10.5 The cumulative hierarchy

We can define a cumulative hierarchy $V$ of all sets in a given universe $𝒰$ as a higher inductive type, in such a way that $V$ is again a set (in a larger universe ${𝒰}^{\prime }$), equipped with a binary “membership” relation $x\in y$ which satisfies the usual laws of set theory.

Definition 10.5.1.

The cumulative hierarchy$V$ relative to a type universe $\mathrm{U}$ is thehigher inductive type generated by the following constructors.

1. For every $A:𝒰$ and $f:A\to V$, there is an element $\mathrm{𝗌𝖾𝗍}(A,f)$ : V.

2. For all $A,B:𝒰$, $f:A\to V$ and $g:B\to V$ such that

   $$(\forall (a:A).\exists (b:B).f(a){=}_{V}g(b))\wedge (\forall (b:B).\exists (a:A).f(a){=}_{V}g(b))$$(10.5.1)

   there is a path $\mathrm{𝗌𝖾𝗍}(A,f){=}_V\mathrm{𝗌𝖾𝗍}(B,g)$.

3. The 0-truncation constructor: for all $x,y:V$ and $p,q:x=y$, we have $p=q$.

In set-theoretic language, $\mathrm{𝗌𝖾𝗍}(A,f)$ can be understood as the set (in the sense of classical set theory) that is the image of $A$ under $f$, i.e. $f(a)|a\in A$.However, we will avoid this notation, since it would clash with our notation for subtypes (but see (10.5.2) and def:TypeOfElements below).

The hierarchy $V$ isbootstrapped from the empty map ${\mathrm{𝗋𝖾𝖼}}_{\mathrm{𝟎}}(V):\mathrm{𝟎}\to V$, which gives the empty set as $\mathrm{\varnothing }=\mathrm{𝗌𝖾𝗍}(\mathrm{𝟎},{\mathrm{𝗋𝖾𝖼}}_{\mathrm{𝟎}}(V))$.Then the singleton $\{\mathrm{\varnothing }\}$ enters $V$ through $\mathrm{𝟏}\to V$, defined as $\star \mapsto \mathrm{\varnothing }$, and soon. The type $V$ lives in the same universe as the base universe $𝒰$.

The second constructor of $V$ has a form unlike any we have seen before: it involves not only paths in $V$ (which in sec:hittruncations we claimed were slightly fishy) but truncations of sums of them.It certainly does not fit the general scheme described in sec:naturality, and thus it may not be obvious what its induction principle should be.Fortunately, like our first definition of the 0-truncation in sec:hittruncations, it can be re-expressed using auxiliary higher inductive types.We leave it to the reader to work out the details (see ex:cumhierhit).

At the end of the day, the induction principle for $V$ (written in pattern matching language) says that given $P:V\to$, in order to construct $h:{\prod }_{(x:V)}P(x)$, it suffices to give the following.

1. For any $f:A\to V$, construct $h(\mathrm{𝗌𝖾𝗍}(A,f))$, assuming as given $h(f(a))$ for all $a:A$.

2. Verify that if $f:A\to V$ and $g:B\to V$ satisfy (10.5.1), then $h(\mathrm{𝗌𝖾𝗍}(A,f))=h(\mathrm{𝗌𝖾𝗍}(B,g))$, assuming inductively that $h(f(a))=h(g(b))$ whenever $f(a)=g(b)$.

The second clause checks that the map being defined must respect the paths introduced in (10.5.1).As usual when we state higher induction principles using pattern matching, it may seem tautologous, but is not.The point is that “$h(f(a))$” is essentially a formal symbol which we cannot peek inside of, which $h(\mathrm{𝗌𝖾𝗍}(A,f))$ must be defined in terms of. Thus, in the second clause, we assume equality of these formal symbols when appropriate, and verify that the elements resulting from the construction of the first clause are also equal.Of course, if $P$ is a family of mere propositions, then the second clause is automatic.

