Mathematics and Computation

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The Burali-Forti argument in HoTT/UF

This is joint work with Marc Bezem, Thierry Coquand, Peter Dybjer.

We use the Burali-Forti argument to show that, in homotopy type theory and univalent foundations, the embedding $$ \mathcal{U} \to \mathcal{U}^+$$ of a universe $\mathcal{U}$ into its successor $\mathcal{U}^+$ is not an equivalence. We also establish this for the types of sets, magmas, monoids and groups. The arguments in this post are also written in Agda.

Ordinals in univalent type theory

The Burali-Forti paradox is about the collection of all ordinals. In set theory, this collection cannot be a set, because it is too big, and this is what the Burali-Forti argument shows. This collection is a proper class in set theory.

In univalent type theory, we can collect all ordinals of a universe $\mathcal{U}$ in a type $\operatorname{Ordinal}\,\mathcal{U}$ that lives in the successor universe $\mathcal{U}^+$: $$ \operatorname{Ordinal}\,\mathcal{U} : \mathcal{U}^+.$$ See Chapter 10.3 of the HoTT book, which uses univalence to show that this type is a set in the sense of univalent foundations (meaning that its equality is proposition valued).

The analogue in type theory of the notion of proper class in set theory is that of large type, that is, a type in a successor universe $\mathcal{U}^+$ that doesn't have a copy in the universe $\mathcal{U}$. In this post we show that the type of ordinals is large and derive some consequences from this.

We have two further uses of univalence, at least:

  1. to adapt the Burali-Forti argument from set theory to our type theory, and

  2. to resize down the values of the order relation of the ordinal of ordinals, to conclude that the ordinal of ordinals is large.

There are also a number of uses of univalence via functional and propositional extensionality.

Propositional resizing rules or axioms are not needed, thanks to (2).

An ordinal in a universe $\mathcal{U}$ is a type $X : \mathcal{U}$ equipped with a relation $$ - \prec - : X \to X \to \mathcal{U}$$

required to be

  1. proposition valued,

  2. transitive,

  3. extensional (any two points with same lower set are the same),

  4. well founded (every element is accessible, or, equivalently, the principle of transfinite induction holds).

The HoTT book additionally requires $X$ to be a set, but this follows automatically from the above requirements for the order.

The underlying type of an ordinal $\alpha$ is denoted by $\langle \alpha \rangle$ and its order relation is denoted by $\prec_{\alpha}$ or simply $\prec$ when we believe the reader will be able to infer the missing subscript.

Equivalence of ordinals in universes $\mathcal{U}$ and $\mathcal{V}$, $$ -\simeq_o- : \operatorname{Ordinal}\,\mathcal{U} \to \operatorname{Ordinal}\,\mathcal{V} \to \mathcal{U} \sqcup \mathcal{V},$$ means that there is an equivalence of the underlying types that preserves and reflects order. Here we denote by $\mathcal{U} \sqcup \mathcal{V}$ the least upper bound of the two universes $\mathcal{U}$ and $\mathcal{V}$. The precise definition of the type theory we adopt here, including the handling of universes, can be found in Section 2 of this paper and also in our Midlands Graduate School 2019 lecture notes in Agda form.

For ordinals $\alpha$ and $\beta$ in the same universe, their identity type $\alpha = \beta$ is canonically equivalent to the ordinal-equivalence type $\alpha \simeq_o \beta$, by univalence.

The lower set of a point $x : \langle \alpha \rangle$ is written $\alpha \downarrow x$, and is itself an ordinal under the inherited order. The ordinals in a universe $\mathcal{U}$ form an ordinal in the successor universe $\mathcal{U}^+$, denoted by $$ \operatorname{OO}\,\mathcal{U} : \operatorname{Ordinal}\,\mathcal{U}^+,$$ for ordinal of ordinals.

Its underlying type is $\operatorname{Ordinal}\,\mathcal{U}$ and its order relation is denoted by $-\triangleleft-$ and is defined by $$\alpha \triangleleft \beta = \Sigma b : \langle \beta \rangle , \alpha = (\beta \downarrow b).$$

This order has type $$-\triangleleft- : \operatorname{Ordinal}\,\mathcal{U} \to \operatorname{Ordinal}\,\mathcal{U} \to \mathcal{U}^+,$$ as required for it to make the type $\operatorname{\operatorname{Ordinal}} \mathcal{U}$ into an ordinal in the next universe.

By univalence, this order is equivalent to the order defined by $$\alpha \triangleleft^- \beta = \Sigma b : \langle \beta \rangle , \alpha \simeq_o (\beta \downarrow b).$$ This has the more general type $$ -\triangleleft^-- : \operatorname{\operatorname{Ordinal}}\,\mathcal{U} \to \operatorname{\operatorname{Ordinal}}\,\mathcal{V} \to \mathcal{U} \sqcup \mathcal{V},$$ so that we can compare ordinals in different universes. But also when the universes $\mathcal{U}$ and $\mathcal{V}$ are the same, this order has values in $\mathcal{U}$ rather than $\mathcal{U}^+$. The existence of such a resized-down order is crucial for our corollaries of Burali-Forti, but not for Burali-Forti itself.

