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In J.D. Hamkins' multiverse view of set theory, every universe has an ill-founded $\mathbb{N}$ from the perspective of another universe. Does this mean that every proof in our universe can be seen as a nonstandard length proof from the perspective of some other universe, so that there is no real "truth" in the Multiverse?

Could this happen even for a simple concrete proof, like the fundamental theorem of arithmetic? Or are some some proofs standard-length in all universes?

Mikhail Katz
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Guest154
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    Length 35253586543 is always length 35253586543. And what proof of the fundament theorem of arithmetic uses properties of a special model of $\mathbb{N}$? – Michael Greinecker Oct 30 '17 at 02:16
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    The actual proofs that we use have standard lengths. The non-standard natural numbers, which account for any ill-foundedness in those systems, are larger than all the standard natural numbers. So you shouldn't expect to ever see anything of such size. – Andreas Blass Oct 30 '17 at 02:17
  • @Guest154: You might find Viteslav Svedar's paper, "Infinite natural numbers: an unwanted phenomenon, or a useful concept?" (look under Author and Title on the Web), of some help in understanding Hamkins' Section 5, "Multiverse Response to the Categoricity Arguments" in his paper, "The Set-Theoretic Multiverse". Pay particular attention to the following concept found in Section 5: "...within any fixed set-theoretic background concept, any set concept that has all the sets [how to define 'all'?--my comment] agrees with that background concept; and hence any two of them agree with – Thomas Benjamin Oct 30 '17 at 06:36
  • (cont.) each other. But such a claim seems far from categoricity, should one entertain the idea that there can be different incompatible set-theoretic backgrounds". (Hamkins does entertain that idea, by the way.) – Thomas Benjamin Oct 30 '17 at 06:42
  • Also, Victoria Gitman's boolesrings blogpost "An Introduction to Nonstandard Models of Arithmetic" (and the comments thereto) and Haim Gaifman's "Non-Standard Models in a Broader Perspective" might be helpful to you also (both are found on the Web as well). – Thomas Benjamin Oct 30 '17 at 07:14
  • Note also that the Fundamental Theorem of Arithmetic (since it can be derived in $PA$) holds for both standard and nonstandard integers. A proof of this can be found in Todd Trimble's answer to Elmore's mathoverflow question, "$FTA$ in first order setting" ($FTA$ stands for Fundamantal Theorem of Arithmetic, of course). – Thomas Benjamin Oct 30 '17 at 12:07
  • The fact that given any model of geometry one can build inside it various models of spherical, Euclidean and hyperbolic geometry does not imply that curvature of our own world is undefined. It may be variable, true, and unknown to us too, but locally at each point it is rigorously defined (I think). – მამუკა ჯიბლაძე Oct 30 '17 at 12:49
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    Every naive integer is present in the instantiation of N in each universe of the multiverse. The point is that the naive ones don't form a set in any of them. The length of any concrete proof is certainly a naive integer. – Mikhail Katz Oct 31 '17 at 10:22

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Thank you for your interest in my views on the set-theoretic multiverse.

Yes, indeed, the well-foundedness mirage axiom you mention is probably the most controversial of my multiverse axioms, and so allow me to explain a little about it.

The axiom expresses in a strong way the idea that we don't actually have a foundationally robust absolute concept of the finite in mathematics. Specifically, the axiom asserts that every universe of set theory is ill-founded even in its natural numbers from the perspective of another, better universe. Thus, every set-theoretic background in which we might seek to undertake our mathematical activity is nonstandard with respect to another universe.

My intention in posing the axiom so provocatively was to point out what I believe is the unsatisfactory nature of our philosophical account of the finite.

You might be interested in the brief essay I wrote on the topic, A question for the mathematics oracle, published in the proceedings of the Singapore workshop on Infinity and Truth. For an interesting and entertaining interlude, the workshop organizers had requested that everyone at the workshop pose a specific question that might be asked of an all-knowing mathematical oracle, who would truthfully answer. My question was whether in mathematics we really do have a absolute concept of the finite.

To explain a bit more, the naive view of the natural numbers in mathematics is that they are the numbers, $0$, $1$, $2$, and so on. The natural numbers, with all the usual arithmetic structure, are taken by many to have a definite absolute nature; arithmetic truth assertions are taken to have a definite absolute nature, in comparison for example with the comparatively less sure footing of set-theoretic truth assertions.

To be sure, many mathematicians and philosophers have proposed a demarcation between arithmetic and analysis, where the claims of number theory and arithmetic are said to have a definite absolute nature, while the assertions of higher levels of set theory, beginning with claims about the set of sets of natural numbers, are less definite. Nik Weaver, for example, has suggested that classical logic is appropriate for the arithmetic realm and intuitionistic logic for the latter realm, and a similar position is advocated by Solomon Feferman and others.

But what exactly does this phrase, "and so on" really mean in the naive account of the finite? It seems truly to be doing all the work, and I find it basically inadequate to the task. The situation is more subtle and problematic than seems to me to be typically acknowledged. Why do people find their conception of the finite to be so clear and absolute? It seems hopelessly vague to me.