Observe that, by induction, for each $v:V$ there merely exist $A:𝒰$ and $f:A\to V$ such that $v=\mathrm{𝗌𝖾𝗍}(A,f)$.Thus, it is reasonable to try to define the membership relation$x\in v$ on $V$ by setting:

$$(x\in \mathrm{𝗌𝖾𝗍}(A,f)):\equiv (\exists (a:A).x=f(a)).$$

To see that the definition is valid, we must use the recursion principle of $V$. Thus, suppose we have a path $\mathrm{𝗌𝖾𝗍}(A,f)=\mathrm{𝗌𝖾𝗍}(B,g)$constructed through (10.5.1). If $x\in \mathrm{𝗌𝖾𝗍}(A,f)$ then there merely is $a:A$ suchthat $x=f(a)$, but by (10.5.1) there merely is $b:B$ such that $f(a)=g(b)$, hence$x=g(b)$ and $x\in \mathrm{𝗌𝖾𝗍}(B,f)$. The converse is symmetric.

The subset relation$x\subseteq y$ is defined on $V$ as usual by

$$(x\subseteq y):\equiv \forall (z:V).z\in x\Rightarrow z\in y.$$

A classmay be taken to be a mere predicate on $V$. We can say that a class $C:V\to \mathrm{𝖯𝗋𝗈𝗉}$ is a$V$-setif there merely exists $v\in V$ such that

$$\forall (x:V).C(x)\iff x\in v.$$

We may also use the conventional notation for classes, which matches our standard notation for subtypes:

$$x|C(x):\equiv \lambda x.C(x).$$(10.5.2)

A class $C:V\to \mathrm{𝖯𝗋𝗈𝗉}$ will be called $𝒰$-smallif all of its values $C(x)$ lie in $𝒰$, specifically $C:V\to {\mathrm{𝖯𝗋𝗈𝗉}}_{𝒰}$.Since $V$ lives in the same universe ${𝒰}^{\prime }$ as does the base universe $𝒰$ from which it is built, the same is true for the identity types $v{=}_{V}w$ for any $v,w:V$. To obtain a well-behaved theory in the absence of propositional resizing,therefore, it will be convenient to have a $𝒰$-small “resizing” of the identity relation, which we can define by induction as follows.

Definition 10.5.3.

Define the bisimulationrelation

$$\mathrm{\sim }:V\times V⟶{\mathrm{𝖯𝗋𝗈𝗉}}_{𝒰}$$

by double induction over $V$, where for $\mathrm{set}\mathit{}\mathrm{(}A\mathrm{,}f\mathrm{)}$ and $\mathrm{set}\mathit{}\mathrm{(}B\mathrm{,}g\mathrm{)}$ we let:

$$\mathrm{𝗌𝖾𝗍}(A,f)\sim \mathrm{𝗌𝖾𝗍}(B,g):\equiv (\forall (a:A).\exists (b:B).f(a)\sim g(b))\wedge (\forall (b:B).\exists (a:A).f(a)\sim g(b)).$$

To verify that the definition is correct, we just need to check that it respects paths $\mathrm{𝗌𝖾𝗍}(A,f)=\mathrm{𝗌𝖾𝗍}(B,g)$ constructed through (10.5.1), but this is obvious, and that ${\mathrm{𝖯𝗋𝗈𝗉}}_{𝒰}$ is a set, which it is. Note that $u\sim v$ is in ${\mathrm{𝖯𝗋𝗈𝗉}}_{𝒰}$ by construction.

Lemma 10.5.4.

For any $u\mathrm{,}v\mathrm{:}V$ we have $\mathrm{(}u{\mathrm{=}}_{V}v\mathrm{)}\mathrm{=}\mathrm{(}u\mathrm{\sim }v\mathrm{)}$.

Proof.

