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Asaf's answer here reminded me of something that should have been bothering me ever since I learned about it, but which I had more or less forgotten about. In first-order logic, there is a convention to only work with non-empty models of a theory $T$. The reason usually given is that the sentences $(\forall x)(x = x)$ and $(\forall x)(x \neq x)$ both hold in the "empty model" of $T$, so if we want the set of sentences satisfied by a model to be consistent, we need to disallow the empty model.

This smells fishy to me. I can't imagine that a sufficiently categorical setup of first-order logic (in terms of functors $C_T \to \text{Set}$ preserving some structure, where $C_T$ is the "free model of $T$" in an appropriate sense) would have this defect, or if it did it would have it for a reason. So something is incomplete about the standard setup of first-order logic, but I don't know what it could be.

The above looks like an example of too simple to be simple, except that I can't explain it to myself in the same way that I can explain other examples.

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Qiaochu Yuan
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  • I don't think I could give you a satisfactory answer, however I can say that categorical way of thinking is not always compatible with the logical way of thinking. This is a good example for such incompatibility. A nice way to spend an hour is to read and contemplate Mathias' "The Ignorance of Bourbaki" which speaks of this incompatibility. – Asaf Karagila Jun 13 '11 at 22:02
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    @Asaf: I have read Mathias' critique, and from what I can recall of it, Mathias is not criticizing the categorical way of thinking; he is criticizing a lack of understanding of Godel's work, which I don't think is at all synonymous. – Qiaochu Yuan Jun 13 '11 at 22:08
  • @Qiaochu: I agree, however the critique is even further about the fact that logic is arithmetic and categories are geometric, and the two are very incompatible in many cases. This is at least what I got from the paper. – Asaf Karagila Jun 13 '11 at 22:16
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    @Asaf: I don't really see this incompatibility you're claiming exists. In fact logic and geometry are very closely related via topos theory. – Qiaochu Yuan Jun 13 '11 at 22:24
  • @Qiaochu: Just like the Highlander movies, in foundations of mathematics and way of thinking - "There can be only one". You either consider categories in a set-like world, or consider sets in a category-like world. I don't really see how the two can co-exists while being foundational at the same time. – Asaf Karagila Jun 13 '11 at 22:28
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    @Asaf: I don't really understand what how you're saying is relevant to this discussion. I can't imagine that the distinction between allowing and disallowing the empty model comes down to a genuine philosophical distinction between logic / set theory and category theory; as far as I can tell (see my answer) it comes from the use of an incorrect axiom. – Qiaochu Yuan Jun 13 '11 at 22:34
  • @Qiaochu: Judging by the comments and answers I have seen over the past year, it seems very much that you're a "categories kind of mathematician". However if I consider an answer (or comment?) of yours from some while ago, sometimes it is better to have categories with an initial object. In logic it is simpler to have no empty models. Since the empty model is "unique" to some extent, and has no "real" [logical/model theoretic] application other than being a counter example, it is simpler to ditch it outside the rules and keep only the non-empty models. – Asaf Karagila Jun 13 '11 at 22:42
  • @Qia You might find useful the references I gave here. See also the following discussion in that thread. – Bill Dubuque Jun 13 '11 at 23:41
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    Perhaps we should turn this apparent paradox on its head and say that the empty model is the unique model for any inconsistent theory. After all, $\lnot ( \exists x . , \top )$ is equivalent (intuitionistically!) to $\forall x . , \bot$, and if we can deduce $\bot$ we can certainly deduce $\forall x . , \bot$. – Zhen Lin Jun 14 '11 at 00:15
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    Addendum: The empty model, in that sense, is like the zero ring. It has a rather unpleasant property which we ignore in most situations, but it is useful to have it around as e.g. the initial object. – Zhen Lin Jun 14 '11 at 10:13
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    @ZhenLin The fact that $\forall x (x=x)$ and $\forall x (x\neq x)$ are both true in the empty model is not a paradox. They are vacuously true. In addition, no inconsistent theory has a model. For example, the theory $\left{ \forall x (x=x), \lnot \forall x (x=x) \right}$ is inconsistent and is NOT satisfied by the empty model. This is because $\lnot \forall x (x=x)$ (equivalent to $\exists x (x \neq x)$) is false in the empty model. – mareoraft Aug 22 '14 at 08:04
  • @QiaochuYuan: http://en.wikipedia.org/wiki/Free_logic – Martin Brandenburg Apr 11 '15 at 11:57
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    One nice thing about allowing the empty model is that you can then joke that ZF's Empty Set axiom is a "large" cardinal axiom. (For that matter, why do people even state Empty Set if they aren't allowing the empty model? It's just a theorem.) – Oscar Cunningham Sep 19 '18 at 23:44

6 Answers6

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Both $(\forall x)(x = x)$ and $(\forall x)(x \not = x)$ do hold in the empty model, and it's perfectly consistent. What we lose when we move to empty models, as Qiaochu Yuan points out, are certain inference rules that we're used to.