Of course, within the axiomatic system of ZFC or other systems, we have a clear definition of what it means to be finite. The issue is not that, but rather the extent to which these internal accounts of finiteness agree with the naive pre-reflective accounts of the finite as used in the meta-theory.

Some mathematicians point to the various categoricity arguments as an explanation of why it is meaningful to speak of the natural numbers as a definite mathematical structure. Dedekind proved, after all, that there is up to isomorphism only one model $\langle\mathbb{N},S,0\rangle$ of the second-order Peano axioms, where $0$ is not a successor, the successor function $S$ is one-to-one, and $\mathbb{N}$ is the unique subset of $\mathbb{N}$ containing $0$ and closed under successor.

But to my way of thinking, this categoricity argument merely pushes off the problem from arithmetic to set theory, basing the absoluteness of arithmetic on the absoluteness of the concept of an arbitrary set of natural numbers. But how does that give one any confidence?

We already know very well, after all, about failures of absoluteness in set theory. Different models of set theory can disagree about whether the continuum hypothesis holds, whether the axiom of choice holds, and so with innumerable examples of non-absoluteness. Different models of set theory can disagree on their natural number structures, and even when they agree on their natural numbers, they can still disagree on their theories of arithmetic truth (see Satisfaction is not absolute). So we know all about how mathematical truth assertions can seem to be non-absolute in set theory.

Skolem pointed out that there are models of set theory $M_1$, $M_2$ and $M_3$ with a set $A$ in common, such that $M_1$ thinks $A$ is finite; $M_2$ thinks $A$ is countably infinite and $M_3$ thinks $A$ is uncountable. For example, let $M_3$ be any countable model of set theory, and let $M_1$ be an ultrapower by a ultrafilter on $\mathbb{N}$ in $M_3$, and let $A$ be a nonstandard natural number of $M_1$. So $M_1$ thinks $A$ is finite, but $M_3$ thinks $A$ has size continuum. If $M_2$ is a forcing extension of $M_3$, we can arrange that $A$ is countably infinite in $M_2$.

No amount of set-theoretic information in our set-theoretic background could ever establish that our current conception of the natural numbers, whatever it is, is the truly standard one, since whatever we assert to be true is also true in some nonstandard models, whose natural numbers are not standard.

The well-foundedness mirage axiom asserts that this phenomenon is universal: all universes are wrong about well-foundedness.

In defense of the mirage axiom, let me point out that whatever attitude toward it one might harbor, nevertheless the axiom cannot be seen as incoherent or inconsistent, because Victoria Gitman and I have proved that all of my multiverse axioms are true in the multiverse consisting of the countable computably saturated models of ZFC. So the axiom is neither contradictory nor incoherent. See A natural model of the multiverse axioms.

I have discussed my multiverse views in several papers.

Hamkins, Joel David, The set-theoretic multiverse, Rev. Symb. Log. 5, No. 3, 416-449 (2012). Doi:10.1017/S1755020311000359, ZBL1260.03103.

<p><em>Hamkins, Joel David</em>, <a href="http://jdh.hamkins.org/multiverse-perspective-on-constructibility/" rel="noreferrer"><strong>A multiverse perspective on the axiom of constructibility</strong></a>, Chong, Chitat (ed.) et al., Infinity and truth. Based on talks given at the workshop, Singapore, July 25--29, 2011. Hackensack, NJ: World Scientific (ISBN 978-981-4571-03-6/hbk; 978-981-4571-05-0/ebook). Lecture Notes Series. Institute for Mathematical Sciences. National University of Singapore 25, 25-45 (2014). <a href="http://dx.doi.org/10.1142/9789814571043_0002" rel="noreferrer">DOI:10.1142/9789814571043_0002</a>,
 <a href="https://zbmath.org/?q=an:1321.03061" rel="noreferrer">ZBL1321.03061</a>.</p>

<p><em>Gitman, Victoria; Hamkins, Joel David</em>, <a href="http://jdh.hamkins.org/naturalmodelofmultiverseaxioms/" rel="noreferrer"><strong>A natural model of the multiverse axioms</strong></a>, Notre Dame J. Formal Logic 51, No. 4, 475-484 (2010). <a href="http://dx.doi.org/10.1215/00294527-2010-030" rel="noreferrer">DOI:10.1215/00294527-2010-030</a>, <a href="https://zbmath.org/?q=an:1214.03035" rel="noreferrer">ZBL1214.03035</a>.</p>

<p><em>Hamkins, Joel David; Yang, Ruizhi</em>, <a href="http://jdh.hamkins.org/satisfaction-is-not-absolute/" rel="noreferrer"><strong>Satisfaction is not absolute</strong></a>, to appear in the Review of Symbolic Logic.</p>

But finally, to address your specific question. Of course, there are specific finite numbers that will be finite with respect to any alternative set-theoretic background. As Michael Greinecker points out in the comments, the number 35253586543 has that value regardless of your meta-mathematical position. So of course, there are many proofs that are standard finite with respect to any of the alternative foundations.

Meanwhile, I find it very interesting to consider the situation where different foundational systems disagree on what is provable. In very recent work of mine, for example, we are looking at the theory of set-theoretic and arithmetic potentialism, where different foundational systems disagree on what is true or provable.

