Coming to Agreement, a logic puzzle for Oxford admissions interviews

Let me dive right in with the main puzzle.

Main puzzle. You are a contestant on a game show, known for having perfectly logical contestants. There is another contestant, whom you’ve never met, but whom you can count on to be perfectly logical, just as logical as you are.

The game is cooperative, so either you will both win or both lose, together. Imagine the stakes are very high—perhaps life and death. You and your partner are separated from one another, in different rooms. The game proceeds in turns—round 1, round 2, round 3, as many as desired to implement your strategy.

On each round, each contestant may choose either to end the game and announce a color (any color) to the game host or to send a message (any kind of message) to their partner contestant, to be received before the next round. Messages are sent simultaneously, crossing in transit.

You win the game if on some round both players opt to end the game and announce a color to the host and furthermore they do so with exactly the same color. That is, you win if you both halt the game on the same round with the same color. lf only one player announces a color, or if both do but the colors don’t match, then the game is over, but you have lost.

Round 1 is about to begin. What do you do?

Before getting to solutions, I should also like to mention several variations of the puzzle.

Alternation variation. In this variation of the puzzle, the contestants alternate in their right to send messages—only contestant 1 can send on round 1, then contestant 2 on round 2 and so forth, but still they aim to announce the same color on a round. You are contestant 1—what do you do?

Collision variation. In this variation, players may opt on each round either to end the game and announce a color, to send a message, or to do nothing. But the new thing is that if both players opt to send a message, then the messages collide and are not delivered, although an error message is generated (so the players know what happened). What do you do?

Pigeon variation. This version is like the alternating turn variation, except that now the contestants are separated at much greater distance, and the messages are sent by carrier pigeon, so neither can be sure that the messages actually arrive. You are contestant 1—what do you do?

Post your solutions in the comments! Please do so before reading the rest of this post, and then come back to read more. There are lively discussions occuring on the Twitter thread about this puzzle, on Hacker News, on www.reddit/r/mathriddles, and on www.reddit/r/philosophy.

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We had used these puzzles in our admissions interviews of candidates for a place at Oxford University in the degree courses Math/philosophy, CS/philosophy, and PPE at University College, Oxford. The candidates were high-school students, mostly 17 years old, having made the interview shortlist on the basis of the rest of their applications, including A-level scores, university entrance exam, essay, and recommendations (all considered in light of relative advantage/disadvantage). The interviews were an in-depth back-and-forth discussion, as much as could be had in about 25 minutes. These interviews, of course, are just one component among many in regard to the difficult admissions decision, a chance for the candidate to show us how they think through a problem, how well they can explain their ideas, how well they take hints and suggestions. In every interview, we had paused at a certain stage, when the candidate had a fully formed argument for one of the variations, and asked them to undertake an integrative exercise, summarizing as clearly as they could the entire problem and solution and how they expected it to play out. Since these were interviews for joint philosophy degrees, in my view this step was a key part of the interview, measuring the ability of the candidate to integrate what they had learned from the discussion and to present a complete, coherent argument.

Several of these puzzles are rather open-ended, not necessarily with a clear-cut objectively correct answer, although there are certain important issues that arise and that we had hoped the candidates would realize. So let me now discuss the puzzles in detail and the aspects of them that I find interesting.

For the main puzzle, it seems clear that one should not announce a color to the host straight away in round 1, because it seems not likely enough that the other contestant would also do so with exactly the same color. Rather, one should use the messaging capability somehow to agree on the color and round number on which to end the game. We should only announce, if we have both clearly expressed a plan to announce a specific color on a specific round.

At first, it might seem reasonable to send a message along the lines of “Let’s both announce red on round 3; please confirm in round 2.” The hope would be that the other player would indeed confirm on round 2, and then both would announce the final color of red on round 3.

That idea would work, if we were taking turns in sending messages, as in the alternation version of the puzzle. But in the main puzzle, the players are sending their messages simultaneously, and there is a difficulty for the previous proposal that can be realized by considering what kind of message we might expect to be receiving from our partner on round 1. Namely, perhaps they had a similar idea. It would be a lucky case indeed, if they had had exactly the same idea, also proposing to announce red in round 3. In that event, we would both confirm in round 2 and win with red in round 3.

But a more problematic case would occur if our partner had had a very similar idea, but happened to have made the proposal with a different color. Perhaps in round 1 our partner would have sent the message, “Let’s both announce blue on round 3; please confirm in round 2.” What shall we do then in round 2? If I decide to try to stick with my original red color, I might say, “Let’s use red in round 3”, then the other player might similarly decide to stick with blue, and we would still be at an impasse. Alternatively, if I decide to defer and switch to blue, perhaps they also defer and switch to red. What is clear is that we can’t be sure of confirming on the same color in round 2, and clearly we shouldn’t announce in round 3 with the color choice still unsettled. So the procedure we have discussed so far seems unsuitable.

A deeper contemplation of the problem reveals that this issue of the other contestant doing something very similar, but perhaps with a different color, is a fundamental obstacle to many solution attempts. There is a fundamental symmetry in play between the contestants. Since the messages are sent and received at the same time, and both players are equally logical, it could be that we both end up sending the same kind of message each time, except with different colors. We need somehow to agree on a color and a round number, but it seems that however we decide to make a proposal, it would be equally logical for our partner to make a similar proposal, but with colors permuted somehow.

We seem to need to break the symmetry somehow. Perhaps one might hope to use the alphabetically least color, amongst the two colors that have been proposed so far. That would seem to be a perfectly reasonable way to break the symmetry. But the problem here is that there are also many other perfectly reasonable ways to break the symmetry—the shortest wavelength color, the shorter color word, the color proposed by the younger person, by the older person, and so on. The difficulty is that any given proposal might be faced with a similar but equally reasonable proposal about how to break the symmetry. If I had proposed one such criterion, perhaps my partner proposes another, and we haven’t made any progress, but rather only replaced the difficulty of agreeing on the color with the difficulty of agreeing on the decision criterion of how to decide.

A key insight comes with the idea to use randomness. Suppose I had said in round 1, “I suggest we both announce red next round; I shall do so, if you also suggest this,” but you had said, “I suggest we both announce blue next round; I shall do so, if you also suggest this.” But I have got a coin in my pocket, and from now on, I intend each round to flip the coin, sending the red message on heads and the blue message on tails. If you do likewise, then we are very likely in a few rounds to hit upon the same message, and then we shall win on the next round by following through, announcing the agreed-upon color. That is, if in the first round we each happen to mention a color (whether or not my partner’s message was of the specific form I had mentioned), then I am going on every round to send a message as above, but using one of the two colors randomly. If you logically also choose to do this, then we are very likely to happen to choose the same color on some round eventually, and then win on the following round. This randomzing strategy thus seems fairly sound as a means to come to agreement, and if the partner also realizes this we should expect to win quickly, in just a few rounds with high likelihood. For this reason, I find this to be the best strategy.

