What is meant by a "diagonalization argument"?

Question

From Wikipedia:

A variety of diagonal arguments are used in mathematics.

• Cantor's diagonal argument
• Cantor's theorem
• Halting problem
• Diagonal lemma

Besides the above four examples, there is another one I found in a blog. When proving that "if a sequence of measurable mappings converges in measure, then there is a subsequence converging a.e.", the construction of the subsequence is also called diagonalization:

The method for establishing this result is fairly typical of such arguments: we rely on diagonalization along with the control that Borel-Cantelli gives us. Let $f_n$ be a sequence that converges in measure to $f$. This means that for any n we have a $f_{m_n}$ with $\mu(|f_{m_n} - f| > 1/n) < 2^{-n} $. Applying Borel-Cantelli to the sequence of sets $A_n = \{x | |f_{m_n}(x) - f(x)| > 1/n\}$ yields $\mu(\limsup_m \cup_{n=m}^\infty A_n) = 0$. But this is simply saying that the set of points on which $f_{m_n}$ doesn’t converge to $f$ has measure $0$.

As someone who has not been very much exposed to "diagonalization arguments", I wonder if the examples above have something in common, so that we may answer what "diagonalization argument" is and what kinds of problems it may help to solve?

Answer 1

Someone with more experience might be better equipped to address this question, but here's my perspective. "Diagonal arguments" are often invoked when dealings with functions or maps. In order to show the existence or non-existence of a certain sort of map, we create a large array of all the possible inputs and outputs. Moving along the diagonal, we use a self-referential construction to introduce an object that cannot possibly be on the list, and the desired object is obtained.

Cantor's Diagonal Argument: The maps are elements in $\mathbb{N}^{\mathbb{N}} = \mathbb{R}$. The diagonalization is done by changing an element in every diagonal entry.

Halting Problem: The maps are partial recursive functions. The killer $K$ program encodes the diagonalization.

Diagonal Lemma / Fixed Point Lemma: The maps are formulas, with input being the codes of sentences. The diagonalization is accomplished by a somewhat confusing self-referential construction.

Cantor's Theorem ($|A| < |P(A)|$) : The maps are the correspondences between elements of $A$ and elements in $P(A)$. Diagonalization is accomplished by considering a self-referential subset in $P(A)$ that nothing can map to.

Answer 2

Yes, I would consider the "diagonal argument" a general proof-construction technique, similar to, say, the pigeonhole principle or other combinatorial principles.

Perhaps the most general form of the diagonal argument is the following:

Theorem: Let $X$ be a set with more than one member, let $I$ be an arbitrary set, and let $X^I$ denote the set of functions from $I$ to $X$. Then there exists no surjection from $I$ to $X^I$.

Proof: Let $f$ be a function from $I$ to $X^I$, and let $a$ and $b$ be distinct members of $X$. For notational convenience, let $f_i$ denote the function $f(i)$. We construct a function $g \in X^I$ as follows:

$$g(i) = \begin{cases} a & \text{if }f_i(i) \ne a \\ b & \text{if }f_i(i) = a \end{cases}$$

By construction, for every $i \in I$, $g(i) \ne f_i(i)$, and thus $g \ne f_i$. Therefore $g$ is not in the image of $f$, and thus $f$ is not a surjection. $\square$

^{(Correction: Contrary to what I originally wrote, the proof above does work also for $I = \emptyset$. In that case, the only function $f: I \to X^I$ is the empty function, which is indeed not surjective, and the construction above vacuously but correctly yields a counterexample $g \in X^I$, where $g$ is itself the empty function from $I$ to $X$.)}

By setting up a bijection between the powerset $\mathcal P(I)$ and the set $\{0,1\}^I$ (sometimes written as $2^I$), we also obtain the following corollary:

Corollary: There exists no surjection from $I$ to $\mathcal P(I)$.

A nice feature of this particular form of the diagonal argument is that it makes very few logical or set-theoretical assumptions: for instance, it doesn't use proof by contradiction or make any reference to cardinalities. Thus, it works even in many oddball versions of logic and set theory than some people occasionally consider. However, in standard set theory, the theorem above can be given a more succint form:

Corollary: If $X$ has more than one member, then $X^I$ has strictly greater cardinality than $I$. Also, $\mathcal P(I)$ has strictly greater cardinality than $I$.

Answer 3

The Arzelà Ascoli theorem is an instance of this. See http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem. What is interesting about this example is that this theorem has applications in differential equations. One example of an application is the theorem that in a hyperbolic surface, every free homotopy class of loops has a unique geodesic in that class.