 100 meters for the folds in the walls, so about 300 meters around the base of the building. The building is two stories tall, about 10 meters tall, making the area of the walls about 3000 square meters. Of course, there are windows and doors that cut out of these walls, but let’s say that they are approximately accounted for by the extra bricks used in the trimming flourishes at the corners and top of the building. So we have 3000 square meters of brick. Each brick is approximately 10 cm by 30 cm, and so one square meter of brick would have about ten rows of a little more than three bricks across, or about 20 bricks. So each building has about $3000\times 20=60,000$ bricks. And with 30 buildings, this amounts to $30\times 60,000=180,000$ bricks in the buildings on campus. There are also some bricks in the central fountain, a few small brick walls and some bricks lining some of the walkways. So let’s add another ten percent for those extra bricks, arriving at about 200,000 bricks on campus in all.

Positive test result for a rare disease. Suppose that as part of your check-up, your doctor randomly administers a clinical test for a certain rare medical condition. The test is 99 percent accurate, in terms of false positives and false negatives, in the sense that 99 percent of people taking the test have an accurate result with the test. Suppose also that the disease itself is rare, held by only one in a million people. If your test comes back positive, what is the likelihood that you actually have the disease?

This is a subtle question. It might seem to make sense initially that there is a 99 percent chance that you have the disease, since that is the accuracy of the test. But this isn’t actually right, because it doesn’t account for the fact that the disease itself is extremely rare, and so the total number of false positives will actually far outweigh the true positive results. For example, suppose that one million people take the test. About one of these people actually has the disease, and that person is 99 percent likely to test positive. So we expect about one true positive result. And for the others, who don’t have the disease, 99 percent of them will test negative. In other words, about 990,000 people will test negative. The remaining 10,000 people, however, one percent of the total, will have false positive results! So you are in this group of 10,000 people who tested positive, with only one of them actually having the disease. So the odds that you actually have the disease are only about one in ten thousand.

Quarters to the moon. How many quarters would you have to stack to reach the moon? How much would it be worth? how much would it weigh? More than the earth? More than the Empire state building? If you put all those quarters into a pile, would it fit in Central Park?

Well, OK, the question is a little absurd, and there are all kinds of problematic issues: we couldn’t ever actually make such a tower of quarters, as it would topple over; it doesn’t make sense to build a tower to the moon, since the moon is in orbit around the earth and it is moving; the quarters would begin to orbit the earth themselves, or fall back down to the earth or to the moon. But let’s just try to ignore all those problematic issues, and just try to answer how many quarters it would take to make a pile that high.

How far is the moon? I don’t really know, and we could look it up to see what the astronomers say about it, but that would go against the back-of-the-envelope spirit of these problems. So I am just going to use two facts that I picked up years ago: first, I know that the speed of light is about 300,000 km per second; and second, I happen to know that it takes radio signals traveling at the speed of light about one second to get to the moon (eight minutes to the sun), and so the moon is about one light-second away. I remember this fact from having learned it in my childhood, because it was important in a science fiction story I had read, in which radio communication from earth to the moon base had a two-second delay: you send a message, it takes a second to get there, and then a second for the answer to come back. So the moon is one light second away, and light travels 300,000 km in one second. So the moon is about 300,000 kilometers distant (actual value: 387400 km).

Let’s now stack up the quarters. By eyeballing a quarter I had in my pocket, I’d say each quarter is about 2 mm thick, which means 50 quarters per meter, or 50,000 quarters per km. So altogether, we have $300,000\times50,000=15,000,000,000$ many quarters in the stack. Fifteen billion quarters to the moon! It takes four quarters to make a dollar, so that is worth about $4 billion dollars, which may be much less than you would have expected initially.

Let’s put those quarters in a big pile, packed as tightly as possible. Each quarter is a little over 2 cm in diameter, and so the volume of the rectangular solid bounding the quarter is a little more than 2 cm by 2 cm by 2 mm, or about 800 cubic mm, which we can round up to 1 cubic centimeter. That makes sense, because we can imagine folding a quarter up into a little 1 cm cube. So 1000 quarters takes up about one liter. Thus, our quarters-to-the-moon have a volume of about 15 million liters. There are 1000 liters in a cubic meter, and so this is 15,000 cubic meters. Since a cube 10 meters on a side, has a volume of 1000 cubic meters, we can fit all those quarters into fifteen such cubes. I can easily imagine an art instalation, with fifteen such 10 meter cubes placed in Sheep Meadow in Central Park. They would definitely fit.