A slightly more abstract way to describe the strategy I am proposing is that the symmetry between the players will need to be broken by us agreeing which of us will be leading the process and which of us will follow. I can flip a coin each round to decide if I should try to be leader or follower on that round. On heads, I shall propose a specific plan, “let’s say red next round, if your message indicates that you will follow my suggestion.” On tails, I shall send a message saying, “I am going to follow your plan of action, if you send one this round.” In this way, we are likely in a few rounds to have established a leader and follower, and win shortly afterward.

Some student candidates and commentators had proposed an interesting idea of trying to blend the two colors. This would be a way of coming to an agreement, but without needing to break the symmetry between the players. If I had proposed initially that we should announce red and you had proposed blue, then the idea is that logically we should both try to average these colors and say purple. I like this idea a lot, but it seems problematic in light of the fact that we don’t have such a clear and unambiguous means of combining colors. For example, should we mix the colors in the manner of mixing paint or rather in the manner of mixing colored light, which produces completely different results? Even when mixing red+blue, I might say purple and you might say violet. And what of stranger color combinations, such as orange+green? If we were using rgb colors, then some colors are simply adjacent in the color space (or an even distance away), and so they have no exact average mixing. Furthermore, why should we use rgb rather than cmyk or another color space system? And color mixing does not work identically for these two systems. For these reasons, I find the color mixing idea ultimately less than completely successful.

The alternation variation of the puzzle admits an easy solution, since the symmetry is broken already by the rules of the game. Player 1 will be the first to send a message, and can simply say, “I shall announce red on round 2; if you do so as well then we shall win.” There would be no reason for player 2 not to follow along, and so the players can expect to win on round 2.

Some candidates had suggested a confirmation round, having player 1 say, “let’s say red on round 3, confirm if that is agreeable.” Then both players would confirm the intention on round 2, and win on round 3. This is also successful, but it seems to me that the confirmation step is not strictly needed.

The collision variation is an interesting hybrid, because although there is a symmetry between the players in terms of how the rules apply to them, the symmetry is broken in the event that one sends a message and the other does not. The best solution here seems to be a random solution. Namely, flip a coin to decide whether to send a message or stay silent. On heads, send the message “Announce red on next round, if this message gets through.” On tails, do nothing, and plan to follow whatever is suggested on the message that might be received. Because of the randomness, it is very likely that in a few rounds a message gets through one way or the other, with a quick win straight afterward.

The pigeon variation is simply a version of the two-generals problem. The first player can try to propose a specific plan, naming a round and color, and ask for confirmation. But the confirmation itself will need to be confirmed, if the other player is to want certainty. But in this case the confirmation of the confirmation will need to be confirmed, and so on ad infinitum. No finite number of confirmations of receipt will be enough, even if all are received, since if $n$ confirmations suffice to attain mutual certainty, then the last confirmation needn’t be sent, since the protocol would work even if it didn’t arrive, and so $n-1$ should also suffice, a contradiction.

Meanwhile, one can attain very high degree of confidence in the plan. If on round 1 the first player sends the message, “We shall say red on round 1000, please confirm,” then he might hope it gets through and confirmation received. On each subsequent round, he should send this message again, together with a complete record of all prior messages sent/received. Player 2, if receiving the message on round 1 should simply follow along and send confirmations; but if nothing was received on round 1, he should not start a competing plan, which would only reintroduce the priority issue of the main puzzle, since the symmetry is already clearly broken for this version with player 1 in the leading role. So if no message was received, player 2 should simply reply with “sorry, no message received; please resend.” Eventually, messages are likely to get through and the two players will be on track to win with high probability. The players should plan on announcing red on round 1000 according to the plan, even if confirmations are missed. One can increase the probability by increasing round 1000 to a much higher number, presuming that the players can keep careful track of the round numbers. It would seem to be a mistake ever to try to change the round number or the color after some messages are sent, since perhaps the originals were received and only the cofirmations were missed, and there would be abundant confusion to have two or more plans in play.

In the admissions interviews, which were generally less than 25 minutes, we were happy if a candidate got to the realization of the symmetry issue in the main puzzle, before going on to the alternating version, which most students got quickly, and then the collision version, where the randomness idea seemed to arise more naturally. The best candidates were then able to realize how to apply the randomness idea to the main version. With the pigeon version of the puzzle, successful candidates realized the need to achieve confirmation and confirmation of confirmation, and a few put this together to mount the impossibility argument for the case of certainty, and most realized how to increase the likelihood of success by picking a distant round and repeating messages. It was quite enjoyable for me to discuss these problems with so many very sharp student candidates.

Let me close by mentioning a few observations that surprised me about using the puzzles in an interview setting.

The first observation is the remarkable amount of personality that was revealed by the candidate’s choices in the puzzles. Some candidates tended to follow what might be called a leader’s approach (or the bully strategy?), attempting in the main puzzle to achieve agreement by conveying obstinateness in the color choice, to convince the other person to change sides as a way of coming to agreement. An equal number of other candidates tried instead to be deferential, sending messages that they would agree to use whichever color the other person wanted. Of course, each of these strategies works fine when paired with the other, but when paired with exactly the same personality, the methods face the symmetry problem we discussed earlier. Some brilliant candidates pointed out that the role that these personality differences played in the puzzle—it was very unlikely that the two contestants would be perfectly balanced in their personalities, and so the symmetry would be broken simply because one candidate would be slightly more insistent or slightly more deferential. And I have to say that realistically, this is how the puzzle would actually be solved in practice.

Another observation was that the candidates overwhelmingly chose red as their color, whenever they mentioned a specific color. About 2/3 or more did so. Far behind this was blue, in second place, and then we had a very small number of mentions of orange, green, yellow, and black.

I really enjoyed using this puzzle for the interviews, and I feel it helped us to choose a really great incoming class.

Art by Erin Carmody.

The hierarchy of geometric constructibility: can we go back?

I have recently become interested in the hierarchy of relative constructibility in geometry (see the discussion on my Twitter thread). From a given collection of points in the plane, we may proceed to construct other points using the classical tools of straightedge and compass. This induces a hierarchy of relative constructiblity, namely, for any two sets of points $X$ and $Y$ in the plane, we may say that $X$ is constructible from $Y$, written as $$X\trianglelefteq Y$$ if for every point $A\in X$, there is a construction procedure using only straightedge and compass proceeding from points in $Y$ and constructing $A$. And so we are faced with a hierarchy of constructibility.

Perhaps one might ask immediately whether this is genuinely a hierarchy. A central case occurs with pairs of points, since one often proceeds in geometry with the idea of a unit segment having already been fixed.

Question 1. Given two distinct points A and B, suppose that distinct points C and D are constructible by straightedge and compass. Can we go back? That is, must A and B both be constructible from C and D?

The answer, surprisingly, is Yes! This is a sense in which the relative constructibility notion is an equivalence relation rather than a hierarchy, since it shows that if AB can construct CD, than it is also true conversely. Since constructibility is also transitive, this shows that the relative constructibility is an equivalence relation on pairs of points in the plane.

Before addressing the question, let me first dispense with a distracting line of reasoning that suggests a wrong answer to this question. Suppose that we have a segment AB of length $\sqrt[3]2$, a nonconstructible number. Since the constructible numbers are closed under multiplication, this might suggest that from AB we can construct a segment of length 2, and therefore also construct a segment CD of length 1. But from a segment of length 1, we can never construct a segment of length $\sqrt[3]2$, because the duplication of the cube is impossible. So this would suggest a negative answer to the question. Is it right?

