What exactly is the semantic difference between set and type?

Question

EDIT: I've now asked a similar question about the difference between categories and sets.

Every time I read about type theory (which admittedly is rather informal), I can't really understand how it differs from set theory, concretely.

I understand that there is a conceptual difference between saying "x belongs to a set X" and "x is of type X", because intuitively, a set is just a collection of objects, while a type has certain "properties". Nevertheless, sets are often defined according to properties as well, and if they are, then I am having trouble understanding how this distinction matters in any way.

So in the most concrete way possible, what exactly does it imply about $x$ to say that it is of type $T$, compared to saying that it is an element in the set $S$?

(You may pick any type and set that makes the comparison most clarifying).

Answer 1

To understand the difference between sets and types, ones has to go back to pre-mathematical ideas of "collection" and "construction", and see how sets and types mathematize these.

There is a spectrum of possibilities on what mathematics is about. Two of these are:

1. We think of mathematics as an activity in which mathematical objects are constructed according to some rules (think of geometry as the activity of constructing points, lines and circles with a ruler and a compass). Thus mathematical objects are organized according to how they are constructed, and there are different types of construction. A mathematical object is always constructed in some unique way, which determines its unique type.

2. We think of mathematics as a vast universe full of pre-existing mathematical objects (think of the geometric plane as given). We discover, analyze and think about these objects (we observe that there are points, lines and circles in the plane). We collect them into set. Usually we collect elements that have something in common (for instance, all lines passing through a given point), but in principle a set may hold together an arbitrary selection of objects. A set is specified by its elements, and only by its elements. A mathematical object may belong to many sets.

We are not saying that the above possibilities are the only two, or that any one of them completely describes what mathematics is. Nevertheless, each view can serve as a useful starting point for a general mathematical theory that usefully describes a wide range of mathematical activities.

It is natural to take a type $T$ and imagine the collection of all things that we can construct using the rules of $T$. This is the extension of $T$, and it is not $T$ itself. For instance, here are two types that have different rules of construction, but they have the same extension:

1. The type of pairs $(n, p)$ where $n$ is constructed as a natural number, and $p$ is constructed as a proof demonstrating that $n$ is an even prime number larger than $3$.

2. The type of pairs $(m, q)$ where $m$ is constructed as a natural number, and $q$ is constructed as a proof demonstrating that $m$ is an odd prime smaller than $2$.

Yes, these are silly trivial examples, but the point stands: both types have nothing in their extension, but they have different rules of construction. In contrast, the sets $$\{ n \in \mathbb{N} \mid \text{$n$ is an even prime larger than $3$} \}$$ and $$\{ m \in \mathbb{N} \mid \text{$m$ is an odd prime smaller than $2$} \}$$ are equal because they have the same elements.

Note that type theory is not about syntax. It is a mathematical theory of constructions, just like set theory is a mathematical theory of collections. It just so happens that the usual presentations of type theory emphasize syntax, and consequently people end up thinking type theory is syntax. This is not the case. To confuse a mathematical object (construction) with a syntactic expression that represents it (a term former) is a basic category mistake that has puzzled logicians for a long time, but not anymore.

Answer 2

In practice, claiming that $x$ being of type $T$ usually is used to describe syntax, while claiming that $x$ is in set $S$ is usually used to indicate a semantic property. I will give some examples to clarify this difference in usage of types and sets. For the difference in what types and sets actually are, I refer to Andrej Bauer's answer.

An example

To clarify this distinction, I will use the example given in Herman Geuvers' lecture notes. First, we look at an example of inhabiting a type:

$$3+(7*8)^5:\mathrm{Nat},$$ and an example of being member of a set: $$3\in \{n\in\mathbb{N}\mid \forall x,y,z\in\mathbb{N}^+ (x^n+y^n\neq z^n)\}$$

The main difference here is that to test whether the first expression is a natural number, we don't have to compute some semantic meaning, we merely have to 'read off' the fact that all literals are of type Nat and that all operators are closed on the type Nat.

However, for the second example of the set, we have to determine the semantic meaning of the $3$ in the context of the set. For this particular set, this is quite hard: the membership of $3$ for this set is equivalent to proving Fermat's last theorem! Do note that, as stated in the notes, the distinction between syntax and semantics cannot always be drawn that clearly. (and you might even argue that even this example is unclear, as Programmer2134 mentions in the comments)

Algorithms vs Proofs

To summarize, types are often used for 'simple' claims on the syntax of some expression, such that membership of a type can be checked by an algorithm, while to test membership of a set, we would in usually require a proof.

To see why this distinction is useful, consider a compiler of a typed programming language. If this compiler has to create a formal proof to 'check types', the compiler is asked to do an almost impossible task (automated theorem proving is, in general, hard). If on the other hand the compiler can simply run an (efficient) algorithm to check the types, then it can realistically perform the task.

A motivation for a strict(er) interpretation

There are multiple interpretations of the semantic meaning of sets and types. While under the distinction made here extensional types and types with undecidable type-checking (such as those used in NuPRL, as mentioned in the comments) would not be 'types', others are of course free to call them as such (just as free as they are as to call them something else, as long as their definitions fit).

However, we (Herman Geuvers and I), prefer to not throw this interpretation out of the window, for which I (not Herman, although he might agree) have the following motivation:

First of all, the intention of this interpretation isn't that far from that of Andrej Bauer. The intention of a syntax is usually to describe how to construct something and having an algorithm to actually construct it is generally useful. Furthermore, the features of a set are usually only needed when we want a semantic description, for which undecidability is allowed.

So, the advantage of our more stricter description is to keep the separation simpler, to get a distinction more directly related to common practical usage. This works well, as long as you don't need or want to loosen your usage, as you would for, e.g. NuPRL.