For example, recently with Hugh Woodin, I have proved that there is a universal finite set $\{x\mid\varphi(x)\}$, a set that ZFC proves is finite, and which is empty in any transitive model of set theory, but if the set is $y$ in some countable model of set theory $M$ and $z$ is any finite set in $M$ with $y\subset z$, then there is a top-extension of $M$ to a model $N$ inside of which the set is exactly $z$. The key to the proof is playing with the non-absolute nature of truth between $M$ and its various top-extensions.

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    I've been away from MO for about a month, but came back to give an answer to this question. – Joel David Hamkins Nov 03 '17 at 01:50
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    Doesn't the Church-Turing Thesis give a strong argument for the absoluteness for what you call the pre-reflective account of the finite? What fundamentally new questions are raised by the application of non-absoluteness results in set theory to provability that don't already occur with much lower strength Gödelian phenomena? – Cameron Zwarich Nov 03 '17 at 04:47
  • @CameronZwarich: " 'Every function can be computable!'... there is a universal algorithm, capable of computing any function...What I mean is that there a universal algorithm, a Turing machine program capable of computing any desired function, if only one should run the program in the right universe. There is a Turing machine program $p$ with the property that for any function $f$: $\mathbb N$ $\rightarrow$ $\mathbb N$ on the natural numbers, including non-computable functions, there is a model of arithmetic or set theory inside of which the function computed by $p$ agrees with $f$ on all – Thomas Benjamin Nov 03 '17 at 05:49
  • (cont.) standard finite input. You have to run the program in a different universe in order that it will compute your desired function $f$." (This from Prof. Hamkins' March 25, 2016 blogpost, "Every function can be computable!") What this seems to mean (if I am understanding the concept correctly) is that (to use the 'multiverse' of universes of $PA$ as an example) if one fixes one of these universes satisfying $PA$ as the 'standard finite input', then, assuming Church's Thesis holds for all universes satisfying $PA$ (i.e. is '$PA$-absolute'?), there is a universe $M$ such that – Thomas Benjamin Nov 03 '17 at 06:24
  • (cont.) for every $f$: $M$ $\rightarrow$ $M$, $f$ is effectively computable in $M$. So Church's Thesis does not seem to provide a strong argument for the "pre-reflective account of the finite" since within any given universe $M$ $\vDash$ $PA$, every element of $M$ is deemed 'finite'. – Thomas Benjamin Nov 03 '17 at 06:36
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    @ThomasBenjamin The Church-Turing Thesis is not a mathematical statement, so if you are discussing it in the same breath as statements off the form "within any given universe $M \models PA$", you are already reflecting the mathematical notion of sets and satisfiability into the metatheory. (cont.) – Cameron Zwarich Nov 03 '17 at 06:49
  • The proof of the result you cite is basically just a careful analysis of the proof of the second incompleteness theorem, and all of the models involved satisfy $\neg \mathrm{Con}(PA)$. Do you interpret the second incompleteness theorem as implying that the consistency of PA is actually a nonabsolute notion, or do you interpret it as implying that the first-order theory PA, if consistent, is not capable of proving its own consistency? – Cameron Zwarich Nov 03 '17 at 06:49
  • Perhaps what I tried to say in the above comments would be better said as: "Let universes $M$, $N$ $\vDash$ $PA$. Then for every $f$: $M$ $\rightarrow$ $M$, there exists a universe $N$ such that $M$ $\subset$ $N$ and $f$ is effectively recursive in $N$. – Thomas Benjamin Nov 03 '17 at 06:50
  • @CameronZwarich: Does the Second Incompleteness Theorem imply the independence of $CON(PA)$, $\lnot$$CON(PA)$ from $PA$, that is, can one can consistently assume either $PA$ + $CON(PA)$ or $PA$ + $\lnot$$CON(PA)$? Also, how does the Church-Turing thesis (a non-mathematical statement) give a strong argument for the absoluteness of what is called the pre-reflective account of the finite? – Thomas Benjamin Nov 03 '17 at 07:18
  • @ThomasBenjamin It might be that $PA$ is consistent, but proves $\neg Con(PA)$, so $PA+Con(PA)$ is inconsistent. Godel's theorems only guarantee that if $PA$ is consistent, so is $PA+\neg Con(PA)$. – Wojowu Nov 03 '17 at 09:11
  • @CameronZwarich: I interpret the Second Incompleteness theorem as implying that the first-order theory $PA$, if consistent, is not capable of proving its own consistency (though I hold that $PRA$ + $TI_{\epsilon_0}$ implies $CON(PA)$, so that both $PA$ and $PA$ + $\lnot$$CON(PA)$ have models). Your move. – Thomas Benjamin Nov 03 '17 at 11:14
  • @ThomasBenjamin My point was that these results supposedly establishing arithmetic non-absoluteness don't appear to have substantially different content than the incompleteness theorems, even if they are dressing it up in a more enticing semantic veneer. Therefore, if you don't interpret the second incompleteness theorem as establishing the actual non-absoluteness of the consistency of PA, then how can you interpret these other results as establishing arithmetic non-absoluteness? – Cameron Zwarich Nov 03 '17 at 16:44
  • This is great. I completely agree with you. For example, there are many ways to add using primes depending on one’s choice. –  Nov 03 '17 at 19:14
  • @CameronZwarich: So would I be correct in saying that Godel's second incompleteness theorem (in your view) implies "arithmetic non-absoluteness" for universes satisfying $PA$ (and what would be the precise mathematical definition of 'arithmetic non-absoluteness', just for the record)? Also, I am in sympathy with the view your latest comment states, just so you know. – Thomas Benjamin Nov 03 '17 at 20:45
  • @ThomasBenjamin The second incompleteness theorem (when combined with the completeness theorem, which can be formalized in relatively weak theories) implies "arithmetic non-absoluteness" as a mathematical statement about models of first-order theories.