But no, this distracting counter-argument is not correct. The reason is that although the constructible numbers are indeed closed under multiplication, the construction methods requires the consultation of a fixed unit segment. That is, to prove that from a segment of length $a$ and a segment of length $b$ one can construct a segment of length $ab$, we require the use of fixed unit segment of length $1$. Indeed, it will follow from our analysis here that one cannot omit this consultation of a unit segment from the construction—we cannot in general construct the cube of a segment just from the segment itself.

So let me prove the answer to question 1 is positive. Suppose that from distinct points AB we can construct distinct points CD. Performing excactly the same construction process again, but proceeding from points CD as input, we construct points EF. From points CD we can see how CD are related to EF, forming a quadrilateral CDFE.

The point is that we may now construct the points A and B from C and D by simply inverting this similarity. (This was also observed by user @ZenoRogue on Twitter.) Namely, the two quadrilaterals ABDC and CDFE are similar, and so have all the same angles. So the line AC is constructible from CD by the angle ACD, which is the same as CEF, and the distance AC is scaled from CE by the same factor that CD is related to EF. So from CD we can construct this similar quadrilateral, and thereby construct both A and B.

Question 2. Given three distinct points forming a triangle ABC, suppose that we can construct triangle DEF by straightedge and compass. Can we go back? That is, must A,B, and C be constructible from D,E,F?

The question is of course similar to question 1, which had a positive answer. Nevertheless, the answer here is negative. The reason is that the original idea we had with question 1 concerning $\sqrt[3]2$ now works here to refute question 2. Namely, let ABC be a right triangle where AB has length 1 but AC has length $\sqrt[3]2$.

From the triangle ABC, we may easily construct the triangle ABD, where AD has length 1, since in fact the point D is constructible from the segment AB alone. But from ABD, which is a right triangle with both legs of length 1, we shall be unable to construct the number $\sqrt[3]2$, and consequently unable to construct the point C. So although ABD is constructible from ABC, the point C is not constructible from points ABD. This is therefore a counterexample to question 2, which therefore must be answered in the negative.

Another argument, offered by Ed Wagstaff, is that from an angle trisection we can construct the original angle, but we can’t necessarily construct the trisection from the original angle.

Finally, I should like to understand the nature of the relative hierarchy of constructibility better. $$X\trianglelefteq Y$$ What is the nature of relative constructibility of sets of points in the plane? For example, one question I didn’t know how to answer at first is the following:

Question 3. Is the hierarchy of relative constructibility a dense order? That is, if the set $X$ is constructible from $Y$ but not conversely, must there be a set of points $Z$ strictly in between? $$X\lhd Y\quad\implies\quad\exists Z\quad X\lhd Z\lhd Y?$$

In an exchange on Twitter with David Madore, I had thought that we answered the question negatively, but now I am not at all sure of that argument (and thanks to Gabin for the comments below!), and so I have deleted the wrong argument here. Question 3 seems currently to be open.

I have now asked question 3 on MathOverflow: Is the hierarchy of relative geometric constructibility by straightedge and compass a dense order?

The surprising strength of reflection in second-order set theory with abundant urelements, Konstanz, December 2021

This will be talk for the workshop Philosophy of Set Theory held at the University of Konstanz, 3 – 4 December 2021 — in person!

Update: Unfortunately, the workshop has been cancelled (perhaps postponed to next year) in light of the Covid resurgence.

Abstract. I shall analyze the roles and interaction of reflection and urelements in second-order set theory. Second-order reflection already exhibits large cardinal strength even without urelements, but recent work shows that in the presence of abundant urelements, second-order reflection is considerably stronger than might have been expected—at the level of supercompact cardinals. This is joint work with Bokai Yao (Notre Dame).

Infinite draughts and the logic of infinitary games, Oslo, November 2021

This will be a talk 11 November 2021 for the Oslo Seminar in Mathematical Logic, meeting online via Zoom at 10:15am CET (9:15am GMT) at Zoom: 671 7500 0197

Abstract. I shall give an introduction to the logic of infinite games, including the theory of transfinite game values, using the case of infinite draughts as a principal illustrative instance. Infinite draughts, also known as infinite checkers, is played like the finite game, but on an infinite checkerboard stretching without end in all four directions. In recent joint work with Davide Leonessi, we proved that every countable ordinal arises as the game value of a position in infinite draughts. Thus, there are positions from which Red has a winning strategy enabling her to win always in finitely many moves, but the length of play can be completely controlled by Black in a manner as though counting down from a given countable ordinal. This result is optimal for games having countably many options at each move—in short, the omega one of infinite draughts is true omega one.

Transfinite game values in infinite draughts

A joint paper with Davide Leonessi, in which we prove that every countable ordinal arises as the game value of a position in infinite draughts, and this result is optimal for games having countably many options at each move. In short, the omega one of infinite draughts is true omega one.

  • J. D. Hamkins and D. Leonessi, “Transfinite game values in infinite draughts,” Mathematics arXiv, 2021.
    [Bibtex]
    @ARTICLE{HamkinsLeonessi:Transfinite-game-values-in-infinite-draughts,
    author = {Joel David Hamkins and Davide Leonessi},
    title = {Transfinite game values in infinite draughts},
    journal = {Mathematics arXiv},
    year = {2021},
    volume = {},
    number = {},
    pages = {},
    month = {},
    note = {Under review},
    abstract = {},
    keywords = {under-review},
    source = {},
    doi = {},
    eprint = {2111.02053},
    archivePrefix = {arXiv},
    primaryClass = {math.LO},
    url = {http://jdh.hamkins.org/transfinite-game-values-in-infinite-draughts},
    }

Download the paper at arXiv:2111.02053

Abstract. Infinite draughts, or checkers, is played just like the finite game, but on an infinite checkerboard extending without bound in all four directions. We prove that every countable ordinal arises as the game value of a position in infinite draughts. Thus, there are positions from which Red has a winning strategy enabling her to win always in finitely many moves, but the length of play can be completely controlled by Black in a manner as though counting down from a given countable ordinal.

Is the twin prime conjecture independent of Peano Arithmetic?

  • A. Berarducci, A. Fornasiero, and J. D. Hamkins, “Is the twin prime conjecture independent of Peano Arithmetic?,” Mathematics arXiv, 2021.
    [Bibtex]
    @ARTICLE{BerarducciFornasieroHamkins:Is-the-twin-prime-conjecture-independent-of-PA,
    author = {Alessandro Berarducci and Antongiulio Fornasiero and Joel David Hamkins},
    title = {Is the twin prime conjecture independent of Peano Arithmetic?},
    journal = {Mathematics arXiv},
    year = {2021},
    volume = {},
    number = {},
    pages = {},
    month = {},
    note = {Under review},
    abstract = {},
    keywords = {under-review},
    source = {},
    doi = {},
    eprint = {2110.08640},
    archivePrefix = {arXiv},
    primaryClass = {math.LO},
    url = {http://jdh.hamkins.org/is-the-twin-prime-conjecture-independent-of-peano-arithmetic/},
    }

Download the article at arXiv:2110.08640

Abstract. We show that there is an arithmetical formula $\varphi$ such that ZF proves that $\varphi$ is independent of PA and yet, unlike other arithmetical independent statements, the truth value of $\varphi$ cannot at present be established in ZF or in any other trusted metatheory. In fact we can choose an example of such a formula $\varphi$ such that ZF proves that $\varphi$ is equivalent to the twin prime conjecture. We conclude with a discussion of notion of trustworthy theory and a sharper version of the result.