How much would the quarters weigh? Some of the students in my lecture had worked as cashiers, and so they were familiar with quarter rolls from the bank, which fit in your palm and have ten dollars worth of quarters, or forty quarters. I can imagine a quarter roll in one hand balancing the weight of a half-pound of gouda cheese at the deli in my other hand. So 40 quarters is .5 pound, which makes 100 quarters about 2.2 pounds, which is about 1 kg. We had fifteen billion quarters, which is 150 million times 100 quarters, so 150 million kilograms.

Since we already imagined the quarters in Sheep Meadow in those fifteen large boxes, clearly they weigh far less than the earth and also much less than the Empire State Building.

How much does the Empire State Building weigh? After posing this question, I found out that it is also evidently a famous Google interview question, and you can easily find other blog posts about it. But let me tell you how I thought about it. The Empire State Building is about 100 stories tall, and if we assume 12-15 feet per floor, or about 4 meters, this makes the total height (without tower) about 400 meters. The CUNY Graduate Center where I work is across the corner on Fifth Avenue and 34th street, and I have walked past the Empire State Building many times. I estimate that the base is about 75 meters on Fifth Avenue by 150 meters on 34th Street. But because of the step-backs, the middle part of the tower is considerably less than that, probably 30 by 80 near the top. Let’s say the average cross section of the building is 50 by 100 meters. Now, how much does that weigh? Of course, there is a lot of steel and masonary in the floors and walls, and concrete floors, and also all the contents of the building, with desks and paper files and books and interior walls and doors. Those things are heavy. I really have no idea how much those things weigh, but let me imagine that we build a virtual copy of a complete floor from the Empire State Building, without any of those floor beams or concrete or desks or walls or anything, and instead we flood that virtual room with water, until it has the same weight. Of course, the metal and concrete and masonary is much heavier than water, but the actual space is mostly empty air. How much water would it take to have the same weight? Well, let me just guess that we’d have to flood the virtual room about one-quarter deep with water to have the same weight as the actual materials. At 50 by 100 meters by 4 meters tall for each floor, this would mean a pool of water 50 by 100 meters, one meter deep, for a volume of 5000 cubic meters of water. We already said that one cubic meter of water is 1000 liters, each of which weighs one kilogram, so we are talking about $5000\times 1000$ kilograms or 5 million kilograms per floor. With 100 floors, this is 500 million kilograms, or 500,000 metric tons. So according to this estimate, the Empire State Building with all its contents weighs about 500,000 metric tons.

So it weighs more than the quarters to the moon (150 million kilograms), but not as much more as I would have thought based on the art exhibition in Sheep Meadow! The Empire state building weighs about three or four times as much as the quarters to the moon.

How many babies are born each day in New York City? The population of New York is approximately ten million people. And those people live about 70 years. Let’s imagine spreading those births out uniformly over the 70 year period. That means one million people born in 7 years. This is about 150,000 people born in one year, which is about 3000 births per week, or about 400 births per day. This estimate would be accurate if the city population were in a steady state, with births balancing deaths, but of course the population is increasing. On the other hand, most of that increase is from people moving here, not just being born here. But then again, the people moving to New York I expect are mainly young adults looking for a career, and so they will contribute to a heightened birth rate as they have children. Shall we add 25 percent more to account for the fact that the city is growing and not in a steady state? In that case, our estimate is that there are about 500 births each day in New York City.

There is an endless supply of such questions that can be calculated. I have been talking about them in the lectures for my course Math for Liberal Arts, an undergraduate course aimed at non-math students at my college. In this course, we are fairly free to cover some imaginative topics, and I’ve covered some game theory, some elementary graph theory, a little bit logic, and now this number sense topic, which I use to try to develop the students ability to attack a technical problem by breaking it down into more managable problems. But meanwhile, I also look upon it just as plain fun.

So here are a few more questions that you might enjoy thinking about. And please make your own.

