Functions vs. Operators

Question

Let $\mathsf{I}$ be an operator that takes as input $x$ and returns $x$. Then $\mathsf{I}$ is not a function.

1 Why is this not a function?

According to Lambda Calculus and Combinators, $\mathsf{I}$ is not a function in the set theoretic sense (that is, $\mathsf{I}$ cannot be represented as a set of ordered pairs):

In ZF, each set $S$ has an identity function $\mathsf{I}_S$ with domain $S$, but there is no 'universal' identity which can be applied to everything.

Question 1: Is this because if $\mathsf{I}$ were a function, then its domain would be some universal class $U$ that contained everything (and hence would run into Russell's paradox, implying $U$ cannot be a set)?

2 Why can't set-theoretic functions be applied to themselves?

Consider that $\mathsf{I}x = x$ implies that $\mathsf{I} \mathsf{I} = \mathsf{I}$. This seems to make sense.

But this implies that $\mathsf{I}$ is in its own "domain", which can evidently never happen in ZFC:

[T]he axiom of foundation ... prevents functions from being applied to themselves.

Now the axiom of foundation, according to Wikipedia, states in English that

[E]very non-empty set $A$ contains an element that is disjoint from $A$.

But it hardly seems evident from this why this means a function cannot be applied to itself.

Question 2: Why can a set theoretic function $f$ not be in its own domain?

3 Modeling operators in ZFC

The operator concept can be modelled in standard ZF set theory if, roughly speaking, we interpret operators as infinite sequences of functions (satisfying certain conditions), instead of as single functions.

The author does not explain how this works.

Question 3: How do we model operators like $\mathsf{I}$ in ZFC?

Answer 1

This is a very good question because it highlights a fundamental problem in arguments over the foundations of mathematics.

When Church introduced the lambda calculus, he had been clear that it falls under the domain of "intensional mathematics". The function you are asking about can be applied to itself for this reason.

With regard to ZF, you must consider which axioms account for those identity functions over sets as they exist in models. Iterations of power sets along with unions generate sets that are interpretable as functions over other sets. But no such set may be formed for the model domain, itself, because that would require further applications of power sets and unions. So, this may only exist within a model that takes the first model as a set.

If a function is a set interpretable as a function over a set, how can that function be an element of the domain unless it occurs as an element of one of the ordered pairs. The axiom of foundation excludes infinite descending chains of membership relations. If the function existed as an element in one of the ordered pairs comprising it, one could form such an infinite descending chain.

I should leave question 3 to more technically proficient responders.

The distinction between intension and extension is an important one. The lambda calculus is certainly significant to the foundations of mathematics with respect to computability. But, if one accepts criticisms of set theory based on alternatives because they, too, can "represent" mathematical systems, then one is adhering to the claim that mathematics is intensional rather than extensional.

The original distinction between intension and extension arises with Leibniz and had been portrayed in order-theoretic terms. When nineteenth century philosophers such as Mill objected Kantian accounts of space and time as ideal rather than actual, logicians began to focus solely on grammatical forms. This, in turn, led to Frege's work. Frege's identity puzzles ask whether or not identity is a relation between objects or a relation between denotations of objects. In his analysis, Frege had been forced to distinguish between grammatical definitions of objects and the object to which they refer. This is the famous distinction between "sense" and "referent". It is also the modern distinction between intension and extension. We say that $f(x)=x+x$ and $f(x)=2x$ are "the same" function because they have the same extensional representation. This is easy to forget when one is learning intensional systems. Intensional mathematics arises on the "syntax" side of the syntax/semantics dichotomy.