This work grows in part out of an answer I posted on MathOverflow in 2012.

Frege’s philosophy of mathematics—Interview with Nathan Ormond, December 2021

I was interviewed by Nathan Ormond for a discussion on Frege’s philosophy of mathematics for his YouTube channel, Digital Gnosis, on 10 December 2021 at 4pm.

The interview concludes with a public comment and question & answer session.

Davide Leonessi, MSc MFoCS, Oxford, September 2021

Mr. Davide Leonessi successfully defended his dissertation for the Masters of Science degree in Mathematics and Foundations of Computer Science, entitled “Transfinite game values in infinite games,” on 15 September 2021. Davide earned a distinction for his thesis, an outstanding result.

Davide Leonessi | Google scholar | Dissertation | arXiv

Abstract. The object of this study are countably infinite games with perfect information that allow players to choose among arbitrarily many moves in a turn; in particular, we focus on the generalisations of the finite board games of Hex and Draughts.

In Chapter 1 we develop the theory of transfinite ordinal game values for open infinite games following [Evans-Hamkins 2014], and we focus on the properties of the omega one, that is the supremum of the possible game values, of classes of open games; we moreover design the class of climbing-through-$T$ games as a tool to study the omega one of given game classes.

The original contributions of this research are presented in the following two chapters.

In Chapter 2 we prove classical results about finite Hex and present Infinite Hex, a well-defined infinite generalisation of Hex.

We then introduce the class of stone-placing games, which captures the key features of Infinite Hex and further generalises the class of positional games already studied in the literature within the finite setting of Combinatorial Game Theory.

The main result of this research is the characterization of open stone-placing games in terms of the property of essential locality, which leads to the conclusion that the omega one of any class of open stone-placing games is at most $\omega$. In particular, we obtain that the class of open games of Infinite Hex has the smallest infinite omega one, that is $\omega_1^{\rm Hex}=\omega$.

In Chapter 3 we show a dual result; we define the class of games of Infinite Draughts and explicitly construct open games of arbitrarily high game value with the tools of Chapter 1, concluding that the omega one of the class of open games of Infinite Draughts is as high as possible, that is $\omega_1^{\rm Draughts}=\omega_1$.

The full dissertation is available:

A deflationary account of Fregean abstraction in Zermelo-Fraenkel ZF set theory, Oxford, November 2021

This will be a talk for the Oxford Seminar in the Philosophy of Mathematics, 1 November, 4:30-6:30 GMT. The talk will be held on Zoom (contact the seminar organizers for the Zoom link).

Abstract. The standard treatment of sets and classes in Zermelo-Fraenkel set theory instantiates in many respects the Fregean foundational distinction between objects and concepts, for in set theory we commonly take the sets as objects to be considered under the guise of diverse concepts, the definable classes, each serving as a predicate on that domain of individuals. Although it is often asserted that there can be no association of classes with objects in a way that fulfills Frege’s Basic Law V, nevertheless, in the ZF framework I have described, it turns out that Basic Law V does hold, and provably so, along with other various Fregean abstraction principles. These principles are consequences of Zermelo-Fraenkel ZF set theory in the context of all its definable classes. Namely, there is an injective mapping from classes to objects, definable in senses I shall explain, associating every first-order parametrically definable class $F$ with a set object $\varepsilon F$, in such a way that Basic Law V is fulfilled: $$\varepsilon F =\varepsilon G\iff\forall x\ (Fx\leftrightarrow Gx).$$ Russell’s elementary refutation of the general comprehension axiom, therefore, is improperly described as a refutation of Basic Law V itself, but rather refutes Basic Law V only when augmented with powerful class comprehension principles going strictly beyond ZF. The main result leads also to a proof of Tarski’s theorem on the nondefinability of truth as a corollary to Russell’s argument.

My favorite theorem

What a pleasure it was to be interviewed by Evelyn Lamb and Kevin Knudson for their wonderful podcast series, My Favorite Theorem, available on Apple, Spotify, and any number of other aggregators.

I had a chance to talk about one my most favorite theorems, the fundamental theorem of finite games.

Theorem.(Zermelo 1913) In any two-player finite game of perfect information, one of the players has a winning strategy, or both players have drawing strategies.

Listen to the podcast here: My Favorite Theorem. A transcript is also available.

Lectures on the Philosophy of Mathematics, Michaelmas term 2021, Oxford

Philosophy of Mathematics, Exam Paper 122, Oxford University

Wednesdays 12-1 during term, Radcliffe Humanities Lecture Room

Joel David Hamkins, Professor of Logic

Lucy, Charles — Personifications of Mathematics and Geometry

This series of self-contained lectures on the philosophy of mathematics is intended for students preparing for Oxford Philosophy exam paper 122. All interested parties from the Oxford University community, however, are welcome to attend, whether or not they intend to sit the exam. The lectures will be organized loosely around mathematical themes, in such a way that brings various philosophical issues naturally to light. Lectures will loosely follow the instructor’s book Lectures on the Philosophy of Mathematics (MIT Press 2021), with supplemental suggested readings each week.

Previously recorded lectures from last year are available on the lecturer’s YouTube channel, below.

In light of the earlier lectures being available online, the plan for the lectures this year will be to feel somewhat more free occasionally to focus on narrower topics, and also to entertain at times a discussion format. Therefore kindly bring questions and well-thought-out opinions to the lecture.

The lectures this term will be held in person. The lecturer requests that students be vaccinated, wear masks, and observe social distancing as practicable. If this proves impossible or unsustainable, we shall regretably revert to online lectures on short notice.

Lecture 1. Numbers

Numbers are perhaps the essential mathematical idea, but what are numbers? There are many kinds of numbers—natural numbers, integers, rational numbers, real numbers, complex numbers, hyperreal numbers, surreal numbers, ordinal numbers, and more—and these number systems provide a fruitful background for classical arguments on incommensurability and transcendentality, while setting the stage for discussions of platonism, logicism, the nature of abstraction, the significance of categoricity, and structuralism.

Lecture 2. Rigour

Let us consider the problem of mathematical rigor in the development of the calculus. Informal continuity concepts and the use of infinitesimals ultimately gave way to the epsilon-delta limit concept, which secured a more rigorous foundation while also enlarging our conceptual vocabulary, enabling us to express more refined notions, such as uniform continuity, equicontinuity, and uniform convergence. Nonstandard analysis resurrected the infinitesimals on a more secure foundation, providing a parallel development of the subject. Meanwhile, increasing abstraction emerged in the function concept, which we shall illustrate with the Devil’s staircase, space-filling curves, and the Conway base 13 function. Finally, does the indispensability of mathematics for science ground mathematical truth? Fictionalism puts this in question.