  • What is the total volume of air that you have breathed in your lifetime? If you filled a balloon with all your hot air, how big would it be?
  • If you used that balloon as a hot-air balloon (with the hot air at your body temperature), would it have enough bouyency to lift you and all your possessions? See my post A lifetime of hot air for a detailed answer.
  • How many NYC metro cards exist? (Each lasts about a year or two; they get thrown out, but still exist in the landfill; the system has used metro cards for about twenty years; there is a large stock of new not-yet-used cards.)
  • How many doorknobs are there in your building?
  • How many subway tiles are there in the NYC subway system?
  • How many riders were on my crowded 4 train downtown this morning?
  • How many lightswitches are there on your university campus?
  • How much water does NYC use each day?
  • What are the odds that you drank a molecule of water that was once also drank by Julius Caesar?
  • How many people have there been? What fraction are currently alive?

Please try to figure them out, and post your solutions in the comments below!

Are all Gödel sentences equivalent?

By fdecomiteAnetode at en.wikipedia (Tunnels of Time Transferred from en.wikipedia) [CC BY 2.0 (], from Wikimedia CommonsI should like to consider a family of natural questions concerning the Gödel sentence — the sentence asserting its own non-provability, used by Gödel to prove the incompleteness theorem — and specifically the question whether we are really entitled to speak of the Gödel sentence. Is it unique? Is it unique up to equivalence? Up to provable equivalence? Could there be more than one Gödel sentence, perhaps for different manners of arithmetizing the syntax? Are they all provably equivalent? These questions have come up a number of times in various contexts, and since James Propp just sent me an email inquiry about it yesterday, let me address the questions here in reply (he is planning an upcoming post in a few weeks about incompleteness and self-reference — check back later for the link). I shall make use only of elementary classical incompleteness ideas, and I assume this has all been known for some time.

For definiteness, let us take first-order Peano Arithmetic ($\newcommand\PA{\text{PA}}\PA$) as the base theory over which we are considering provability, although similar observations can be made concerning other base theories. By formalizing the syntax in arithmetic using Gödel coding, we may easily produce an arithmetically expressible provability predicate $\newcommand\Prov{\text{Prov}}\Prov(x)$, which asserts that $x$ is the Gödel code of a $\PA$-provable sentence. By using this predicate and the fixed-point lemma, we may construct a sentence $\psi$ that asserts its own non-provability, meaning precisely: $\PA\vdash\psi\leftrightarrow\neg\Prov(\ulcorner\psi\urcorner)$. Such a sentence $\psi$ is known as the Gödel sentence, and we may use it to prove Gödel’s first incompleteness theorem as follows. Namely, it is easy to see that $\psi$ cannot be provable in $\PA$, for if it were, then in virtue of what $\psi$ asserts, we will have also proved that $\psi$ is not provable, contrary to fact. So $\psi$ is not provable, and therefore we see that indeed $\psi$ is actually true. So the sentence $\psi$ is true, yet unprovable. By paying a little closer attention to the argument, what we have actually argued is that $\PA$ itself proves that if $\newcommand\Con{\text{Con}}\Con(\PA)$, then $\psi$.

Of course, there are many fixed points, and for example any sentence that is provably equivalent to $\psi$, such as $\psi\wedge\psi$, is also provably equivalent to its own non-provability, and therefore serves as yet another Gödel sentence. In this trivial sense, there are infinitely many different Gödel sentences. But these particular sentences, by construction, are provably equivalent. Are all the Gödel sentences equivalent?

Indeed, a bit more generally, suppose that we have two implementations of the provability predicate, using perhaps different formalizations of the syntax, with $\varphi$ and $\psi$ being fixed points of non-provability-in-$\PA$, so that $\varphi$ and $\psi$ each assert their own non-provability with respect to those predicates. Must $\PA$ prove that $\varphi$ and $\psi$ are equivalent?