Answer 3

To start, sets and types aren't even in the same arena. Sets are the objects of a first-order theory, such as ZFC set theory. While types are like overgrown sorts. To put it a different way, a set theory is a first-order theory within first-order logic. A type theory is an extension of logic itself. Martin-Löf Type Theory, for example, is not presented as a first-order theory within first-order logic. It's not that common to talk about sets and types at the same time.

As Discrete lizard states, types (and sorts) serve a syntactic function. A sort/type behaves as a syntactic category. It lets us know what expressions are well-formed. For a simple example using sorts, let's say we described the theory of vector spaces over an arbitrary field as a 2-sorted theory. We have a sort for scalars, $\mathsf S$, and a sort for vectors, $\mathsf V$. Among many other things, we'd have an operation for scaling: $\mathsf{scale}:\mathsf S\times \mathsf V\to \mathsf V$. This lets us know that $\mathsf{scale}(\mathsf{scale}(s,v),v)$ is simply not a well-formed term. In a type theoretic context, an expression like $f(x)$ requires $f$ to have a type $X\to Y$ for some types $X$ and $Y$. If $f$ does not have the type of a function, then $f(x)$ is simply not a well-formed expression. Whether an expression is of some sort or has some type is a meta-logical statement. It makes no sense to write something like: $(x:X)\implies y=3$. First, $x : X$ is simply not a formula, and second, it doesn't even conceptually make sense as sorts/types are what let us know which formulas are well-formed. We only consider the truth value of well-formed formulas, so by the time we're considering whether some formula holds, we better already know that it is well-formed!

In set theory, and particularly ZFC, the only non-logical symbol at all is the relation symbol for set membership, $\in$. So $x\in y$ is a well-formed formula with a truth value. There are no terms other than variables. All the usual notation of set theory is a definitional extension to this. For example, a formula like $f(x)=y$ is often taken to be shorthand for $(x,y)\in f$ which itself may be taken as shorthand for $\exists p.p\in f\land p=(x,y)$ which is shorthand for $$\exists p.p\in f\land(\forall z.z\in p\iff [z = x\lor(\forall w.w\in z\Leftrightarrow w = y)])$$ At any rate, any set can take the place of $f$ and everything is a set! As I pointed out in a different question recently, $\pi(7)=3$ where $\pi$ is the real number is a completely legitimate and meaningful (and potentially even true) set theoretic expression. Basically, anything you write that parses in set theory can be given some meaning. It may be a completely spurious meaning, but it has one. Sets are also "first-class" objects in set theory. (They better be as they are the only objects usually.) A function like $$f(x)=\begin{cases}\mathbb N, &\text{if }x=1\\7,&\text{if }x=\mathbb Q\\x\cap \mathbb R^\mathbb R,&\text{if }x=(\mathbb Z,\mathbb N)\end{cases}$$ is a completely legitimate function in set theory. There is nothing even remotely analogous to this in type theory. The closest would be to use codes for a Tarskian universe. Sets are the objects of set theory; types are not the object of type theory.

A type is not a collection of things (neither is a set for that matter...), and it is not defined by a property. A type is a syntactic category that lets you know what operations are applicable to terms of that type and which expressions are well-formed. From a propositions-as-types perspective, what types are classifying are the valid proofs of the proposition to which the type corresponds. That is, the well-formed (i.e. well-typed) terms of a given type correspond to the valid proofs (which are also syntactic objects) of the corresponding proposition. Nothing like this is happening in set theory.

Set theory and type theory are really not anything alike.

Answer 4

I believe that one of the most concrete differences about sets and types is the difference in the way the "things" in your mind are encoded into the formal language.

Both sets and types allow you to speak about things, and collections of things. The main difference is that with sets, you can ask any question you want about things and it will maybe be true, maybe not; while with types, you first have to prove that the question makes sense.

For example, if you have booleans $\mathbb B=\{\operatorname{true},\operatorname{false}\}$ and natural numbers $\mathbb N=\{0, 1,\dots \}$, with types, you can not ask if $\operatorname{true}=1$ which you can with sets.

One way to interpret this is that with sets, everything is encoded into a single collection: the collection of all sets. $0$ is encoded as $[0]=\{\}$, $n+1$ is encoded as $[n+1]=\{[n]\}\cup [n]$ and $\operatorname{true}$ and $\operatorname{false}$ can be encoded by any two distinct sets. So that it actually makes sense to ask if $\operatorname{true}=1$, as it can be understood as asking if "the encoding chosen for $\operatorname{true}$ is the same as the encoding chosen for $1$". But the answer to this question could change if we chose another encoding: it is about the encodings and not really about the things.

You can then think of types as describing the encoding of the things inside it. With types, to ask the question of whether $a=b$, you first have to show that $a$ and $b$ have the same type, i.e. that they were encoded in the same way, which prohibits questions such as $\operatorname{true}=1$. You could still want to have a big type $S$ in which both $\mathbb B$ and $\mathbb N$ could be encoded, and then given two encodings $\iota_\mathbb B:\mathbb B\to S$ and $\iota_\mathbb N:\mathbb N\to S$, you could ask whether $\iota_\mathbb B(\operatorname{true})=\iota_\mathbb N(1)$ but the fact that this question depends on the encodings (and the choice of encodings) is now explicit.

Note that in those cases, whether the question made sense was actually easy to see but it could be much harder as in, for example, $(\operatorname{if} \text{very_hard_question} \operatorname{then} 1 \operatorname{else} \operatorname{true})=1$.

In summary, sets let you ask any question you want, but types force you to make encodings explicit when the answer may depend on them.