Lecture 3. Infinity

We shall follow the allegory of Hilbert’s hotel and the paradox of Galileo to the equinumerosity relation and the notion of countability. Cantor’s diagonal arguments, meanwhile, reveal uncountability and a vast hierarchy of different orders of infinity; some arguments give rise to the distinction between constructive and nonconstructive proof. Zeno’s paradox highlights classical ideas on potential versus actual infinity. Furthermore, we shall count into the transfinite ordinals.

Lecture 4. Geometry

Classical Euclidean geometry is the archetype of a mathematical deductive process. Yet the impossibility of certain constructions by straightedge and compass, such as doubling the cube, trisecting the angle, or squaring the circle, hints at geometric realms beyond Euclid. The rise of non-Euclidean geometry, especially in light of scientific theories and observations suggesting that physical reality is not Euclidean, challenges previous accounts of what geometry is about. New formalizations, such as those of David Hilbert and Alfred Tarski, replace the old axiomatizations, augmenting and correcting Euclid with axioms on completeness and betweenness. Ultimately, Tarski’s decision procedure points to a tantalizing possibility of automation in geometrical reasoning.

Lecture 5. Proof

What is proof? What is the relation between proof and truth? Is every mathematical truth true for a reason? After clarifying the distinction between syntax and semantics and discussing various views on the nature of proof, including proof-as-dialogue, we shall consider the nature of formal proof. We shall highlight the importance of soundness, completeness, and verifiability in any formal proof system, outlining the central ideas used in proving the completeness theorem. The compactness property distills the finiteness of proofs into an independent, purely semantic consequence. Computer-verified proof promises increasing significance; its role is well illustrated by the history of the four-color theorem. Nonclassical logics, such as intuitionistic logic, arise naturally from formal systems by weakening the logical rules.

Lecture 6. Computability

What is computability? Kurt Gödel defined a robust class of computable functions, the primitive recursive functions, and yet he gave reasons to despair of a fully satisfactory answer. Nevertheless, Alan Turing’s machine concept of computability, growing out of a careful philosophical analysis of the nature of human computability, proved robust and laid a foundation for the contemporary computer era; the widely accepted Church-Turing thesis asserts that Turing had the right notion. The distinction between computable decidability and computable enumerability, highlighted by the undecidability of the halting problem, shows that not all mathematical problems can be solved by machine, and a vast hierarchy looms in the Turing degrees, an infinitary information theory. Complexity theory refocuses the subject on the realm of feasible computation, with the still-unsolved P versus NP problem standing in the background of nearly every serious issue in theoretical computer science.

Lecture 7. Incompleteness

David Hilbert sought to secure the consistency of higher mathematics by finitary reasoning about the formalism underlying it, but his program was dashed by Gödel’s incompleteness theorems, which show that no consistent formal system can prove even its own consistency, let alone the consistency of a higher system. We shall describe several proofs of the first incompleteness theorem, via the halting problem, self-reference, and definability, showing senses in which we cannot complete mathematics. After this, we shall discuss the second incompleteness theorem, the Rosser variation, and Tarski’s theorem on the nondefinability of truth. Ultimately, one is led to the inherent hierarchy of consistency strength rising above every foundational mathematical theory.

Lecture 8. Set Theory

We shall discuss the emergence of set theory as a foundation of mathematics. Cantor founded the subject with key set-theoretic insights, but Frege’s formal theory was naive, refuted by the Russell paradox. Zermelo’s set theory, in contrast, grew ultimately into the successful contemporary theory, founded upon a cumulative conception of the set-theoretic universe. Set theory was simultaneously a new mathematical subject, with its own motivating questions and tools, but it also was a new foundational theory with a capacity to represent essentially arbitrary abstract mathematical structure. Sophisticated technical developments, including in particular, the forcing method and discoveries in the large cardinal hierarchy, led to a necessary engagement with deep philosophical concerns, such as the criteria by which one adopts new mathematical axioms and set-theoretic pluralism.

Book review, Catarina Dutilh Novaes, The dialogical roots of deduction

In this insightful and remarkable work, Professor Novaes defends and explores at length the philosophical thesis that mathematical proof and deduction generally has a fundamentally dialogical nature, proceeding in a back-and-forth dialogue between two semi-adversarial but collaborative actors, the Prover and the Skeptic, who together aim to find mathematical insight. This view of proof-as-dialogue, she argues, carries explanatory power for the philosophy of mathematical practice, explaining diverse aspects of proof-writing, refereeing, and more, including the multifaceted roles of proof, including proof as verification, proof as certification, proof for communication and persuasion, proof as explanation, and proof as a driver of mathematical innovation.

In extensive, refined scholarly work, Novaes explores the historical and intellectual roots of the dialogical perspective on deduction, tracing the idea from ancient times through medieval philosophy and into the present day, including case studies of current mathematical developments, such as Mochizuki’s claimed proof of the abc conjecture, as well as recent psychological experiments on the role of group reasoning in resolving certain well-known disappointing failures of rationality, such as in the Wason card experiment. Truly fascinating.

On the basis of her work, I have nominated Novaes for the Lakatos Award (given annually “for an outstanding contribution to the philosophy of science, widely interpreted, in the form of a book published in English during the current year or the previous five years”). Lakatos himself, of course, was a friend of dialogical mathematics—his famous Proofs and Refutations proceeds after all in a dialogue between mathematicians of different philosophical outlooks. Novaes engages with Lakatos’s work explicitly, pointing out the obvious parallels, but also highlighting important differences between her Prover/Skeptic dialogical account and the kind of proof dialogues appearing in Proofs and Refutations. In light of this connection with Lakatos, I would find it especially fitting for Novaes to win the Lakatos Award.

My review on GoodReads

Counting to Infinity and Beyond

Anyone can learn to count in the ordinals, even a child, and so let us learn how to count to $\omega^2$, the first compound limit ordinal.

The large-format poster is available:

Some close-up views:

I would like to thank the many people who had made helpful suggestions concerning my poster, including Andrej Bauer and especially Saul Schleimer, who offered many detailed suggestions.

Infinite sets and Foundations—Interviewed on the Daniel Rubin Show

I was interviewed 26 August 2021 by mathematician Daniel Rubin on his show, and we had a lively, wideranging discussion spanning mathematics, infinity, and the philosophy of mathematics. Please enjoy!

Contents

0:00 Intro

2:11 Joel’s background. Interaction between math and philosophy

9:04 Joel’s work; infinite chess.

14:45 Infinite ordinals

22:27 The Cantor-Bendixson process

29:41 Uncountable ordinals

32:10 First order vs. second order theories

41:16 Non-standard analysis

46:57 The ZFC axioms and well-ordering of the reals

58:11 Showing independence of statements. Models and forcing.

1:04:38 Sets, classes, and categories

1:19:22 Is there one true set theory? Are projective sets Lebesgue measurable?

1:30:20 What does set theory look like if certain axioms are rejected?

1:36:06 How to judge philosophical positions about math

1:42:01 Concrete math where set theory becomes relevant. Tarski-Seidenberg on positive polynomials.