For any $\alpha : \operatorname{Ordinal}\,\mathcal{U}$ we have $$ \alpha \simeq_o (\operatorname{OO}\,\mathcal{U} \downarrow \alpha),$$ so that $\alpha$ is an initial segment of the ordinal of ordinals, and hence $$ \alpha \triangleleft^- \operatorname{OO}\,\mathcal{U}.$$

The Burali-Forti theorem in HoTT/UF

We adapt the original formulation and argument from set theory.

Theorem. No ordinal in a universe $\mathcal{U}$ can be equivalent to the ordinal of all ordinals in $\mathcal{U}$.

Proof. Suppose, for the sake of deriving absurdity, that there is an ordinal $\alpha \simeq_o \operatorname{OO}\,\mathcal{U}$ in the universe $\mathcal{U}$. By the above discussion, $\alpha \simeq_o \operatorname{OO}\,\mathcal{U} \downarrow \alpha$, and, hence, by symmetry and transitivity, $\operatorname{OO}\,\mathcal{U} \simeq_o \operatorname{OO}\,\mathcal{U} \downarrow \alpha$. Therefore, by univalence, $\operatorname{OO}\,\mathcal{U} = \operatorname{OO}\,\mathcal{U} \downarrow \alpha$. But this means that $\operatorname{OO}\,\mathcal{U} \triangleleft \operatorname{OO}\,\mathcal{U}$, which is impossible as any accessible relation is irreflexive. $\square$

Some corollaries follow.

The type of ordinals is large

We say that a type in the successor universe $\mathcal{U}^+$ is small if it is equivalent to some type in the universe $\mathcal{U}$, and large otherwise.

Theorem. The type of ordinals of any universe is large.

Proof. Suppose the type of ordinals in the universe $\mathcal{U}$ is small, so that there is a type $X : \mathcal{U}$ equivalent to the type $\operatorname{Ordinal}\, \mathcal{U} : \mathcal{U}^+$. We can then transport the ordinal structure from the type $\operatorname{Ordinal}\, \mathcal{U}$ to $X$ along this equivalence to get an ordinal in $\mathcal{U}$ equivalent to the ordinal of ordinals in $\mathcal{U}$, which is impossible by the Burali-Forti theorem.

But the proof is not concluded yet, because we have to say how we transport the ordinal structure. At first sight we should be able to simply apply univalence. However, this is not possible because the types $X : \mathcal{U}$ and $\operatorname{Ordinal}\,\mathcal{U} :\mathcal{U}^+$ live in different universes. The problem is that only elements of the same type can be compared for equality.

  1. In the cumulative universe hierarchy of the HoTT book, we automatically have that $X : \mathcal{U}^+$ and hence, being equivalent to the type $\operatorname{Ordinal}\,\mathcal{U} : \mathcal{U}^+$, the type $X$ is equal to the type $\operatorname{Ordinal}\,\mathcal{U}$ by univalence. But this equality is an element of an identity type of the universe $\mathcal{U}^+$. Therefore when we transport the ordinal structure on the type $\operatorname{Ordinal}\,\mathcal{U}$ to the type $X$ along this equality and equip $X$ with it, we get an ordinal in the successor universe $\mathcal{U}^+$. But, in order to get the desired contradiction, we need to get an ordinal in $\mathcal{U}$.

  2. In the non-cumulative universe hierarchy we adopt here, we face essentially the same difficulty. We cannot assert that $X : \mathcal{U}^+$ but we can promote $X$ to an equivalent type in the universe $\mathcal{U}^+$, and from this point on we reach the same obstacle as in the cumulative case.

So we have to transfer the ordinal structure from $\operatorname{Ordinal}\,\mathcal{U}$ to $X$ manually along the given equivalence, call it $$f : X \to \operatorname{Ordinal}\,\mathcal{U}.$$ We define the order of $X$ from that of $\operatorname{Ordinal}\,\mathcal{U}$ by $$ x \prec y = f(x) \triangleleft f(y). $$ It is laborious but not hard to see that this order satisfies the required axioms for making $X$ into an ordinal, except that it has values in $\mathcal{U}^+$ rather than $\mathcal{U}$. But this problem is solved by instead using the resized-down relation $\triangleleft^-$ discussed above, which is equivalent to $\triangleleft$ by univalence. $\square$

There are more types and sets in $\mathcal{U}^+$ than in $\mathcal{U}$

By a universe embedding we mean a map $$f : \mathcal{U} \to \mathcal{V}$$ of universes such that, for all $X : \mathcal{U}$, $$f(X) \simeq X.$$ Of course, any two universe embeddings of $\mathcal{U}$ into $\mathcal{V}$ are equal, by univalence, so that there is at most one universe embedding between any two universes. Moreover, universe embeddings are automatically type embeddings (meaning that they have propositional fibers).

So the following says that the universe $\mathcal{U}^+$ is strictly larger than the universe $\mathcal{U}$:

Theorem. The universe embedding $\mathcal{U} \to \mathcal{U}^+$ doesn't have a section and therefore is not an equivalence.