The first, main answer is that yes, indeed, if we have used a natural provability predicate, then all the Gödel sentences are provably equivalent, and this remains true even for different natural manners of formalizing the syntax. In this sense, therefore, there really is only one Gödel sentence and we are entitled to speak of the Gödel sentence. The reason is that every such Gödel sentence is provably equivalent to the assertion $\Con(\PA)$, and for natural formalizations of the syntax, meaning implementations where we can provably translate proofs from one system to another and the meta-theory, these assertions are all equivalent. To see this, suppose that $\psi$ is a Gödel sentence, which means that $\psi$ asserts its own non-provability. Since $\psi$ implies that there is a non-provable sentence, namely $\psi$ itself, it follows immediately that $\psi$ implies $\Con(\PA)$; conversely, if $\Con(\PA)$ holds, then the argument of the first incompleteness theorem that I mentioned above shows that $\psi$ is true. So $\PA\vdash\Con(\PA)\leftrightarrow\psi$. So all the fixed points are equivalent to the assertion $\Con(\PA)$. For systems where we can provably translate proofs from one system to another, the corresponding consistency assertions $\Con(\PA)$ are equivalent. And so it is fine to speak of the Gödel sentence.

But let me now discuss what happens if we should implement a more exotic form of the provability predicate. For example, consider the Rosser provability predicate, where we say that a sentence $\sigma$ is Rosser provable, if there is a proof of $\sigma$, but no shorter proof of $\neg\sigma$, meaning no proof of $\neg\sigma$ with a smaller Gödel code. The Rosser sentence $\rho$ is a non-provability fixed point of this notion, saying, “I am not Rosser provable.” Since Rosser provability is intuitively a stronger notion of provability, it would be reasonable to expect that the Rosser non-provability assertion is weaker than the Gödel sentence, and indeed that is the case. If the base theory is consistent, then the Rosser sentence cannot be provable, since if it were, then there would have to be a smaller proof of $\neg\rho$, violating consistency; but also, $\neg\rho$ cannot be provable, since this sentence implies immediately that $\rho$ is also provable with a shorter proof, in light of what $\rho$ asserts. So $\PA$ proves that $\Con(\PA)$ implies both $\Con(\PA+\rho)$ and $\Con(\PA+\neg\rho)$, and moreover, $\PA\vdash\Con(\PA)\to\rho$. But since $\PA\vdash\Con(\PA)\to\Con(\PA+\rho)$, it follows that $\PA$, if consistent, cannot prove the converse, $\rho\to\Con(\PA)$, since otherwise $\PA+\rho$ would prove its own consistency, in violation of the second incompleteness theorem. So the Rosser sentence $\rho$ is strictly weaker than $\Con(\PA)$ over $\PA$, and hence also strictly weaker than the usual Gödel sentence. But since the Rosser sentence $\rho$ is itself a fixed point of non-provability with respect to the Rosser formalization of provability, it shows that exotic formalizations of the provability predicate can indeed give rise to inequivalent fixed-point assertions.

Note that the Rosser provability predicate does not sustain provable translations to any of the usual (natural) formalizations of provability, because we cannot prove in $\PA$ as a general statement that whenever a statement is provable, then it is Rosser provable. We can, however, generally translate specific proofs to Rosser proofs, and to the meta-theory, assuming that the base theory is consistent.

Lastly, let’s consider a somewhat more exotic provability predicate, arising from the Feferman-style enumerations of $\PA$, which are arithmetically definable but not computably axiomatizable. Specifically, consider the axiomatization of $\PA$, where we continue to add the usual $\PA$ axioms, but only so long as the resulting theory remains consistent. This way of describing the theory gives rise to a corresponding provability predicate, which expresses provability in that enumerated theory. So sentence $\theta$ is Feferman provable, if it is provable using axioms that come from a consistent initial segment of the $\PA$ axiomatization. If $\phi$ is a Feferman-non-provability fixed point, then $\PA$ proves that $\phi$ is equivalent to the assertion that $\phi$ is not Feferman provable. Since $\PA$ easily proves that the Feferman theory is consistent, and also that any finite collection of $\PA$ axioms are part of the Feferman theory, it follows that the negation of any particular theorem of $\PA$ is a Feferman-non-provability fixed point, since $\PA$ proves that it is false and also that it is not provable in the Feferman theory. But those statements are not equivalent to the usual Gödel sentence, since they are inconsistent with $\PA$.