1:48:48 Goodstein sequences and the use of infinite ordinals

1:58:43 The state of set theory today

2:01:41 Joel’s recent books

Go check out the other episodes on Daniel’s channel!

An algebra of orders

Did you know that you can add and multiply orders? For any two order structures $A$ and $B$, we can form the ordered sum $A+B$ and ordered product $A\otimes B$, and other natural operations, such as the disjoint sum $A\sqcup B$, which make altogether an arithmetic of orders. We combine orders with these operations to make new orders, often with interesting properties. Let us explore the resulting algebra of orders!

$\newcommand\Z{\mathbb{Z}}\newcommand\N{\mathbb{N}}\newcommand\Q{\mathbb{Q}}\newcommand\P{\mathbb{P}}\newcommand\iso{\cong}\newcommand\of{\subseteq}$
One of the most basic operations that we can use to combine two orders is the disjoint sum operation $A\sqcup B$. This is the order resulting from placing a copy of $A$ adjacent to a copy of $B$, side-by-side, forming a combined order with no instances of the order relation between the two parts. If $A$ is the orange $\vee$-shaped order here and $B$ is the yellow linear order, for example, then $A\sqcup B$ is the combined order with all five nodes.

Another kind of addition is the ordered sum of two orders $A+B$, which is obtained by placing a copy of $B$ above a copy of $A$, as indicated here by adding the orange copy of $A$ and the yellow copy of $B$. Also shown is the sum $B+A$, with the summands reversed, so that we take $B$ below and $A$ on top. It is easy to check that the ordered sum of two orders is an order. One notices immediately, of course, that the resulting ordered sums $A+B$ and $B+A$ are not the same! The order $A+B$ has a greatest element, whereas $B+A$ has two maximal elements. So the ordered sum operation on orders is not commutative. Nevertheless, we shall still call it addition. The operation, which has many useful and interesting features, goes back at least to the 19th century with Cantor, who defined the addition of well orders this way.

In order to illustrate further examples, I have assembled here an addition table for several simple finite orders. The choices for $A$ appear down the left side and those for $B$ at the top, with the corresponding sum $A+B$ displayed in each cell accordingly.

We can combine the two order addition operations, forming a variety of other orders this way.

The reader is encouraged to explore further how to add various finite orders using these two forms of addition. What is the smallest order that you cannot generate from $1$ using $+$ and $\sqcup$? Please answer in the comments.

We can also add infinite orders. Displayed here, for example, is the order $\N+(1\sqcup 1)$, the natural numbers wearing two yellow caps. The two yellow nodes at the top form a copy of $1\sqcup 1$, while the natural numbers are the orange nodes below. Every natural number (yes, all infinitely many of them) is below each of the two nodes at the top, which are incomparable to each other. Notice that even though we have Hasse diagrams for each summand order here, there can be no minimal Hasse diagram for the sum, because any particular line from a natural number to the top would be implied via transitivity from higher such lines, and we would need such lines, since they are not implied by the lower lines. So there is no minimal Hasse diagram.

This order happens to illustrate what is called an exact pair, which occurs in an order when a pair of incomparable nodes bounds a chain below, with the property that any node below both members of the pair is below something in the chain. This phenomenon occurs in sometimes unexpected contexts—any countable chain in the hierarchy of Turing degrees in computability theory, for example, admits an exact pair.

Let us turn now to multiplication. The ordered product $A\otimes B$ is the order resulting from having $B$ many copies of $A$. That is, we replace each node of $B$ with an entire copy of the $A$ order. Within each of these copies of $A$, the order relation is just as in $A$, but the order relation between nodes in different copies of $A$, we follow the $B$ relation. It is not difficult to check that indeed this is an order relation. We can illustrate here with the same two orders we had earlier.

In forming the ordered product $A\otimes B$, in the center here, we take the two yellow nodes of $B$, shown greatly enlarged in the background, and replace them with copies of $A$. So we have ultimately two copies of $A$, one atop the other, just as $B$ has two nodes, one atop the other. We have added the order relations between the lower copy of $A$ and the upper copy, because in $B$ the lower node is related to the upper node. The order $A\otimes B$ consists only of the six orange nodes—the large highlighted yellow nodes of $B$ here serve merely as a helpful indication of how the product is formed and are not in any way part of the product order $A\otimes B$.

Similarly, with $B\otimes A$, at the right, we have the three enlarged orange nodes of $B$ in the background, which have each been replaced with copies of $A$. The nodes of each of the lower copies of $A$ are related to the nodes in the top copy, because in $B$ the two lower nodes are related to the upper node.

I have assembled a small multiplication table here for some simple finite orders.

So far we have given an informal account of how to add and multiply ordered ordered structures. So let us briefly be a little more precise and formal with these matters.

In fact, when it comes to addition, there is a slightly irritating matter in defining what the sums $A\sqcup B$ and $A+B$ are exactly. Specifically, what are the domains? We would like to conceive of the domains of $A\sqcup B$ and $A+B$ simply as the union the domains of $A$ and $B$—we’d like just to throw the two domains together and form the sums order using that combined domain, placing $A$ on the $A$ part and $B$ on the $B$ part (and adding relations from the $A$ to the $B$ part for $A+B$). Indeed, this works fine when the domains of $A$ and $B$ are disjoint, that is, if they have no points in common. But what if the domains of $A$ and $B$ overlap? In this case, we can’t seem to use the union in this straightforward manner. In general, we must disjointify the domains—we take copies of $A$ and $B$, if necessary, on domains that are disjoint, so that we can form the sums $A\sqcup B$ and $A+B$ on the union of those nonoverlapping domains.

What do we mean precisely by “taking a copy” of an ordered structure $A$? This way of talking in mathematics partakes in the philosophy of structuralism. We only care about our mathematical structures up to isomorphism, after all, and so it doesn’t matter which isomorphic copies of $A$ and $B$ we use; the resulting order structures $A\sqcup B$ will be isomorphic, and similarly for $A+B$. In this sense, we are defining the sum orders only up to isomorphism.

Nevertheless, we can be definite about it, if only to verify that indeed there are copies of $A$ and $B$ available with disjoint domains. So let us construct a set-theoretically specific copy of $A$, replacing each individual $a$ in the domain of $A$ with $(a,\text{orange})$, for example, and replacing the elements $b$ in the domain of $B$ with $(b,\text{yellow})$. If “orange” is a specific object distinct from “yellow,” then these new domains will have no points in common, and we can form the disjoint sum $A\sqcup B$ by using the union of these new domains, placing the $A$ order on the orange objects and the $B$ order on the yellow objects.

Although one can use this specific disjointifying construction to define what $A\sqcup B$ and $A+B$ mean as specific structures, I would find it to be a misunderstanding of the construction to take it as a suggestion that set theory is anti-structuralist. Set theorists are generally as structuralist as they come in mathematics, and in light of Dedekind’s categorical account of the natural numbers, one might even find the origin of the philosophy of structuralism in set theory. Rather, the disjointifying construction is part of the general proof that set theory abounds with isomorphic copies of whatever mathematical structure we might have, and this is part of the reason why it serves well as a foundation of mathematics for the structuralist. To be a structuralist means not to care which particular copy one has, to treat one’s mathematical structures as invariant under isomorphism.