Proof. A section would give a type in the universe $\mathcal{U}$ equivalent to the type of ordinals in $\mathcal{U}$, but we have seen that there is no such type. $\square$

(However, by Theorem 29 of Injective types in univalent mathematics, if propositional resizing holds then the universe embedding $\mathcal{U} \to \mathcal{U}^+$ is a section.)

The same argument of the above theorem shows that there are more sets in $\mathcal{U}^+$ than in $\mathcal{U}$, because the type of ordinals is a set. For a universe $\mathcal{U}$ define the type $$\operatorname{hSet}\,\mathcal{U} : \mathcal{U}^+$$ by $$ \operatorname{hSet}\,\mathcal{U} = \Sigma A : \mathcal{U} , \text{$A$ is a set}.$$ By an hSet embedding we mean a map $$f : \operatorname{hSet}\,\mathcal{U} → \operatorname{hSet}\,\mathcal{V}$$ such that the underlying type of $f(\mathbb{X})$ is equivalent to the underlying type of $\mathbb{X}$ for every $\mathbb{X} : \operatorname{hSet}\,\mathcal{U}$, that is, $$ \operatorname{pr_1} (f (\mathbb{X})) ≃ \operatorname{pr_1}(\mathbb{X}). $$ Again there is at most one hSet-embedding between any two universes, hSet-embeddings are type embeddings, and we have:

Theorem. The hSet-embedding $\operatorname{hSet}\,\mathcal{U} \to \operatorname{hSet}\,\mathcal{U}^+$ doesn't have a section and therefore is not an equivalence.

There are more magmas and monoids in $\mathcal{U}^+$ than in $\mathcal{U}$

This is because the type of ordinals is a monoid under addition with the ordinal zero as its neutral element, and hence also a magma. If the inclusion of the type of magmas (respectively monoids) of one universe into that of the next were an equivalence, then we would have a small copy of the type of ordinals.

Theorem. The canonical embeddings $\operatorname{Magma}\,\mathcal{U} → \operatorname{Magma}\,\mathcal{U}^+$ and $\operatorname{Monoid}\,\mathcal{U} → \operatorname{Monoid}\,\mathcal{U}^+$ don't have sections and hence are not equivalences.

There are more groups in $\mathcal{U}^+$ than in $\mathcal{U}$

This case is more interesting.

The axiom of choice is equivalent to the statement that any non-empty set can be given the structure of a group. So if we assumed the axiom of choice we would be done. But we are brave and work without assuming excluded middle, and hence without choice.

It is also the case that the type of propositions can be given the structure of a group if and only if the principle of excluded middle holds. And the type of propositions is a retract of the type of ordinals, which makes it unlikely that the type of ordinals can be given the structure of a group without excluded middle.

So our strategy is to embed the type of ordinals into a group, and the free group does the job.

  1. First we need to show that the inclusion of generators, or the universal map into the free group, is an embedding.

  2. But having a large type $X$ embedded into a type $Y$ is not enough to conclude that $Y$ is also large. For example, if $P$ is a proposition then the unique map $P \to \mathbb{1}$ is an embedding, and the unit type $\mathbb{1}$ is small but $P$ may be large.

  3. So more work is needed to show that the group freely generated by the type of ordinals is large. We say that a map is small if each of its fibers is small, and large otherwise. De Jong and Escardo showed that if a map $X \to Y$ is small and the type $Y$ is small, then so is the type $X$, and hence if $X$ is large then so is $Y$. Therefore our approach is to show that the universal map into the free group is small. To do this, we exploit the fact that the type of ordinals is locally small (its identity types are all equivalent to small types).

But we want to be as general as possible, and hence work with a spartan univalent type theory which doesn't include higher inductive types other than propositional truncation. We include the empty type, the unit type, natural numbers, list types (which can actually be constructed from the other type formers), coproduct types, $\Sigma$-types, $\Pi$-types, identity types and a sequence of universes. We also assume the univalence axiom (from which we automatically get functional and propositional extensionality) and the axiom of existence of propositional truncations.

  1. We construct the free group as a quotient of a type of words following Mines, Richman and Ruitenburg. To prove that the universal map is an embedding, one first proves a Church-Rosser property for the equivalence relation on words. It is remarkable that this can be done without assuming that the set of generators has decidable equality.

  2. Quotients can be constructed from propositional truncation. This construction increases the universe level by one, but eliminates into any universe.

  3. To resize back the quotient used to construct the group freely generated by the type of ordinals to the original universe, we exploit the fact that the type of ordinals is locally small.

  4. As above, we have to transfer manually group structures between equivalent types of different universes, because univalence can't be applied.

Putting the above together, and leaving many steps to the Agda code, we get the following in our spartan univalent type theory.

Theorem. For any large, locally small set, the free group is also large and locally small.

Corollary. In any successor universe $\mathcal{U}^+$ there is a group which is not isomorphic to any group in the universe $\mathcal{U}$.

Corollary. The canonical embedding $\operatorname{Group}\,\mathcal{U} → \operatorname{Group}\,\mathcal{U}^+$ doesn't have a section and hence is not an equivalence.

Can we formulate and prove a general theorem of this kind that specializes to a wide variety of mathematical structures that occur in practice?

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