You can't model $\lambda$ functions as set functions because the domain of a $\lambda$ function includes that function itself. This violates the foundation axiom.

However, they clearly are some sort of arrow. Are the arrows of $\lambda$ calculus their own thing or can they be represented in some other way?

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    $\begingroup$ The text you are reading probably defined $\lambda$ terms, so you should know what they are, in the technical sense. So what are you asking precisely'? $\endgroup$ Oct 7, 2016 at 5:51
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    $\begingroup$ The text I'm reading defined $\lambda$ as terms with rewriting rules. However $\lambda$ functions also behave like they are (partial) mappings $ \lambda \to \lambda $. So is there a way of treating lambda functions as if they are mappings/arrows? $\endgroup$ Oct 7, 2016 at 6:19

4 Answers 4


Actually it is possible to interpret $\lambda$-terms (I assume you use $\lambda$-function as a synonym for $\lambda$-term) as some sort of functions.

This is due to a result of Dana Scott, who introduced domain theory exactly for the purpose of giving a function-like based semantics for the (untyped) $\lambda$-calculus.

The idea is to model the universe of the $\lambda$-terms as a very special domain $D$, which is a poset satisfying some properties. One of these particular properties of this domain is that there is an isomorphism between the domain $D$ and the domain $[D,D]$ of the scott continuous functions from $D$ into itself.

Through the said isomorphism is possible to interpret each $\lambda$-term (i.e. an element of $D$) as a function (as an element of $[D,D]$) and, on the other hand, every function in $[D,D]$ is (i.e. corresponds via the isomorphism to) a $\lambda$-term.

The critical point here is that $[D,D]$ is not the set of functions from $D$ into itself: it would not be possible to represent all these functions through $\lambda$-terms due to cardinality reasons, but it is possible to represent $\lambda$-terms as a very well suited subset of functions over some set.

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    $\begingroup$ I feel like this answer has things exactly backwards -- the importance of this, as I understand it is not about interpreting $\lambda$-terms as functions (which is relatively easy), but about finding semantics where every function on the domain of the interpretation turns out to (uniquely) correspond to an element of the domain. That is, having a map $D \to D^D$ (satisfying the obvious properties) is a very easy thing to construct from an interpretation; having a map $D^D \to D$ (satisfying the obvious properties) is hard. Having it also be an isomorphism is harder. $\endgroup$
    – user14972
    Oct 8, 2016 at 10:45
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    $\begingroup$ @Hurkyl, apologize, but I do not understand your objection. The answer provides (in a vague way, I'll give you that) a semantics that models/interprets $\lambda$-terms as functions (of a specific type). So I have to ask, are you saying that there is something wrong in the content of the answer or in the form? (Any suggestion to improve the answer is very welcome). $\endgroup$ Oct 8, 2016 at 15:59

I've added a section to the bottom of my answer on a more general approach that shows how you can treat functions that can be applied to themselves within set theory, representing them as something other than sets of ordered pairs.

First, in spite of the name, a $\lambda$-function isn't a function in the mathematical sense. It's a string (or some equivalent mathematical object) that represents an algorithm — in other words, it's a computer program. (This is the similar to the sense in which the word "function" is used in computer languages, which is different from the usual mathematical usage.)

Mathematical functions are extensional; in other words, if $f$ and $g$ have the same values on all their arguments, then $f=g.$ This is why functions can be considered to be sets of ordered pairs. Only the argument-value pairs matter, not how they're computed or determined.

Computer programs, in contrast, are intensional. Two different programs that compute the same values on all their arguments are still different programs. In fact we have no way of even determining (algorithmically) whether two computer programs compute the same function.

A better name than "$\lambda$-function" is "$\lambda$-expression" because these things are expressions or strings in a language, not actual functions themselves.

Formally, if you have an alphabet $A$ and if $S$ is the set of strings on that alphabet, a $\lambda$-expression $e$ is a member of $S$ satisfying certain syntactical requirements. It isn't a function, but it does describe an algorithm that computes a certain function $\phi_e\colon D_e\to S$ for some set $D_e \subseteq S$ (the domain $D_e$ may not be all of $S$ because the computer program that $e$ describes may not halt on all possible input strings). Since $e\in S,$ you can compute $\phi_e(e)$ just as you can compute $\phi_e(x)$ for any $x\in S$ (of course, any such computation may or may not halt, so it's possible that $\phi_e(x)$ doesn't happen to have a value).

Now $\phi_e$ is just a normal mathematical function; you can regard it as a set of ordered pairs. From that point of view, $\phi_e \subseteq S\times S.$ Notice that $e$ itself is a member of just $S.$ You can have $e_1\ne e_2$ (different programs) but $\phi_{e_1}=\phi_{e_2}$ (the different programs happen to always compute the same output if given the same input).

So we never apply a function to itself. We apply a function to a string, and we also happen to use strings as computer programs that can be used to compute functions.

Incidentally, you can also define the application function $\alpha$ mapping some subset of $S\times S$ to $S$ by setting $\alpha(x,y)=\phi_x(y);$ the ordered pair $\langle x,y \rangle$ belongs to the domain of $\alpha$ iff the string $x$ is a $\lambda$-expression and the computation $\phi_x(y)$ halts.

And there's no problem coding all of this inside ZF.

There's a somewhat different approach to all this which is more general. Let's view the question as asking: How can functions that can be applied to themselves be handled, in light of the fact that a function in set theory is never a member of its own domain?