But let me mention a certain regrettable consequence of defining the operations by means of a specific such disjointifying construction in the algebra of orders. Namely, it will turn out that neither the disjoint sum operation nor the ordered sum operation, as operations on order structures, are associative. For example, if we use $1$ to represent the one-point order, then $1\sqcup 1$ means the two-point side-by-side order, one orange and one yellow, but really what we mean is that the points of the domain are $\set{(a,\text{orange}),(a,\text{yellow})}$, where the original order is on domain $\set{a}$. The order $(1\sqcup 1)\sqcup 1$ then means that we take an orange copy of that order plus a single yellow point. This will have domain
$$\set{\bigl((a,\text{orange}),\text{orange}\bigr),\bigl((a,\text{yellow}),\text{orange}\bigr),(a,\text{yellow})}$$
The order $1\sqcup(1\sqcup 1)$, in contrast, means that we take a single orange point plus a yellow copy of $1\sqcup 1$, leading to the domain
$$\set{(a,\text{orange}),\bigl((a,\text{orange}),\text{yellow}\bigr),\bigl((a,\text{yellow}),\text{yellow}\bigr)}$$
These domains are not the same! So as order structures, the order $(1\sqcup 1)\sqcup 1$ is not identical with $1\sqcup(1\sqcup 1)$, and therefore the disjoint sum operation is not associative. A similar problem arises with $1+(1+1)$ and $(1+1)+1$.

But not to worry—we are structuralists and care about our orders here only up to isomorphism. Indeed, the two resulting orders are isomorphic as orders, and more generally, $(A\sqcup B)\sqcup C$ is isomorphic to $A\sqcup(B\sqcup C)$ for any orders $A$, $B$, and $C$, and similarly with $A+(B+C)\cong(A+B)+C$, as discussed with the theorem below. Furthermore, the order isomorphism relation is a congruence with respect to the arithmetic we have defined, which means that $A\sqcup B$ is isomorphic to $A’\sqcup B’$ whenever $A$ and $B$ are respectively isomorphic to $A’$ and $B’$, and similarly with $A+B$ and $A\otimes B$. Consequently, we can view these operations as associative, if we simply view them not as operations on the order structures themselves, but on their order-types, that is, on their isomorphism classes. This simple abstract switch in perspective restores the desired associativity. In light of this, we are free to omit the parentheses and write $A\sqcup B\sqcup C$ and $A+B+C$, if care about our orders only up to isomorphism. Let us therefore adopt this structuralist perspective for the rest of our treatment of the algebra of orders.

Let us give a more precise formal definition of $A\otimes B$, which requires no disjointification. Specifically, the domain is the set of pairs $\set{(a,b)\mid a\in A, b\in B}$, and the order is defined by $(a,b)\leq_{A\otimes B}(a’,b’)$ if and only if $b\leq_B b’$, or $b=b’$ and $a\leq_A a’$. This order is known as the reverse lexical order, since we are ordering the nodes in the dictionary manner, except starting from the right letter first rather than the left as in an ordinary dictionary. One could of course have defined the product using the lexical order instead of the reverse lexical order, and this would give $A\otimes B$ the meaning of “$A$ copies of $B$.” This would be a fine alternative and in my experience mathematicians who rediscover the ordered product on their own often tend to use the lexical order, which is natural in some respects. Nevertheless, there is a huge literature with more than a century of established usage with the reverse lexical order, from the time of Cantor, who defined ordinal multiplication $\alpha\beta$ as $\beta$ copies of $\alpha$. For this reason, it seems best to stick with the reverse lexical order and the accompanying idea that $A\otimes B$ means “$B$ copies of $A$.” Note also that with the reverse lexical order, we shall be able to prove left distributivity $A\otimes(B+C)=A\otimes B+A\otimes C$, whereas with the lexical order, one will instead have right distributivity $(B+C)\otimes^* A=B\otimes^* A+C\otimes^* A$.

Let us begin to prove some basic facts about the algebra of orders.

Theorem. The following identities hold for orders $A$, $B$, and $C$.

  1. Associativity of disjoint sum, ordered sum, and ordered product.\begin{eqnarray*}A\sqcup(B\sqcup C) &\iso& (A\sqcup B)\sqcup C\\ A+(B+C) &\iso& (A+B)+C\\ A\otimes(B\otimes C) &\iso& (A\otimes B)\otimes C \end{eqnarray*}
  2. Left distributivity of product over disjoint sum and ordered sum.\begin{eqnarray*} A\otimes(B\sqcup C) &\iso& (A\otimes B)\sqcup(A\otimes C)\\ A\otimes(B+C) &\iso& (A\otimes B)+(A\otimes C) \end{eqnarray*}

In each case, these identities are clear from the informal intended meaning of the orders. For example, $A+(B+C)$ is the order resulting from having a copy of $A$, and above it a copy of $B+C$, which is a copy of $B$ and a copy of $C$ above it. So one has altogether a copy of $A$, with a copy of $B$ above that and a copy of $C$ on top. And this is the same as $(A+B)+C$, so they are isomorphic.

One can also aspire to give a detailed formal proof verifying that our color-coded disjointifying process works as desired, and the reader is encouraged to do so as an exercise. To my way of thinking, however, such a proof offers little in the way of mathematical insight into algebra of orders. Rather, it is about checking the fine print of our disjointifying process and making sure that things work as we expect. Several of the arguments can be described as parenthesis-rearranging arguments—one extracts the desired information from the structure of the domain order and puts that exact same information into the correct form for the target order.

For example, if we have used the color-scheme disjointifying process described above, then the elements of $A\sqcup(B\sqcup C)$ each have one of the following forms, where $a\in A$, $b\in B$, and $c\in C$:
$$(a,\text{orange})$$
$$\bigl((b,\text{orange}),\text{yellow}\bigr)$$
$$\bigl((c,\text{yellow}),\text{yellow}\bigr)$$
We can define the color-and-parenthesis-rearranging function $\pi$ to put them into the right form for $(A\sqcup B)\sqcup C$ as follows:
\begin{align*}
\pi:(a,\text{orange})\quad&\mapsto\quad \bigl((a,\text{orange}),\text{orange}\bigl) \\
\pi:\bigl((b,\text{orange}),\text{yellow}\bigr)\quad&\mapsto\quad \bigl((b,\text{yellow}),\text{orange}\bigl) \\
\pi:\bigl((c,\text{yellow}),\text{yellow}\bigr)\quad&\mapsto\quad (c,\text{yellow})
\end{align*}
In each case, we will preserve the order, and since the orders are side-by-side, the cases never interact in the order, and so this is an isomorphism.

Similarly, for distributivity, the elements of $A\otimes(B\sqcup C)$ have the two forms:
$$\bigl(a,(b,\text{orange})\bigr)$$
$$\bigl(a,(c,\text{yellow})\bigr)$$
where $a\in A$, $b\in B$, and $c\in C$. Again we can define the desired ismorphism $\tau$ by putting these into the right form for $(A\otimes B)\sqcup(A\otimes C)$ as follows:
\begin{align*}
\tau:\bigl(a,(b,\text{orange})\bigr)\quad&\mapsto\quad \bigl((a,b),\text{orange}\bigr) \\
\tau:\bigl(a,(c,\text{yellow})\bigr)\quad&\mapsto\quad\bigl((a,c),\text{yellow}\bigr)
\end{align*}
And again, this is an isomorphism, as desired.