    I think to most mathematicians (IMO of course, there is no survey) this does not introduce any new doubt about the absoluteness of our concept of finitary mathematics, where absoluteness in interpreted in an ordinary philosophical sense, not in a mathematical sense. (cont.)

    – Cameron Zwarich Nov 03 '17 at 21:48
  • For an ordinary set-theoretic realist, there is also no problem, because set theory proves the uniqueness of the structure of natural numbers. This absoluteness (and quite a bit more) is also true across forcing extensions of $V$ and inner models, assuming you permit yourself a metatheorem or your realism includes the second-order structure of $V$ necessary to express this absoluteness.

    However, if you accept a notion of set-theoretic realism that includes universes that are further from these examples, then you find you find yourself in a bit of a bind. (cont.)

    – Cameron Zwarich Nov 03 '17 at 22:03
  • You can use a construction not too far from the one that shows any function can be made computable to show that you can fix a structure and vary the truth of a statement across universes in your multiverse, and you can pick the minimum rank at which this phenomenon occurs. If the reason you believed in the (real-world) absoluteness of finitary mathematics in the first place was that ZF proves the uniqueness of the naturals, then you now have a problem. (cont.) – Cameron Zwarich Nov 03 '17 at 22:14
  • However, this just seems like a fancier second-order version of establishing $\mathrm{Con}(PA)$ by $\epsilon_0$-induction and then struggling to find a justification for $\epsilon_0$-induction. – Cameron Zwarich Nov 03 '17 at 22:29
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    This may be something of an off-topic comment, but one puzzlement I have (at the philosophical level) about these kinds of doubts about the natural numbers is that if we seriously entertain them, then I am no longer sure what "absolute" means. In other words, to state the challenge precisely, it seems that I have to say something like, "The word 'absolute' has a clear and definite meaning, and according to this clear and definite meaning, our concept of 'finite' is not absolute." But if I'm seriously unsure about the word 'finite' then won't I be similarly unsure about the word 'absolute'? – Timothy Chow Nov 04 '17 at 03:19
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    For example, Joel writes, "Of course, within the axiomatic system of ZFC or other systems, we have a clear definition of what it means to be finite." But if I am not sure what "finite" means then I will not be sure what ZFC is exactly. To specify the axiom schema, we have to invoke "and so on" at some point. If it is not completely clear what ZFC is, then it can't be used to provide a clear definition of what "finite" is. – Timothy Chow Nov 04 '17 at 03:24
  • @JoelDavidHamkins: Regarding your Skolem example ("...there are models of set theory $M_1$, $M_2$, and $M_3$ with a set $A$ in common..."), do you have a cite for that, and what was Skolem's original example (i.e., did he know what an ultraproduct was...)? – Thomas Benjamin Nov 04 '17 at 05:05
  • (cont.) oops, I meant, "( ...did he know what an 'ultrapower by a ultrafilter on $\mathbb N$' was)?" Sorry. – Thomas Benjamin Nov 04 '17 at 05:18
  • @TimothyChow: What, exactly, is your "clear and definite meaning" of the term "absolute"? For example (and you might not consider this a good example, considering the source), the Wikipedia entry "Absoluteness" states, "In mathematical logic, a formula is said to be absolute if it has the same truth value in each of some class of structures (also called models)" (so that the term "absolute" is defined over a given class of models). This seems to be (at least to my way of thinking--possibly bad....) an adequate intuitive definition of "absolute" for the set-theoretic multiverse where there – Thomas Benjamin Nov 10 '17 at 11:39
  • (cont.) is no fixed 'real world' $V$. Given the aforementioned definition of absolute in the Wikipedia entry, how does this definition (for example) contribute to any lack of clarity of what $ZFC$ is, given that the definition of the term "finite" may not be fixed in the set-theoretic multiverse? – Thomas Benjamin Nov 10 '17 at 11:52
  • @ThomasBenjamin : If "finite" is hopelessly vague, then even "ZFC", let alone "models of ZFC", is hopelessly vague as well. Which ZFC are you talking about? All definitions of ZFC use "and so on" to describe axiom schemata, and if "and so on" is hopelessly vague then I am unclear what these axiom schemata are exactly. And if you're defining "absolute" in terms of hopelessly vague concepts such as "ZFC" and "model" then "absolute" is hopelessly vague as well. – Timothy Chow Nov 10 '17 at 15:33
  • @TimothyChow: So you would say that since $ZFC$ has, say, the following variants, e.g., $ZFC$ + $V$= $L$ and $ZFC$ + $\mathfrak M$[$G$] = $V$ (where $\mathfrak M$[$G$] is, say, a forcing extension in which $2^{\aleph_0}$=$\aleph_2$) then $ZFC$ is hopelessly vague? – Thomas Benjamin Nov 11 '17 at 07:14