To avoid confusion, I'll call these hypothetical new functions *-functions just to distinguish them from the actual functions of set theory.

To do this, we need to start with (1) a set $F$ (which is intended to serve eventually as both the domain of these *-functions and the set of all *-functions), (2) a subset $D$ of $F\times F,$ and (3) a function $\alpha\colon D\to F.$ Note that these are defined in set theory; in particular, $\alpha$ is a function, not a *-function. The function $\alpha$ is intended to represent the application of a *-function to an argument.

So now we'll define a *-function to be a member of $F.$ If $f$ and $g$ are *-functions, we'll define $f(g) = \alpha(f, g)$ if $\langle f,g \rangle \in D;$ if $\langle f,g \rangle \not\in D,$ then $f(g)$ is undefined. This essentially means that we can use each *-function as a partial "function" from $F$ to $F,$ that is, a "function" from some subset of $F$ to $F.$ (I've put "function" in quotes here since these aren't sets of ordered pairs, as functions are usually defined, but they can be used just like functions because we have a definition of what $f(x)$ means for any $f$ and $x$ in $F$. On the other hand, for each $f\in F,$ the function $x \mapsto \alpha(f,x)$ is just a regular mathematical function; we could call it, say, $\varphi_f,$ and treat it as a set of ordered pairs, as usual. We could even have started with the function $f\mapsto \varphi_f$ instead of $\alpha.$ But notice that, in line with the extensionality/intensionality distinction in the first part of this answer, it's possible to have $\varphi_f=\varphi_g$ even when $f\ne g.)$

Incidentally, if you want to avoid having partial "functions," just distinguish one element in $F,$ which I'll call $\uparrow.$ Make $\alpha$ a total function now, with domain all of $F\times F.$ If $\alpha(x,y)$ was undefined, set it equal to $\uparrow$ now. Also set $\alpha(x,\uparrow)=\alpha(\uparrow,x)=\,\uparrow$ for all $x\in F.$ But I think the approach with partial "functions" is clearer, since it better represents the situation.

What really underlies this entire approach is that the definition of function as a set of ordered pairs isn't sacrosanct; it's just a convenient coding that has been adopted to represent functions as sets. Some other coding would be fine too (just as real numbers can be represented as Cauchy sequences or Dedekind cuts or something else, as long as they have the right properties). The key thing for functions is that we need to have a way to apply a function to an argument.

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    $\begingroup$ Your parenthetical on not halting on all inputs would be a good place to use the term "partial function" and to distinguish it from the unmodified term "function" used by the OP. This is a specific way in which the computation objects of computer science fail to be "functions". $\endgroup$ Oct 7, 2016 at 13:46
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    $\begingroup$ @EricTowers I purposely avoided the term "partial function", since I thought that obscured things. After all, a partial function is a function; it's just that its domain is not all of $S$ (where $S$ is as in my answer). So the distinction between total and partial function (which is really just distinguishing functions based on whether their domain is all of $S$ or is a proper subset of $S)$ is not relevant to OP's question, which is how can these "functions", which can seemingly be applied to themselves, be reconciled with the functions of set theory, which can't be applied to themselves. $\endgroup$ Oct 7, 2016 at 16:34
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    $\begingroup$ @EricTowers I just added a section to my answer on a different approach to this question, in which it's natural to think specifically about partial functions. $\endgroup$ Oct 7, 2016 at 17:31

We can define composition

$$ f \circ g = \lambda x. f(gx) $$

(where $x$ is chosen so that it does not appear in $f$ or $g$)

If you take $\alpha$, $\beta$, and $\eta$ reduction as specifying that two terms are equal, then composition is an associative, binary operation with identity

$$ i = \lambda x.x $$

and thus the lambda terms with this operation form a monoid $\Lambda$, and so you can do all of the usual monoid things with it.

Monoids can act on sets. Any monoid can act on its set of elements (e.g. where the action is given by the monoid operation). Application of lambda terms gives another interesting way for $\Lambda$ to act on itself.

Since you bring up category theory, the simply typed $\lambda$-calculus is very important — in a suitable sense, simply typed $\lambda$ calculi and cartesian closed categories are equivalent notions.


What you are missing is that $f(f)$ is not the same as $f(code(f))$. A function $f$ may be coded as another object $code(f)$, which may be in the domain of $f$. In computation theory, λ-functions are coded as strings of a certain form, which can be thought of as programs of some sort. Applying one program $p$ to another program $q$ requires a machine/interpreter to perform the application, such as a universal Turing machine $U$, such that $U(p,x)$ is the output of executing the program $p$ on input $x$. $U$ itself is not a program, but corresponds to one, namely $code(U)$. Multiple inputs can be coded, for example we can have a function $pair$ such that a program given $pair(x,y)$ can obtain $x$ and $y$. Indeed, we could design $code(U)$ such that $U(code(U),pair(p,x)) = U(p,x)$. Similarly, we have $f(code(f)) = U(code(f),code(f))$.

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    $\begingroup$ At least for computation theory, one needs very little to build a model for programs and Turing machines, because finite strings can themselves be coded as natural numbers, and so any meta-system that has the collection of natural numbers suffices. ZF is much must stronger than necessary, and the issue has really nothing to do with set theory unless you want to talk about things beyond ordinary programs, such as iterating the Turing jump... $\endgroup$
    – user21820
    Oct 7, 2016 at 13:51

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