For first-order languages that include equality, the set $S$ of statements that are true all models (empty or not) is a proper subset of the set $S_N$ of statements that are true in all nonempty models. Because the vast majority of models we are interested in are nonempty, in logic we typically look at sets of inference rules that generate $S_N$ rather than rules that generate $S$.

One particular example where this is useful is the algorithm to put a formula into prenex normal form, which is only correct when we limit to nonempty models. For example, the formula $(\forall x)(x \not = x) \land \bot$ is false in every model, but its prenex normal form $(\forall x)(x \not = x \land \bot)$ is true in the empty model. The marginal benefit of considering the empty model doesn't outweigh the loss of the beautiful algorithm for prenex normal form that works for every other model. In the rare cases when we do need to consider empty models, we realize we have to work with alternative inference rules; it just isn't usually worth the trouble.

From a different point of view, only considering nonempty models is analogous to only considering Hausdorff manifolds. But with the empty model there is only one object being ignored, which we can always treat as a special case if we need to think about it.

Carl Mummert
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    Okay, but can't we just add $(\exists x)$ to the beginning of every prenex normal form where $x$ doesn't occur elsewhere in the formula? I don't want to use inference rules that are false for the empty model. – Qiaochu Yuan Jun 13 '11 at 22:51
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    You could do something like that, based on whether the original formula is true or false in the empty model. But there are other inference rules that are invalid in empty models, like the substitution rule you mentioned, $(\forall x)\phi \to \phi[t/x]$, and $\phi[t/x] \to (\exists x)\phi$. I think it's easier to just work by cases: the empty model, and everything else. – Carl Mummert Jun 13 '11 at 23:17
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    I think this is a bad logical habit. There are situations in ordinary mathematics where one can prove a statement of the form $(\forall x \in S) P$ where $P$ is a false statement not depending on $x$ and $S$ is empty, and if you are in such a situation then the good habit is to check whether $S$ is non-empty before declaring, for example, that you've proven some false statement which implies the Riemann hypothesis. – Qiaochu Yuan Jun 13 '11 at 23:33
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    People certainly write that sort of proof all the time, for example by studying properties of some class of spaces which later turns out to have no spaces in it. But that problem won't translate directly to first-order logic because $(\forall x)P$ is not quite the same as $(\forall x \in S) P$. If we try to make $S$ the universe of a first-order theory that has no non-empty models, that theory will just be inconsistent when we look at it with the normal set of inference rules for non-empty models. And in logic we are already very used to checking that our theories are consistent. – Carl Mummert Jun 14 '11 at 00:13
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    On the other hand if we make $S$ a predicate, then we can still assume that the overall universe (of sets, say) is nonempty, and use our normal inference rules, which will handle the case where $S$ has no elements. So $(\forall x \in S)P$ is an abbreviation for $(\forall x)(x \in S \to P)$ but the overall universe being quantified over is nonempty. – Carl Mummert Jun 14 '11 at 00:14
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    "we have to work with alternative inference rules; it just isn't usually worth the trouble" Is this really true? How much more complex are the alternative inference rules? – user76284 Feb 14 '20 at 23:20
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Isn't this a non-issue?

Many of the most common set-ups for the logical axioms were developed long ago, in a time when mathematicians (not just logicians) thought that they wanted to care only about non-empty structures, and so they made sure that $\exists x\, x=x$ was derivable in their logical system. They had to do this in order to have the completeness theorem, that every statement true in every intended model was derivable. And so those systems continue to have that property today.

Meanwhile, many mathematicians developed a fancy to consider the empty structure seriously. So logicians developed logical systems that handle this, in which $\exists x\, x=x$ is not derivable. For example, this is always how I teach first order logic, and it is no problem at all. But as you point out in your answer, one does need to use a different logical set-up.

So if you care about it, then be sure to use the right logical axioms, since definitely you will not want to give up on the completeness theorem.

Srivatsan
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JDH
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    Thanks for the reassurance. In my logic and set theory course we were not told that one could modify the axioms to handle the empty model, and I forgot to ask before the course ended. So I didn't know if this could be easily done. – Qiaochu Yuan Jun 14 '11 at 00:16
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    It is true that the most common treatments exclude the empty model, which is traditional and allows for certain conveniences as Carl mentions (and in any case, we are not confused about the empty structure). I think one of the main reasons involves a tradition of dealing with formulas (with free variables) in deductions, rather than with their universally quantified forms. This is sensible only in non-empty structures. – JDH Jun 14 '11 at 00:48
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    What do you mean by "in order to have the completeness theorem". As far as I can see with the appropriate trivial modifications of the inference rules the completeness theorem and still holds, namely any consistent first-order theory has a model (which may be the empty model). – user21820 Aug 19 '15 at 05:28
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(This is just a minor addition to the other excellent answers.)