Ord is not definably weakly compact

  • A. Enayat and J. D. Hamkins, “ZFC proves that the class of ordinals is not weakly compact for definable classes.” (manuscript under review)  
    author = {Ali Enayat and Joel David Hamkins},
    title = {ZFC proves that the class of ordinals is not weakly compact for definable classes},
    journal = {},
    year = {},
    volume = {},
    number = {},
    pages = {},
    month = {},
    note = {manuscript under review},
    abstract = {},
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    doi = {},
    eprint = {1610.02729},
    archivePrefix = {arXiv},
    primaryClass = {math.LO},
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In ZFC the class of all ordinals is very like a large cardinal.  Being closed under exponentiation, for example, Ord is a strong limit.  Indeed, it is a beth fixed point. And Ord is regular with respect to definable classes by the replacement axiom.  In this sense, ZFC therefore proves that Ord is definably inaccessible.  Which other large cardinal properties are exhibited by Ord? Perhaps you wouldn’t find it unreasonable for Ord to exhibit, at least consistently with ZFC, the definable proper class analogues of other much stronger large cardinal properties?

Meanwhile, the main results of this paper, joint between myself and Ali Enayat, show that such an expectation would be misplaced, even for comparatively small large cardinal properties. Specifically, in a result that surprised me, it turns out that the class of ordinals NEVER exhibits the definable proper class analogue of weak compactness in any model of ZFC.

Theorem. The class of ordinals is not definably weakly compact. In every model of ZFC:

  1. The definable tree property fails; there is a definable Ord-tree with no definable cofinal branch.
  2. The definable partition property fails; there is a definable 2-coloring of a definable proper class, with no homogeneous definable proper subclass.
  3. The definable compactness property fails for $\mathcal{L}_{\mathrm{Ord,\omega}}$; there is a definable theory in this logic, all of whose set-sized subtheories are satisfiable, but the whole theory has no definable class model.

The proof uses methods from the model theory of set theory, including especially the fact that no model of ZFC has a conservative $\Sigma_3$-elementary end-extension.

Theorem. The definable $\Diamond _{\mathrm{Ord}}$ principle holds in a model of ZFC if and only if the model has a definable well-ordering.

We close the paper by proving that the theory of the spartan models of Gödel-Bernays set theory GB — those equipped with only their definable classes — is $\Pi^1_1$-complete.

Theorem. The set of sentences true in all spartan models of GB is $\Pi_{1}^{1}$-complete.

Recent advances in set-theoretic geology, Harvard Logic Colloquium, October 2016

I will speak at the Harvard Logic Colloquium, October 20, 2016, 4-6 pm.


Abstract. Set-theoretic geology is the study of the set-theoretic universe $V$ in the context of all its ground models and those of its forcing extensions. For example, a bedrock of the universe is a minimal ground model of it and the mantle is the intersection of all grounds. In this talk, I shall explain some recent advances, including especially the breakthrough result of Toshimichi Usuba, who proved the strong downward directed grounds hypothesis: for any set-indexed family of grounds, there is a deeper common ground below them all. This settles a large number of formerly open questions in set-theoretic geology, while also leading to new questions. It follows, for example, that the mantle is a model of ZFC and provably the largest forcing-invariant definable class. Strong downward directedness has also led to an unexpected connection between large cardinals and forcing: if there is a hyper-huge cardinal $\kappa$, then the universe indeed has a bedrock and all grounds use only $\kappa$-small forcing.


Set-theoretic potentialism, CUNY Logic Workshop, September, 2016

This will be a talk for the CUNY Logic Workshop, September 16, 2016, at the CUNY Graduate Center, Room 6417, 2-3:30 pm.

Book 06487 20040730160046 droste effect nevit.jpgAbstract.  In analogy with the ancient views on potential as opposed to actual infinity, set-theoretic potentialism is the philosophical position holding that the universe of set theory is never fully completed, but rather has a potential character, with greater parts of it becoming known to us as it unfolds. In this talk, I should like to undertake a mathematical analysis of the modal commitments of various specific natural accounts of set-theoretic potentialism. After developing a general model-theoretic framework for potentialism and describing how the corresponding modal validities are revealed by certain types of control statements, which we call buttons, switches, dials and ratchets, I apply this analysis to the case of set-theoretic potentialism, including the modalities of true-in-all-larger-$V_\beta$, true-in-all-transitive-sets, true-in-all-Grothendieck-Zermelo-universes, true-in-all-countable-transitive-models and others. Broadly speaking, the height-potentialist systems generally validate exactly S4.3 and the height-and-width-potentialist systems generally validate exactly S4.2. Each potentialist system gives rise to a natural accompanying maximality principle, which occurs when S5 is valid at a world, so that every possibly necessary statement is already true.  For example, a Grothendieck-Zermelo universe $V_\kappa$, with $\kappa$ inaccessible, exhibits the maximality principle with respect to assertions in the language of set theory using parameters from $V_\kappa$ just in case $\kappa$ is a $\Sigma_3$-reflecting cardinal, and it exhibits the maximality principle with respect to assertions in the potentialist language of set theory with parameters just in case it is fully reflecting $V_\kappa\prec V$.