  • @ThomasBenjamin : No. You're not taking seriously enough the notion that "finite" is hopelessly vague. Look at the definition of ZFC. Look in particular at the axiom schema of comprehension. It's an axiom schema because it comprises one axiom for each formula. Tell me, Thomas, what is a formula? When you start trying to explain to me what a formula is, you will find yourself saying "and so on" at some point, just as when you try to explain what an integer is, you will say "and so on" at some point. If "and so on" is hopelessly vague, the so is "formula" and therefore so is ZFC. – Timothy Chow Nov 11 '17 at 23:02
  • @TimothyChow: Would I be correcting in stating that it is some metatheory that determines the number of wffs that form subsets of a given set in the Separation axiom of $ZFC$ and that is that metatheory that determines the length of a formula in the metalanguage? If so, would you counter that if that metatheory is "hopelessly vague", then one could not use that metatheory to 'fix' the length of a formula (since the length of a formula in that metatheory could be 'nonstandard finite') or the number of formulas represented by the Separation axiom? – Thomas Benjamin Nov 12 '17 at 13:50
  • @ThomasBenjamin : Yes, that is pretty close to what I am saying. Mind you, I personally don't think that "finite" is hopelessly vague. But what I am saying is that if "finite" is hopelessly vague then so is "ZFC". – Timothy Chow Nov 12 '17 at 17:58
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    Timothy, yes, of course it is part of my view that different notions of the finite all give rise to different concepts of ZFC. The way I think about it is that it is wrong to think that there is only one meta-theoretic context. Rather, every object theory provides a meta-theoretic context for the theories it has, and those theories provide meta-theoretic contexts for further theories and so on. And also, it goes the other way: one makes stronger assumptions in the meta theory (such as consistency or whatever), and ultimately finds that one needs a meta-meta-theory and so on. – Joel David Hamkins Nov 12 '17 at 22:30
  • @JoelDavidHamkins: Exactly how does a given "object theory provide a meta-theoretic context for the theories it has"? For example (possibly a bad example), since it is known how to syntacticly represent nonstandard integers for $PA$, it is possible to Godel-number $PA^{'}$ ($PA$ with non-standard integers) in $PA$ so that $PA$ 'recognizes' the difference between standard and nonstandard integers, correct? In any case (in practice) researchers studying $PA$ (for example) always are able to distinguish between $PA$ and $PA^{'}$. If $PA$ (say) is 'hopelessly vague', one might – Thomas Benjamin Nov 13 '17 at 08:55
  • (cont.) (philosophically) wonder how they manage to distinguish between $PA$ and $PA^{'}$? – Thomas Benjamin Nov 13 '17 at 08:56
  • When you prove the theorems of elementary model theory in ZFC, then you are essentially using ZFC as a meta-theory for the object theories you are considering. When you prove in ZFC (or much less) that every model of PA has an elementary top-extension, then you are proving something about the current version of the object theory PA. But those theories themselves can prove such theorems also, so we have many levels of meta-theory, and basically every foundational theory can serve as a meta-theory. – Joel David Hamkins Nov 13 '17 at 13:27
  • @JoelDavidHamkins: So then $PA$ can code $PA{'}$ (via Godel-numbering) so $PA$ can act as $PA^{'}$'s meta-theory and can distinguish between non-standard and standard numbers in $PA^{'}$? – Thomas Benjamin Nov 13 '17 at 14:39
  • (cont.) (Actually it can't, for the simple reason that if it could, Tennenbaum's theorem would fail, i.e., there would exist countable recursive non-standard models of $PA$) So perhaps a more suitable metatheory would be second-order $PA$, under the Henkin interpretation. Would second-order $PA$, under the Henkin interpretation, adequately capture the notion of the multiverse of first-order arithmetic universes satisfying $PA$? – Thomas Benjamin Nov 15 '17 at 12:33
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There is a subtle issue here but it is not where the OP thinks it is. Any explicitly written integer is obviously "standard" whereas each new integer arising in the ultrapower of $\mathbb N$ is obviously "nonstandard". The subtle issue is that in the particular $\mathbb N$ we are working with, there may be integers (certainly bigger than the explicitly written ones) that look nonstandard from the viewpoint of another set-theoretic universe.

But this does not change the fact that any concrete proof has "standard" length. And since we are accepting ZFC axiomatics for each instance of a universe in Hamkins' multiverse, this has nothing to do with ultrafinitism.

In this publication (see also here) we proved a theorem that each universe in the Hamkins-Gitman "toy model" actually turns out to be a model of a variant of Edward Nelson's Internal Set Theory. Translated into the terms of Robinson's framework, this means that the integers in each instance of a universe in the multiverse contain not only an initial cut including all explicitly specifiable integers, but also a wealth of integers that are infinite according to a "smaller" universe in the multiverse.