An easy induction shows that $\sim$ is reflexive, so by transport we have $(u{=}_{V}v)\to (u\sim v)$.Thus, it remains to show that $(u\sim v)\to (u{=}_{V}v)$.By induction on $u$ and $v$, we may assume they are $\mathrm{𝗌𝖾𝗍}(A,f)$ and $\mathrm{𝗌𝖾𝗍}(B,g)$ respectively.Then by definition, $\mathrm{𝗌𝖾𝗍}(A,f)\sim \mathrm{𝗌𝖾𝗍}(B,g)$ implies $(\forall (a:A).\exists (b:B).f(a)\sim g(b))$ and conversely.But the inductive hypothesis then tells us that $(\forall (a:A).\exists (b:B).f(a)=g(b))$ and conversely.So by the path-constructor for $V$ we have $\mathrm{𝗌𝖾𝗍}(A,f){=}_V\mathrm{𝗌𝖾𝗍}(B,g)$.∎

Now we can use the resized identity relation to get the following useful principle.

Lemma 10.5.5.

For every $u\mathrm{:}V$ there is a given $A_u\mathrm{:}\mathrm{U}$ and monic $m_u\mathrm{:}{A}_u\mathit{}\mathit{}V$ such that $u\mathrm{=}\mathrm{set}\mathit{}\mathrm{(}{A}_u\mathrm{,}{m}_u\mathrm{)}$.

Proof.

Take any presentation $u=\mathrm{𝗌𝖾𝗍}(A,f)$ and factor $f:A\to V$ as a surjection followed by an injection:

$$f=m_u\circ e_u:A{A}_{u}V.$$

Clearly $u=\mathrm{𝗌𝖾𝗍}(A_u,m_u)$ if only $A_u$ is still in $𝒰$, which holds if the kernel of $e_u:A{A}_u$ is in $𝒰$. But the kernel of $e_u:A{A}_u$ is the pullback along $f:A\to V$ of the identity on $V$, which we just showed to be $𝒰$-small, up to equivalence. Now, this construction of the pair $(A_u,m_u)$ with $m_u:A_{u}V$ and $u=\mathrm{𝗌𝖾𝗍}(A_u,m_u)$ from $u:V$ is unique up to equivalence over $V$, and hence up to identity by univalence. Thus by the principle of unique choice (LABEL:cor:UC) there is a map $c:V\to {\sum }_{(A:𝒰)}(A\to V)$ such that $c(u)=(A_u,m_u)$, with $m_u:A_{u}V$ and $u=\mathrm{𝗌𝖾𝗍}(c(u))$, as claimed.∎

Definition 10.5.6.

For $u\mathrm{:}V$, the just constructed monic presentation $m_u\mathrm{:}{A}_u\mathit{}\mathit{}V$ such that $u\mathrm{=}\mathrm{set}\mathit{}\mathrm{(}{A}_u\mathrm{,}{m}_u\mathrm{)}$ may be called the type of membersof $u$ and denoted $m_u\mathrm{:}\mathrm{[}u\mathrm{]}\mathit{}\mathit{}V$, or even $\mathrm{[}u\mathrm{]}\mathit{}\mathit{}V$. We can think of $\mathrm{[}u\mathrm{]}$ as the “subclass of $V$ consisting of members of $u$”.

Theorem 10.5.7.

The following hold for $\mathrm{(}V\mathrm{,}\mathrm{\in }\mathrm{)}$:

1. extensionality:

   $$\forall (x,y:V).x\subseteq y\wedge y\subseteq x\iff x=y.$$

2. empty set: for all $x:V$, we have $\mathrm{\neg }(x\in \mathrm{\varnothing })$.

3. pairing: for all $u,v:V$, the class $u\cup v:\equiv x|x=u\vee x=v$ is a $V$-set.

4. infinity: there is a $v:V$ with $\mathrm{\varnothing }\in v$ and $x\in v$ implies $x\cup \{x\}\in v$.

5. union: for all $v:V$, the class $\cup v:\equiv x|\exists (u:V).x\in u\in v$ is a $V$-set.