This is current joint work with Øystein Linnebo, in progress, which builds on some of my prior work with George Leibman and Benedikt Löwe in the modal logic of forcing.

CUNY Logic Workshop abstract | link to article will be posted later

Set-theoretic mereology as a foundation of mathematics, Logic and Metaphysics Workshop, CUNY, October 2016

This will be a talk for the Logic and Metaphysics Workshop at the CUNY Graduate Center, GC 5382, Monday, October 24, 2016, 4:15-6:15 pm.

Venn_Diagram_of_sets_((P),(Q),(R))Abstract. In light of the comparative success of membership-based set theory in the foundations of mathematics, since the time of Cantor, Zermelo and Hilbert, it is natural to wonder whether one might find a similar success for set-theoretic mereology, based upon the set-theoretic inclusion relation $\subseteq$ rather than the element-of relation $\in$.  How well does set-theoretic mereological serve as a foundation of mathematics? Can we faithfully interpret the rest of mathematics in terms of the subset relation to the same extent that set theorists have argued (with whatever degree of success) that we may find faithful representations in terms of the membership relation? Basically, can we get by with merely $\subseteq$ in place of $\in$? Ultimately, I shall identify grounds supporting generally negative answers to these questions, concluding that set-theoretic mereology by itself cannot serve adequately as a foundational theory.

This is joint work with Makoto Kikuchi, and the talk is based on our joint article:

J. D. Hamkins and M. Kikuchi, Set-theoretic mereology, Logic and Logical Philosophy, special issue “Mereology and beyond, part II”, pp. 1-24, 2016.

The modal logic of set-theoretic potentialism, Kyoto, September 2016

Kyoto cuisineThis will be a talk for the workshop conference Mathematical Logic and Its Applications, which will be held at the Research Institute for Mathematical Sciences, Kyoto University, Japan, September 26-29, 2016, organized by Makoto Kikuchi. The workshop is being held in memory of Professor Yuzuru Kakuda, who was head of the research group in logic at Kobe University during my stay there many years ago.

Abstract.  Set-theoretic potentialism is the ontological view in the philosophy of mathematics that the universe of set theory is never fully completed, but rather has a potential character, with greater parts of it becoming known to us as it unfolds. In this talk, I should like to undertake a mathematical analysis of the modal commitments of various specific natural accounts of set-theoretic potentialism. After developing a general model-theoretic framework for potentialism and describing how the corresponding modal validities are revealed by certain types of control statements, which we call buttons, switches, dials and ratchets, I apply this analysis to the case of set-theoretic potentialism, including the modalities of true-in-all-larger-$V_\beta$, true-in-all-transitive-sets, true-in-all-Grothendieck-Zermelo-universes, true-in-all-countable-transitive-models and others. Broadly speaking, the height-potentialist systems generally validate exactly S4.3 and the height-and-width-potentialist systems validate exactly S4.2. Each potentialist system gives rise to a natural accompanying maximality principle, which occurs when S5 is valid at a world, so that every possibly necessary statement is already true.  For example, a Grothendieck-Zermelo universe $V_\kappa$, with $\kappa$ inaccessible, exhibits the maximality principle with respect to assertions in the language of set theory using parameters from $V_\kappa$ just in case $\kappa$ is a $\Sigma_3$-reflecting cardinal, and it exhibits the maximality principle with respect to assertions in the potentialist language of set theory with parameters just in case it is fully reflecting $V_\kappa\prec V$.

This is joint work with Øystein Linnebo, which builds on some of my prior work with George Leibman and Benedikt Löwe in the modal logic of forcing. Our research article is currently in progress.

Slides | Workshop program