One should note that the idea of a "standard $\mathbb N$" being an illusion, as implied in these question and answers, was the main driving force behind Petr Vopěnka's approach to the foundations of mathematics, formulated as his Alternative set theory already in the 1970s, according to what I have read of Vopenka's work.

Mikhail Katz
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  • I'm not sure I believe explicitly written integers are "obviously standard" -- or more accurately, I don't think it's a meaningful proposition about the "real world". At least, not unless we do something trite like define "standard" to refer to explicitly written things rather than something resembling its formal usage. –  Nov 06 '17 at 12:17
  • @Hurkyl, the issue here was the length of an existing proof. That's a pretty clear sufficient condition for being explicitly written. Do you have an example of a published proof about which it is unclear whether its length is a standard integer? – Mikhail Katz Nov 06 '17 at 12:23
  • @MikhailKatz: A question. Consider the proof of the totality of a function $f$($x$) recursive in some non-standard model $\mathfrak M$ of $PA$ whose totality was not provable in $<$ $\mathbb N$, $+$, $\cdot$ $>$, where $\mathbb N$ are just the 'naive integers' (refer back to you and your colleagues' paper regarding your view of Reeb's understanding of the 'naive integers'). The aforementioned proof in the extension $PA^{'}$ (containing the constants representing the 'nonstandard integers') would certainly not be in $PA$ but could be explicitly written in $PA^{'}. But Reeb contends that – Thomas Benjamin Nov 09 '17 at 13:44
  • @ThomasBenjamin, there is a problem in punctuation and latex formatting here. Also the sequel is missing. I can't really follow what you are getting at but it sounds interesting. – Mikhail Katz Nov 09 '17 at 13:57
  • (cont.) (at least according to what is written in your paper), "The naive integers are those integers that all members of humanity [in some universe where $PA$ holds whether standard or non-standard--my comment] share before they understand any advanced mathematics", so that the denizens of $\mathfrak M$ would consider their integers 'naive'. Though it is clear that any recursive function provably total in $PA$ would be provably total in $PA^{'}$, the notions 'explicitly written in $PA$' and 'explicitly written in $PA^{'}$' do not coincide (my previous comment should read, – Thomas Benjamin Nov 09 '17 at 14:14
  • (cont.) "...in some non-standard model of $PA$ whose totality was not provable in $PA$ satisfying $<$ $\mathbb N$, $+$, $\cdot$ $>$,..."). Could such a situation actually exist? – Thomas Benjamin Nov 09 '17 at 14:18
  • To make a long story short, you seem to be asking the following. Take a nonstandard integer $H$ in PA'. It is explicitly written in PA' (namely, "H") but not explicitly written in PA. This shows that the notion of "explicitly written" is not well-defined. Is that what you are asking? – Mikhail Katz Nov 09 '17 at 14:20
  • @MikhailKatz: Somewhat. According to (at least my understanding of) Prof. Hamkins' blogpost, "Every function can be computable!", there can exist proofs of the totality of functions recursive in some extension $PA^{'}$ in which non-standard integers exist that cannot be proven in $PA$. Therefore, the notion of "explicitly written" is not well-defined. In fact, since a nonstandard integer $H$ can be explicitly written in $PA^{'}$, this suggests that to the denizens of the universe in which $PA^{'}$ holds, $H$ is (to them) a "naive integer". – Thomas Benjamin Nov 09 '17 at 14:35
  • @ThomasBenjamin, I think you missed the point of my answer. I agree that the notion of "explicitly written" is not well-defined, but there are some sufficient conditions for an integer to have been explicitly written. Note that the collection of standard integers in Nelson's Internal Set Theory is *not a set*. This is the formal counterpart of a "soritic" property of being a "naive" integer or an "explicitly specifiable" integer. More on the sorites and the Robinson/Nelson framework can be found in this publication. – Mikhail Katz Nov 09 '17 at 14:41
  • @MikhailKatz: Thanks. I'll read further in you and your colleagues' paper. – Thomas Benjamin Nov 09 '17 at 14:47
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I think only Joel can answer this question, but I'd like to point out that there are some subtleties here that the commenters are missing. In Joel's multiverse conception, there is no "standard" $\mathbb{N}$, so the standard natural numbers in one model could turn out to be nonstandard in another model. This means that something that appears to be a proof in PA in one model would be revealed to have nonstandard length, and not be a proof, in some other model.

So: "Length 35253586543 is always length 35253586543"? I'm not even sure we know, from Joel's perspective, that 35253586543 is a standard natural number in every model. It would take a long long time to actually count that high. Surely in models with nonstandard $\mathbb{N}$ it seems like nonstandard numbers are standard, and you could count up to them and think that you had executed only a finite number of steps. So how could you know that numbers you think are standard are also seen as standard by everyone else? Maybe for small numbers you can just directly intuit this, but this gets pretty philosophical.

You could argue that this number is only 11 digits long, and maybe we could give a proof that all 11 digit long numbers have to be standard in every universe, such that the proof itself was sufficiently short that we could just directly "see" that it continues to be a proof in any model. Maybe. That would at least require some serious argumentation.