There are categorical foundations for model theory: Makkai and Reyes, First order categorical logic (LNM 611). Here is a quote from page 72:

An important point is that we allow the (partial) domains $M(s)$ of $M$ to be empty. In model theory, usually the domains are stipulated to be non-empty. This difference slightly effects what sequents are considered logically valid; c.f. below.

François G. Dorais
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Okay, if this is the answer, it is quite silly. The axioms of first-order logic in my notes include

$$(\forall x) p \Rightarrow p[t/x]$$

which is manifestly false for the empty model and $p = \perp$ so should just be thrown out and replaced by the correct axiom

$$(\forall x) p \wedge (\exists x) \Rightarrow p[t/x].$$

Nothing changes except for the empty set, where the statement $(\forall x) \perp$ is true but $\perp$ is not, so there is no contradiction.

Qiaochu Yuan
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    What does $p[t/x]$ mean? – Asaf Karagila Jun 13 '11 at 22:18
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    usually something like the formula p with free occurrences of x replaced by t. (edited: you get the point :P ) – matt Jun 13 '11 at 22:19
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    There still seems to be an issue for you with empty models with this suggested axiom schema $(\forall x)p \wedge (\exists x) \rightarrow p[t/x]$. For example, take $p$ (with no occurrences of $x$) to be $\exists y (y=y)$. Now if we had an empty model, it fails the antecedent by non-existence of an $x$, but to be a model must satisfy the consequent $p[t/x]$ which is still $\exists y (y=y)$. – matt Jun 13 '11 at 22:27
  • @matt: I'm not sure I understand why that's a problem. – Qiaochu Yuan Jun 13 '11 at 22:36
  • edit: oop, nevermind. I slipped and was thinking models need to satisfy the consequent when the antecedent failed. That's not true at all :s – matt Jun 13 '11 at 22:48
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    Actually if I understand convention correctly (and I very well might not), this inference $\forall \langle x , ~ Px \rangle \vdash Pt$ is true even in an empty model, since free variables represent a universal quantification over the entire deduction (at least in natural deduction), so the $\forall ~t$ that is tacitly prefixing the statement satisfies the empty model. Some might say that a model has to have an object designated in it as $t$ though, I'm not so sure that's meaningful though. – DanielV Jul 24 '16 at 11:55
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    @DanielV: I know this is late, but for completeness haha.. Intrinsically, the problem actually is because of another rule that allows you to get from $P(t)$ to $\exists x\ ( P(x) )$ for any predicate $P$. Indeed that is the case if you interpret free variables as implicit universally quantified. But some authors define a model to interpret every free variable, which it can't do if it's empty, and hence the rule Qiaochu identified is invalid in this particular case. – user21820 Apr 04 '17 at 08:02
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    @user21820 If it interests you, a source of the problem is how natural deduction defines $\lor$-intro and $\land$-elim. If you look closely you'll see $\exists$-intro and $\forall$-elim are just generalizations of those. A definition that actually fits in empty models is $((a \lor b) \vdash y) \iff {a \vdash y, b\vdash y}$ and $(x \vdash (a \land b)) \iff {x \vdash a, x\vdash b}$. Generalizing that actually leads to empty model consistent definitions of quantification. Jaskowski did research on this back in the 1920s or so, the study is generally referred to as "inclusive logic". – DanielV Apr 04 '17 at 08:14
  • @DanielV: Now that you mention it, I can see the relation between your $\lor$-elim and $\exists$-elim in certain ND calculii like the sequent calculus I devised for ND that works for empty structures, but I still find it more natural to use programming-style instantiation of existentially quantified statements, even though it wouldn't have an elegant corresponding sequent calculus. I guess it's because I think of existential statements as oracles that I can call... =) – user21820 Apr 04 '17 at 08:28
  • @user21820 I've seen that post before and read it over. if I come across (or devise) a programmer friendly inclusive logic I'll try to remember to show it to you. But I tend to agree with the sentiment in a comment I read somewhere, anyone can come up with a set of inference rules, but without a proof of soundness and completeness wrt an established logic, it is rather inconsequential. – DanielV Apr 04 '17 at 08:58
  • @user21820 Oh one thing I haven't really mentioned anywhere else, there is a natural interpretation of $\forall$ as $\min$ and $\exists$ as $\max$ in multivalued logic. In which case, the exclusion of an empty universe is actually preferable, you don't want $\min$ and $\max$ giving values back from an empty set. – DanielV Apr 04 '17 at 09:22
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    @DanielV: Hmm I know the min/max interpretation, but it's most elegant to define min of an empty-set to be $\top$ and max of an empty-set to be $\bot$, such as in boolean algebra or Kleene's 3-valued logic. So I don't see what's the problem having an empty universe. As for logics, yes proof is important (after all that's the point of logic) but for my post I was lazy because it's just mechanically checking bi-interpretability. I hope Qiaochu Yuan doesn't mind this, but you can also come to the logic chat-room to continue. =) – user21820 Apr 04 '17 at 10:56
2

Since nobody else has mentioned the phrase, I think what you are looking for is called free logic by those who adopt the convention about what "first-order logic" means that would imply all models are nonempty.