6. function set: for all $u,v:V$, the class $v^u:\equiv x|x:u\to v$ is a $V$-set.

7. $\in$-induction: if $C:V\to \mathrm{𝖯𝗋𝗈𝗉}$ is a class such that $C(a)$ holds whenever $C(x)$ for all $x\in a$, then $C(v)$ for all $v:V$.

8. replacement: given any $r:V\to V$ and $a:V$, the class

   $$x|\exists (y:V).y\in a\wedge x=r(y)$$

   is a $V$-set.

9. separation: given any $a:V$ and $𝒰$-small $C:V\to {\mathrm{𝖯𝗋𝗈𝗉}}_{𝒰}$, the class

   $$x|x\in a\wedge C(x)$$

   is a $V$-set.

Sketch of proof.

1. Extensionality: if $\mathrm{𝗌𝖾𝗍}(A,f)\subseteq \mathrm{𝗌𝖾𝗍}(B,g)$ then $f(a)\in \mathrm{𝗌𝖾𝗍}(B,g)$for every $a:A$, therefore for every $a:A$ there merely exists $b:B$ such that$f(a)=g(b)$. The assumption $\mathrm{𝗌𝖾𝗍}(B,g)\subseteq \mathrm{𝗌𝖾𝗍}(A,f)$ gives the other halfof (10.5.1), therefore $\mathrm{𝗌𝖾𝗍}(A,f)=\mathrm{𝗌𝖾𝗍}(B,g)$.

2. Empty set: suppose $x\in \mathrm{\varnothing }=\mathrm{𝗌𝖾𝗍}(\mathrm{𝟎},{\mathrm{𝗋𝖾𝖼}}_{\mathrm{𝟎}}(V))$. Then $\exists (a:\mathrm{𝟎}).x={\mathrm{𝗋𝖾𝖼}}_{\mathrm{𝟎}}(V,a)$, which is absurd.

3. Pairing: given $u$ and $v$, let $w=\mathrm{𝗌𝖾𝗍}(\mathrm{𝟐},{\mathrm{𝗋𝖾𝖼}}_{\mathrm{𝟐}}(V,u,v))$.

4. Infinity: take $w=\mathrm{𝗌𝖾𝗍}(\mathbb{N},I)$, where $I:\mathbb{N}\to V$ is given by the recursion $I(0):\equiv \mathrm{\varnothing }$ and $I(n+1):\equiv I(n)\cup \{I(n)\}$.

5. Union: Take any $v:V$ and any presentation $f:A\to V$ with $v=\mathrm{𝗌𝖾𝗍}(A,f)$. Then let $\tilde{A}:\equiv {\sum }_{(a:A)}[fa]$, where $m_{fa}:[fa]V$ is the type of members from def:TypeOfElements. $\tilde{A}$ is plainly $𝒰$-small, and we have $\cup v:\equiv \mathrm{𝗌𝖾𝗍}(\tilde{A},\lambda x.m_{f({\mathrm{𝗉𝗋}}_1(x))}({\mathrm{𝗉𝗋}}_2(x)))$.

6. Function set: given $u,v:V$, take the types of elements $[u]V$ and $[u]V$, and the function type $[u]\to [v]$. We want to define a map

   $$r:([u]\to [v])⟶V$$

   with “$r(f)=(x,f(x))|x:[u]$”, but in order for this to make sense we must first define the ordered pair $(x,y)$, and then we take the map $r^{\prime }:x\mapsto (x,f(x))$, and then we can put $r(f):\equiv \mathrm{𝗌𝖾𝗍}([u],r^{\prime })$. But the ordered pair can be defined in terms of unordered pairing as usual.