"The actual proofs that we use have standard lengths" --- of course it would seem that way to us, but might our proof of the fundamental theorem of arithmetic turn out to have a nonstandard length in some other model? How could we know it wouldn't? You see these questions are a lot more subtle than they seem at first.

Edit: I guess I ought to clarify that I am NOT expressing my own view here. In my view, which is otherwise close to Joel's, there is a standard $\mathbb{N}$ and an absolute notion of arithmetical truth. However, there are no uncountable sets and "sets" like the real line are actually proper classes. So if ZFC is consistent then it has a range of countable models, but no canonical model.

I came to this view as a result of clarifying what I think is the essential difference between sets and proper classes (something philosophers of mathematics have struggled to explain). I have several papers on the arXiv which lay out my position; "Mathematical conceptualism" is probably the best place to start. The fullest account of my ideas about the set/proper class distinction is given in the last chapter of my book Truth & Assertibility.

Nik Weaver
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    Cf. https://en.wikipedia.org/wiki/Ultrafinitism ! – Mike Shulman Oct 30 '17 at 17:08
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    @MikeShulman: exactly. Though I would be surprised to hear Joel describe himself as an ultrafinitist ... – Nik Weaver Oct 30 '17 at 17:23
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    The individual models in the multiverse are supposed to be models of ZF, and it's fairly easy to (express and) prove in an extension by definitions of ZF that 35253586543 exists. And the length of the proof, including the relevant definitions, is (unlike 35253586543 itself) short enough that one could count up to it. So 35253586543 should be "standard" in the sense of each model in the multiverse. – Andreas Blass Oct 30 '17 at 17:24
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    @AndreasBlass: but again, if your model has a nonstandard $\mathbb{N}$ then you could "count up to" nonstandard numbers and think they were standard. – Nik Weaver Oct 30 '17 at 17:34
  • @AndreasBlass I agree it's not exactly ultrafinitism, since there's no debate about whether 35253586543 exists; but if there is debate about whether it's "standard" then I think there is at least an analogy. – Mike Shulman Oct 30 '17 at 17:53
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    @AndreasBlass: The point is that you're observing the multiverse from the perspective of the particular universe in which you've "expressed and proved that 35253586543 exists". Every universe in your observable multiverse will have the integer 35253586543, but you can't say anything about universes outside of it. –  Oct 31 '17 at 07:00
  • I may have to wait for Joel's hopeful comments for this, but I'm a bit confused. I was under the impression that we can (metamathematically) take 'the' $\mathbb{N}$ as the intersection of all $\mathbb{N}$ over all theories; it's certainly true that no specific theory can talk about this $\mathbb{N}$, but there is a theory (if not more) that has this minimal $\mathbb{N}$ as 'its' $\mathbb{N}$, and its theorems can't have nonstandard lengths in any other model. What part of this (if not several) am I confused about? – Steven Stadnicki Oct 31 '17 at 07:39
  • @NikWeaver Is there some paper in which you describe your views? I read your book on forcing and had the impression that you view the idea of standard natural numbers as harmless. – Michael Greinecker Oct 31 '17 at 08:04
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    @StevenStadnicki, one of the "points" of Hamkins' multiverse view is that the position you outlined does not hold there. Namely, in every instance of a universe in the multiverse, the integers of that particular universe will contain nonstandard integers with respect to another universe. – Mikhail Katz Oct 31 '17 at 10:09
  • @MichaelGreinecker: thank you for the question. I've added a clarifying edit. – Nik Weaver Oct 31 '17 at 11:51
  • @NikWeaver, all naive integers including 35253586543 are present in each universe of the multiverse. – Mikhail Katz Oct 31 '17 at 12:17
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    @MikhailKatz: on what basis do you say this? Is it just your opinion? (And what exactly is a "naive" integer?) – Nik Weaver Oct 31 '17 at 14:01
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    @Hurkyl My comment was written not from the viewpoint of some particular universe (in the multiverse), nor from the viewpoint of a ZF model or other mathematical structure, but from the "viewpoint" of the physical world in which we live. The expressing and proving that I have in mind is not about (standard or nonstandard) formulas or Gödel numbers thereof but about what could be written with actual physical pens on actual sheets of paper. – Andreas Blass Oct 31 '17 at 16:31
  • @NikWeaver, naive integers include anything you can write down in the time allotted to our civilisation using a computer as large as the universe and exploiting the fastest growing functions in our logical arsenal. Writing down 35253586543 certainly takes less effort than that. For details you can see my recent publication, All in all this question would have been more appropriate at MSE. – Mikhail Katz Oct 31 '17 at 16:58
  • @MikhailKatz: again, my main question for you is on what basis you make this confident statement of fact about Hamkins' multiverse conception. – Nik Weaver Oct 31 '17 at 17:05
  • Again, on the basis of what I wrote in the publication cited above. – Mikhail Katz Oct 31 '17 at 17:18
  • @MikeShulman, there could not be any debate about whether 35253586543 is "standard" or any other explicitly specified number. The universes in Hamkins' multiverse are models of ZFC and therefore are unrelated to either finitism or ultrafinitism. – Mikhail Katz Nov 01 '17 at 12:41
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    @MikhailKatz: I'm afraid you really don't understand the issues here. Joel does not believe in a standard $\mathbb{N}$. So you have to be careful about what it even means to be a model of ZFC. This matter is absolutely related to finitism and possibly even to ultrafinitism. The best source on this would be Joel. Presuming that you can give a categorical answer, as you have, is quite arrogant and unsupported. – Nik Weaver Nov 01 '17 at 13:13
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    "Not believing in a standard N" does not entail any of the philosophical anxieties you outlined, contrary to what you seem to assume. @NikWeaver – Mikhail Katz Nov 01 '17 at 14:49
  • Is the statement "all naive integers are present in all universes of the multiverse" a metamathematical statement about natural numbers that are constructed as the result of physical processes, or is it a mathematical statement subject about an unbounded finitary process that is itself subject to non-absoluteness? Taking the latter view and importing the non-absoluteness of arithmetic statements about provability while also trying to establish absolute arithmetic statements about provability seems self-defeating and ultimately incoherent. – Cameron Zwarich Nov 03 '17 at 05:18
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    "In Joel's multiverse conception, there is no 'standard' N, so the standard natural numbers in one model could turn out to be nonstandard in another model. This means that something that appears to be a proof in PA in one model would be revealed to have nonstandard length, and not be a proof, in some other model": Weaver's first sentence is correct but his second sentence is not as Joel David's answer conclusively shows. – Mikhail Katz Nov 05 '17 at 10:27
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    It should be noted that just because we can "see" a number doesn't make it standard. For example, in a non-standard universe, there could be a creature with a non-standard number of fingers. If we told them that their "10" doesn't exist, they would laugh at us. – Christopher King Nov 16 '17 at 21:58
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Since the Fundamental Theorem of Arithmetic is a theorem of $PA$, it holds for both standard and nonstandard models of $PA$. Since one can interpret $PA$ in both $ZFC$ and $GBC$ (e.g., for $ZFC$, it is interpretable in $ZFC$ as the fragment $ZFC$ $-$ Infinity--note that in $ZFC$ all definitions of "finite" are equivalent, while in $ZF$ there are an "infinitude of finitudes" but since $V_{\omega}$ $\vDash$ $ZF$ $-$ Infinity also satisfies $ZFC$ $-$ Infinity, all definitions of 'finite' for $V_{\omega}$ are still equivalent), one can see that $PA$ is interpretable in some background set theory so as long as one's model of $PA$ can be interpreted in a model of $ZFC$ (say), then the Fundamental Theorem of Arithmetic will hold.