Since order multiplication is not commutative, it is natural to inquire about the right-sided distributivity laws:
\begin{eqnarray*}
(B+C)\otimes A&\overset{?}{\cong}&(B\otimes A)+(C\otimes A)\\
(B\sqcup C)\otimes A&\overset{?}{\cong}&(B\otimes A)\sqcup(C\otimes A)
\end{eqnarray*}
Unfortunately, however, these do not hold in general, and the following instances are counterexamples. Can you see what to take as $A$, $B$, and $C$? Please answer in the comments.

Theorem.

  1. If $A$ and $B$ are linear orders, then so are $A+B$ and $A\otimes B$.
  2. If $A$ and $B$ are nontrivial linear orders and both are endless, then $A+B$ is endless; if at least one of them is endless, then $A\otimes B$ is endless.
  3. If $A$ is an endless dense linear order and $B$ is linear, then $A\otimes B$ is an endless dense linear order.
  4. If $A$ is an endless discrete linear order and $B$ is linear, then $A\otimes B$ is an endless discrete linear order.

Proof. If both $A$ and $B$ are linear orders, then it is clear that $A+B$ is linear. Any two points within the $A$ copy are comparable, and any two points within the $B$ copy, and every point in the $A$ copy is below any point in the $B$ copy. So any two points are comparable and thus we have a linear order. With the product $A\otimes B$, we have $B$ many copies of $A$, and this is linear since any two points within one copy of $A$ are comparable, and otherwise they come from different copies, which are then comparable since $B$ is linear. So $A\otimes B$ is linear.

For statement (2), we know that $A+B$ and $A\otimes B$ are nontrivial linear orders. If both $A$ and $B$ are endless, then clearly $A+B$ is endless, since every node in $A$ has something below it and every node in $B$ has something above it. For the product $A\otimes B$, if $A$ is endless, then every node in any copy of $A$ has nodes above and below it, and so this will be true in $A\otimes B$; and if $B$ is endless, then there will always be higher and lower copies of $A$ to consider, so again $A\otimes B$ is endless, as desired.

For statement (3), assume that $A$ is an endless dense linear and that $B$ is linear. We know from (1) that $A\otimes B$ is a linear order. Suppose that $x<y$ in this order. If $x$ and $y$ live in the same copy of $A$, then there is a node $z$ between them, because $A$ is dense. If $x$ occurs in one copy of $A$ and $B$ in another, then because $A$ is endless, there will a node $z$ above $x$ in its same copy, leading to $x<z<y$ as desired. (Note: we don’t need $B$ to be dense.)

For statement (4), assume instead that $A$ is an endless discrete linear order and $B$ is linear. We know that $A\otimes B$ is a linear order. Every node of $A\otimes B$ lives in a copy of $A$, where it has an immediate successor and an immediate predecessor, and these are also immediate successor and predecessor in $A\otimes B$. From this, it follows also that $A\otimes B$ is endless, and so it is an endless discrete linear order. $\Box$

The reader is encouraged to consider as an exercise whether one can drop the “endless” hypotheses in the theorem. Please answer in the comments.

Theorem. The endless discrete linear orders are exactly those of the form $\Z\otimes L$ for some linear order $L$.

Proof. If $L$ is a linear order, then $\Z\otimes L$ is an endless discrete linear order by the theorem above, statement (4). So any order of this form has the desired feature. Conversely, suppose that $\P$ is an endless discrete linear order. Define an equivalence relation for points in this order by which $p\sim q$ if and only $p$ and $q$ are at finite distance, in the sense that there are only finitely many points between them. This relation is easily seen to be reflexive, transitive and symmetric, and so it is an equivalence relation. Since $\P$ is an endless discrete linear order, every object in the order has an immediate successor and immediate predecessor, which remain $\sim$-equivalent, and from this it follows that the equivalence classes are each ordered like the integers $\Z$, as indicated by the figure here.

The equivalence classes amount to a partition of $\P$ into disjoint segments of order type $\Z$, as in the various colored sections of the figure. Let $L$ be the induced order on the equivalence classes. That is, the domain of $L$ consists of the equivalence classes $\P/\sim$, which are each a $\Z$ chain in the original order, and we say $[a]<_L[b]$ just in case $a<_{\P}b$. This is a linear order on the equivalence classes. And since $\P$ is $L$ copies of its equivalence classes, each of which is ordered like $\Z$, it follows that $\P$ is isomorphic to $\Z\otimes L$, as desired. $\Box$

(Interested readers are advised that the argument above uses the axiom of choice, since in order to assemble the isomorphism of $\P$ with $\Z\otimes L$, we need in effect to choose a center point for each equivalence class.)

If we consider the integers inside the rational order $\Z\of\Q$, it is clear that we can have a discrete suborder of a dense linear order. How about a dense suborder of a discrete linear order?

Question. Is there a discrete linear order with a suborder that is a dense linear order?

What? How could that happen? In my experience, mathematicians first coming to this topic often respond instinctively that this should be impossible. I have seen sophisticated mathematicians make such a pronouncement when I asked the audience about it in a public lecture. The fundamental nature of a discrete order, after all, is completely at odds with density, since in a discrete order, there is a next point up and down, and a next next point, and so on, and this is incompatible with density.

Yet, surprisingly, the answer is Yes! It is possible—there is a discrete order with a suborder that is densely ordered. Consider the extremely interesting order $\Z\otimes\Q$, which consists of $\Q$ many copies of $\Z$, laid out here increasing from left to right. Each tiny blue dot is a rational number, which has been replaced with an entire copy of the integers, as you can see in the magnified images at $a$, $b$, and $c$.

The order is quite subtle, and so let me also provide an alternative presentation of it. We have many copies of $\Z$, and those copies are densely ordered like $\Q$, so that between any two copies of $\Z$ is another one, like this:

Perhaps it helps to imagine that the copies of $\Z$ are getting smaller and smaller as you squeeze them in between the larger copies. But you can indeed always fit another copy of $\Z$ between, while leaving room for the further even tinier copies of $\Z$ to come.

The order $\Z\otimes\Q$ is discrete, in light of the theorem characterizing discrete linear orders. But also, this is clear, since every point of $\Z\otimes\Q$ lives in its local copy of $\Z$, and so has an immediate successor and predecessor there. Meanwhile, if we select exactly one point from each copy of $\Z$, the $0$ of each copy, say, then these points are ordered like $\Q$, which is dense. Thus, we have proved:

Theorem. The order $\Z\otimes\Q$ is a discrete linear order having a dense linear order as a suborder.

One might be curious now about the order $\Q\otimes\Z$, which is $\Z$ many copies of $\Q$. This order, however, is a countable endless dense linear order, and therefore is isomorphic to $\Q$ itself.


This material is adapted from my book-in-progress, Topics in Logic, drawn from Chapter 3 on Relational Logic, which incudes an extensive section on order theory, of which this is an important summative part.