7. $\in$-induction: let $C:V\to \mathrm{𝖯𝗋𝗈𝗉}$ be a class such that $C(a)$ holds whenever $C(x)$ for all $x\in a$, and take any $v=\mathrm{𝗌𝖾𝗍}(B,g)$. To show that $C(v)$ by induction, assume that $C(g(b))$ for all $b:B$. For every $x\in v$ there merely exists some $b:B$ with $x=g(b)$, and so $C(x)$. Thus $C(v)$.

8. Replacement: the statement “$C$ is a $V$-set” is a mere proposition, so we mayproceed by induction as follows. Supposing $x$ is $\mathrm{𝗌𝖾𝗍}(A,f)$, we claim that $w:\equiv \mathrm{𝗌𝖾𝗍}(A,r\circ f)$ is the set we are looking for. If $C(y)$ then there merely exists$z:V$ and $a:A$ such that $z=f(a)$ and $y=r(z)$, therefore $y\in w$.Conversely, if $y\in w$ then there merely exists $a:A$ such that $y=r(f(a))$, soif we take $z:\equiv f(a)$ we see that $C(y)$ holds.

9. Let us say that a class $C:V\to \mathrm{𝖯𝗋𝗈𝗉}$ is separableif for any $a:V$ the class

   $$a\cap C:\equiv x|x\in a\wedge C(x)$$

   is a $V$-set.We need to show that any $𝒰$-small $C:V\to {\mathrm{𝖯𝗋𝗈𝗉}}_{𝒰}$ is separable. Indeed, given $a=\mathrm{𝗌𝖾𝗍}(A,f)$, let $A^{\prime }=\exists (x:A).C(fx)$, and take $f^{\prime }=f\circ i$, where $i:A^{\prime }\to A$ is the obvious inclusion. Then we can take $a^{\prime }=\mathrm{𝗌𝖾𝗍}(A^{\prime },f^{\prime })$ and we have $x\in a\wedge C(x)\iff x\in a^{\prime }$ as claimed. We needed the assumption that $C$ lands in $𝒰$ in order for $A^{\prime }=\exists (x:A).C(fx)$ to be in $𝒰$.∎

It is also convenient to have a strictly syntactic criterion of separability, so that one can read off from the expression for a class that it produces a $V$-set. One such familiar condition is being “${\mathrm{\Delta }}_0$”, which means that the expression is built up from equality $x{=}_{V}y$ and membership $x\in y$, using only mere-propositional connectives $\mathrm{\neg }$, $\wedge$, $\vee$, $\Rightarrow$ and quantifiers $\forall$, $\exists$ over particular sets, i.e. of the form $\exists (x\in a)$ and $\forall (y\in b)$ (these are called bounded quantifiers).

Corollary 10.5.8.

If the class $C\mathrm{:}V\mathrm{\to }\mathrm{Prop}$ is ${\mathrm{\Delta }}_{\mathrm{0}}$ in the above sense, then it is separable.

Proof.

Recall that we have a $𝒰$-small resizing $x\sim y$ of identity $x=y$. Since $x\in y$ is defined in terms of $x=y$, we also have a $𝒰$-small resizing of membership

$$x\tilde{\in }\mathrm{𝗌𝖾𝗍}(A,f):\equiv \exists (a:A).x\sim f(a).$$

Now, let $\mathrm{\Phi }$ be a ${\mathrm{\Delta }}_0$ expression for $C$, so that as classes $\mathrm{\Phi }=C$ (strictly speaking, we should distinguish expressions from their meanings, but we will blur the difference). Let $\tilde{\mathrm{\Phi }}$ be the result of replacing all occurrences of $=$ and $\in$ by their resized equivalents $\sim$ and $\tilde{\in }$. Clearly then $\tilde{\mathrm{\Phi }}$ also expresses $C$, in the sense that for all $x:V$, $\tilde{\mathrm{\Phi }}(x)\iff C(x)$, and hence $\tilde{\mathrm{\Phi }}=C$ by univalence. It now suffices to show that $\tilde{\mathrm{\Phi }}$ is $𝒰$-small, for then it will be separable by the theorem.