[Addendum: Though Andres is right, i.e., that interpretations are not theories, one can interpret one theory in another (see Richard Kaye's and Tin Lok Wong's paper, "On interpretations of arithmetic and set theory", for the interpretation of $PA$ in $ZF$ and $ZFC$). I am using this interpretation in order to show that, given the right background set theory ($ZFC$), one still has that the Fundamental Theorem of Arithmetic will hold in models of $ZFC$ for its interpretation of $PA$, regardless of whether the model of $PA$ in question is 'standard' or 'nonstandard' (whatever those terms might mean in Hamkins' set-theoretic multiverse).]

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    No, an interpretation is not a theory. – Andrés E. Caicedo Oct 30 '17 at 12:44
  • @AndrésE.Caicedo: Correct. Where did I say it was in my answer or my comments? If you can show me I will correct (except in the comments--all of my comments were made more than five minutes ago). – Thomas Benjamin Oct 30 '17 at 13:08
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    I think this answer and the other comments are missing the point that in Hamkins' multiverse conception (as I understand it) there is no "standard" $\mathbb{N}$. That makes the question much more subtle and philosophical. I'd like to see Joel's response. – Nik Weaver Oct 30 '17 at 13:32
  • @NikWeaver: In Hamkins' multiverse conception, can there be a "standard $\mathbb N$" relative to some given background set theory? As I understand it, the relativity of $\mathbb N$ is due to a relativity of background set theory, as stated in the quote from Hamkins' preprint (I know it has been published a while back) I gave in my comments. – Thomas Benjamin Oct 30 '17 at 13:44
  • Thomas, read what you wrote. – Andrés E. Caicedo Oct 30 '17 at 14:20
  • @AndrésE.Caicedo: I did rewrite what was in the parentheses already. Am I to presume that that editing did not correct the problem completely? Since I have errands to run, I will correct these later--don't fault me for that, please. – Thomas Benjamin Oct 30 '17 at 14:30
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    @ThomasBenjamin: I don't want to speak for Joel, but yes, it is also my understanding that the relativity of $\mathbb{N}$ is due to a relativity of background set theory. But that means that the concept "is a theorem of PA" varies depending on the background set theory. – Nik Weaver Oct 30 '17 at 16:09
  • One concern is that within any model, proofs that are of nonstandard length relative to some other model may seem to be of standard length. If you were to write out the proof of the fundamental theorem of arithmetic in PA, I'm sure it would be very long ... so could we be sure that it's not of nonstandard length in some model? I feel obliged to emphasize again that I don't speak for Joel and would be curious as to how he would answer the question. – Nik Weaver Oct 30 '17 at 16:12