We can construct a $C_T$ that is defective in the same way; we include a bunch of arrows $1 \to T$ to represent indeterminate elements of $T$. Then any lex functor $C_T \to \mathbf{Set}$ must send $T$ to a non-empty set.

Of course, the usual categorical approach of interpreting free variables as generalized elements doesn't have this defect.

0

Much ado about nothing!

Suppose that we admit empty structures. There is no real technical hurdle, we are endlessly ingenious.

However, instead of starting with "Let $\mathbb{A}$ be an $L$-structure," many theorems would have to start with "Let $\mathbb{A}$ be a non-empty $L$-structure." Think of the cumulative waste of resources, whole forests destroyed to produce the additional paper needed. And in this electronic age, are bits a renewable resource?

But one must admit there would also be benefits. There could be a new mathematical specialty, nit-picker, whose task would be to point out with glee the various places where a famous mathematician had blundered by forgetting to deal with empty structures. At a time of economic difficulty, this would boost employment, and contribute to the gross national product.

André Nicolas
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  • I would note that the upcoming SSD technology is especially prone to failure after much rewriting, as this whole process would require ;-) – Asaf Karagila Jun 13 '11 at 23:13
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    I disagree that pointing out a failure to deal with empty structures is nit-picking. There are theorems which genuinely deserve to have "non-empty" as a hypothesis (for example, Zorn's lemma), and if you forget to check that a set is non-empty before you apply them, you will say something which is wrong. More generally, the empty set is an extremal case, and a lot of very fruitful mathematics comes from looking at extremal cases. I think it is silly to ignore this just because it makes some theorems easier to state. – Qiaochu Yuan Jun 13 '11 at 23:18
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    (I once e-mailed Szamuely to inform him that in his excellent book Galois Groups and Fundamental Groups, whenever he claims that a category is isomorphic to a category of $G$-sets for some group $G$, he should actually say "non-empty $G$-sets." He dismissed this as nit-picking, but the category of $G$-sets is a topos and the category of non-empty $G$-sets isn't, so there is a genuine distinction to be made between the two, and geometrically the non-empty condition is quite natural, so why neglect to state it? All of the theorems about transitive $G$-sets, on the other hand, are correct... – Qiaochu Yuan Jun 13 '11 at 23:23
  • ...because the empty $G$-set is not transitive, and again this is geometrically quite natural.) – Qiaochu Yuan Jun 13 '11 at 23:24
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    @Qiaochu Yuan: To be more serious about these things, a notion has to earn its way. The empty set certainly has, a copy of the whole mathematical universe has been built using $\emptyset$ as the fundamental brick. In mathematical linguistics, the empty word makes the algebra more natural. If empty structures were ever to show an even modest usefulness, of course they would be admitted. – André Nicolas Jun 13 '11 at 23:33
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    @QiaochuYuan: I know this is really old, but I don't see where non-emptiness is required for Zorn's lemma. If the empty chain has an upper bound, the set is automatically nonempty. – Michael Greinecker Jul 29 '12 at 21:34
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    @Michael: okay, fair, but students do in practice attempt to apply Zorn's lemma to empty sets and they don't notice that an argument which shows that a non-empty chain has an upper bound might not show that the empty chain has an upper bound. – Qiaochu Yuan Jul 29 '12 at 22:06
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    @AndréNicolas: The empty structure is not interesting, but constructions with structures are interesting, and it is convenient when we can do them without restrictions. For example, the intersection of two subgraphs of a graph - this is not possible in general if you don't consider the empty graph to be a model of your theory. A similar criticism applies to the "definition" that ${0}$ is not a ring with unit. It is absurd to exclude the trivial ring from the class / category of rings with unit. – Martin Brandenburg Apr 11 '15 at 11:51
  • If some people in algebra want empty structures, they are welcome to them, while model theory/logic, for good reason, continues to use the traditional definition. Parenthetically, it is interesting/surprising that "general algebra" and model theory overlap so little. – André Nicolas Apr 11 '15 at 14:23
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    @AndréNicolas The empty model is useful for the exact same reason that zero, the empty string, and the empty set are useful. – user76284 Feb 14 '20 at 23:30