We show that $\tilde{\mathrm{\Phi }}$ is $𝒰$-small by induction on the construction of the expression. The base cases are $x\sim y$ and $x\tilde{\in }y$, which have already been resized into $𝒰$. It is also clear that $𝒰$ is closed under the mere-propositional operations (and $(-1)$-truncation), so it just remains to check the bounded quantifiers $\exists (x\in a)$ and $\forall (y\in b)$. By definition,

$\exists (x\in a)P(x)$$:\equiv \parallel {\displaystyle \sum _{x:V}}(x\tilde{\in }a\wedge P(x))\parallel ,$

$\forall (y\in b)P(x)$$:\equiv {\displaystyle \prod _{x:V}}(x\tilde{\in }a\to P(x)).$

Let us consider $\parallel {\sum }_{(x:V)}(x\tilde{\in }a\wedge P(x))\parallel$. Although the body $(x\tilde{\in }a\wedge P(x))$ is $𝒰$-small since $P(x)$ is so by the inductive hypothesis, the quantification over $V$ need not stay inside $𝒰$. However, in the present case we can replace this with a quantification over the type $[a]V$ of members of $a$, and easily show that

$$\sum _{x:V}(x\tilde{\in }a\wedge P(x))=\sum _{x:[a]}P(x).$$

The right-hand side does remain in $𝒰$, since both $[a]$ and $P(x)$ are in $𝒰$. The case of ${\prod }_{(x:V)}(x\tilde{\in }a\to P(x))$ is analogous, using ${\prod }_{(x:V)}(x\tilde{\in }a\to P(x))={\prod }_{(x:[a])}P(x)$.∎

We have shown that in type theory with a universe $𝒰$, the cumulative hierarchy $V$ is a model of a “constructive set theory”with many of the standard axioms.However, as far as we know, it lacks the strong collectionand subset collectionaxioms which are included in Constructive Zermelo–Fraenkel Set Theory [AczelCZF].In the usual interpretation of this set theory into type theory, these two axioms are consequences of the setoid-like definition of equality; while in other constructed models of set theory, strong collection may hold for other reasons.We do not know whether either of these axioms holds in our model $(V,\in )$, but it seems unlikely.Since $V$ is a higher inductive type inside the system, rather than being an external construction, it is not surprising that it differs in some ways from prior interpretations.

Finally, consider the result of adding the axiom of choice for sets to our type theory, in the form $\mathrm{𝖠𝖢}$ from subsec:emacinsets above. This has the consequence that $\mathrm{𝖫𝖤𝖬}$ then also holds, by thm:1surj_to_surj_to_pem, and so $$ is a topos with subobject classifier $\mathrm{𝟐}$, by thm:settopos. In this case, we have $\mathrm{𝖯𝗋𝗈𝗉}=\mathrm{𝟐}:𝒰$, and so all classes are separable.Thus we have shown:

Lemma 10.5.9.

In type theory with $\mathrm{AC}$, the law of (full) separationholds for $V$: given any class $C\mathrm{:}V\mathrm{\to }\mathrm{Prop}$ and $a\mathrm{:}V$, the class $a\mathrm{\cap }C$ is a $V$-set.

Theorem 10.5.10.

In type theory with $\mathrm{AC}$ and a universe $\mathrm{U}$, the cumulative hierarchy $V$ is a model of Zermelo–Fraenkel set theory with choice, ZFC.

Proof.

We have all the axioms listed in thm:VisCST, plus full separation, so we just need to show that there are power sets $𝒫(a):V$ for all $a:V$. But since we have $\mathrm{𝖫𝖤𝖬}$ these are simply function types $𝒫(a)=(a\to \mathrm{𝟐})$. Thus $V$ is a model of Zermelo–Fraenkel set theory ZF. We leave the verification of the set-theoretic axiom of choice from $\mathrm{𝖠𝖢}$ as an easy exercise.∎