On a proof of Cantor’s theorem

The famous theorem by Cantor states that the cardinality of a powerset $P(A)$ is larger than the cardinality of $A$. There are several equivalent formulations, and the one I want to consider is

Theorem (Cantor): There is no onto map $A \to P(A)$.

In this post I would like to analyze the usual proof of Cantor’s theorem and present an insightful reformulation of it which has applications outside set theory. All of what is written here is quite easy and far from being new, but in my opinion still interesting enough to be presented to a wider audience.

If we open a book on set theory, we will find a proof of Cantor’s theorem which shows explicitly that for every map $e : A \to P(A)$ there is a subset of $A$ outside its image, namely
$$S = \lbrace x \in A \mid x \not\in e(x) \rbrace$$
If we had $S = e(y)$ for some $y \in A$ it would follow both that $y$ is and is not an element of $S$. A first observation is that this is a constructively valid proof, hence Cantor’s theorem holds in intuitionistic set theory just as well. But how wide is the scope of the theorem really? Let us rework it as abstractly as possible, to give it a wider applicability.

First we replace the powerset $P(A)$ with the set of functions $\Omega^A$ where $\Omega$ is the set of truth values. In the case of classical logic $\Omega = \lbrace 0,1 \rbrace$ but there is no need to rely on this fact. We prefer to think of the general situation in which the truth values correspond to subsets of the singleton set $\lbrace 0 \rbrace$, so that $\Omega = P(\lbrace 0 \rbrace)$. The bijection between $P(A)$ and $\Omega^A$ is then just the usual one between subsets and their characteristic maps: a subset $S subseteq A$ corresponds to the map $\chi_S(x) = \lbrace u \in \lbrace 0 \rbrace \mid x \in S\rbrace $, while a map $\chi : A \to \Omega$ corresponds to the subset $\lbrace x \in A \mid 0 \in \chi(x)\rbrace $.

Logical negation $\lnot$ can be seen as a map $N : \Omega \to \Omega$ defined by $N(p) = \lbrace u \in \lbrace 0 \rbrace \mid 0 \not\in p\rbrace $. Note that $N$ does not have a fixed point, for if there were $p \in \Omega$ such that $N(p) = p$ then we would have both $0 \in p$ and $0 \not\in p$.

Now our proof reads as follows: suppose we have a map $e : A \to \Omega^A$. Consider the map $s : A \to \Omega$ defined by $s(x) = N (e(x)(x))$. If there were $y \in A$ such that $s = e(y)$, we would have $e(y)(y) = s(y) = N(e(y)(y))$, a contradiction. Therefore $e$ is not onto. QED. How is this any better than what we had before? It gives us a chance to think about the positive aspects of the situation: if $e$ were onto, then $\Omega$ could not have an endomap without fixed points. Because nothing in the proof specifically relies on $\Omega$ being the set of thruth values we may replace it with a general set to obtain:

Theorem (Lawvere): If there is an onto map $e : A \to B^A$ then every $f : B \to B$ has a fixed point.

We already know how to prove this: consider the map $s : A \to B$ defined by $s(x) = f(e(x)(x))$. Because $e$ is onto, there is $y \in A$ such that $e(y) = s$. Then we have $e(y)(y) = s(y) = f(e(y)(y))$, therefore $e(y)(y)$ is a fixed point of $f$. QED.

Cantor’s theorem is a corollary of Lawvere’s theorem with $B = \Omega$ and the observation than negation does not have a fixed point.

Now consider Lawvere’s theorem in isolation and how one would go about proving it, perhaps something like this: “How can I have such an onto map $e : A \to B^A$? Surely $B$ cannot have too many elements, in fact, this can only happen if $B$ is a singleton or empty. I can see Lawvere’s theorem to be obviously true but is rubbish because it only holds in trivial cases.” There is a mistake in the last sentence: as we shall see shortly, Lawvere’s theorem is true in interesting cases, but you (the imaginary mathematician, not the readers of this blog…) can only imagine it in trivial cases because you did not bother to look outside the narrow set-theoretic scope.

Lawvere’s theorem is a positive reformulation of the diagonalization trick that is at the heart of Cantor’s theorem. It can be formulated in any cartesian closed category, and its proof uses just equational reasoning with a modicum of first-order logic. We should expect it to have a much wider applicability than Cantor’s theorem. Indeed, immediately we see that other well-known proofs by diagonalization are corollaries, for example:

  1. The set of sequences of numbers $\mathbb{N} \to \mathbb{N}$ is uncountable because the successor operation does not have a fixed point.
  2. There is no continuous surjection $\mathbb{R} \to C(\mathbb{R}, \mathbb{R})$ from the real line onto the Banach space of continuous real functions, equipped with the compact-open topology, because the real map $x \mapsto x+1$ is continuous and has no fixed points.

More interestingly, there are positive consequences of Lawvere’s theorem, too:

  1. To contrast the second case above, we ask whether there is a continuous surjection from $\mathbb{R}$ onto $C(\mathbb{R}, [0,1])$, the space of continuous real functions taking values on the closed interval, and equipped with the sup metric. If there is such a map, it follows that the closed interval has the fixed-point property, and moreover that every cube $[0,1]^n$ has the fixed-point property too (exercise). So this might be a nice way to prove Brouwer’s fixed point theorem, and even if it does not work, it is a nice idea that will get you thinking about space filling curves for a while.
  2. In the effective topos the c.e. sets are represented as maps $\Sigma^\mathbb{N}$ where $\Sigma$ is the set of semidecidable truth values. Because there is an effective enumeration of c.e. sets, in the effective topos there is an onto map $W : \mathbb{N} \to \Sigma^\mathbb{N}$, which immediately tells us that $\Sigma$ has the fixed-point property, and so does $\Sigma^\mathbb{N}$ because it is isomorphic to $(\Sigma^\mathbb{N})^\mathbb{N}$. Thus we obtain a theorem in computability theory stating that every enumeration operator has a fixed point.

Lastly, let me comment on a question by Paul Stadtmann on the FOM mailing list. He wonders whether the axiom of separation (a.k.a. the subset axiom) is needed to prove Cantor’s theorem. If we are working just in straight set theory, then bounded separation certainly suffices. (This is the form of separation in which the defining predicate has only bounded quantifiers of the form $\forall x \in A$ and $\exists x \in A$, but none of the form $\forall x$ and $\exists x$.) However, bounded separation is only needed to establish a general fact about the universe of sets, namely that it forms a cartesian closed category. After that Lawvere’s theorem kicks in and gets the job done. So I would say that separation is not used in an essential way here (for example, topos theory directly axiomatizes exponentials and so separation is not needed at all there).

6 thoughts on “On a proof of Cantor’s theorem

  1. ”If there is such a map, it follows that the closed interval has the fixed-point property…”

    I am afraid the premise is false.

Leave a Reply to Andrej Bauer Cancel reply

Your email address will not be published. Required fields are marked *