Franz and Georg
/As far as I know, Kafka and Cantor never met, and there is no reason to believe they did. Still, I can't help wondering if Franz knew about Georg's work, even though he claimed to have great difficulties with all things scientific and mathematical. Here's why: Kafka's Great Wall of China, which in typical Kafka fashion is about all sorts of things and kind of goes nowhere, has elements that immediately make me think of Cantor's work, particularly the so-called Cantor set.
The Cantor set \(C\) is one of those mathematical curiosities that we like to trot out to blow our students' minds. It is constructed as follows. Start with the closed unit interval \([0,1]\). First remove the open middle third \( (1/3,2/3)\). Then remove the open middle thirds from the remaining two intervals: \((1/9,2/9)\) and \((7/9,8/9)\). Then remove the open middle thirds from the remaining four intervals. Iterate this process, at the \(n\)th stage removing \(2^{n-1}\) intervals of length \(1/3^n\). The set \(C\) is what remains at the "end."
The first claim about \(C\) is that it is, remarkably, uncountable. The way to prove this is to use Cantor's diagonal argument (I wrote about this in the previous entry). Here goes: let's first abandon decimal notation and instead represent each number \(x\) in the interval \([0,1]\) using its ternary expansion: \[ x=\frac{a_1}{3} + \frac{a_2}{3^2} + \frac{a_3}{3^3} + \cdots\] where each \(a_i = 0,1,\,\text{or}\, 2\). Now, observe that the elements of \(C\) are precisely those real numbers in the interval \([0,1]\) whose ternary expansions have all \(a_i = 0\,\text{or}\, 2\). (Aside: note that \(1/3\) is in \(C\). Its ternary expansion is \(0.1000\dots\), so you might think that I've told you a lie. But note that we also have \(1/3 = 0.02222\dots\), just like in decimal notation \( 0.9999999\dots = 1\).) If we have a bijection \(f:{\mathbb N}\to C\), then we construct a number \(x\) by taking the \(i\)th digit of \(x\) to be \(2\) if the \(i\)th digit of \(f(i)\) is \(0\) and \(0\) if the \(i\)th digit is \(2\). Then \(x\) isn't in the image of \(f\), contradiction.
But, in typical Cantorian fashion, \(C\) has another weird property. Let's add up the lengths of the intervals we remove from \([0,1]\) to get \(C\): \[\frac{1}{3} + \frac{2}{9} + \frac{4}{27} +\cdots +\frac{2^{n-1}}{3^n} +\cdots = \frac{1/3}{1-(2/3)} = 1.\] You read that correctly: we've removed "everything" yet what remains is an uncountable dust scattered throughout the unit interval.
Compare this with Kafka's description of how the Great Wall was built:
“The Great Wall of China was finished at its northernmost location. The construction work moved up from the south-east and south-west and joined at this point. The system of building in sections was also followed on a small scale within the two great armies of workers, the eastern and western. It was carried out in the following manner: groups of about twenty workers were formed, each of which had to take on a section of the wall, about five hundred metres. A neighbouring group then built a wall of similar length to meet it. But afterwards, when the sections were fully joined, construction was not continued on any further at the end of this thousand-metre section. Instead the groups of workers were shipped off again to build the wall in completely different regions. Naturally, with this method many large gaps arose, which were filled in only gradually and slowly, many of them not until after it had already been reported that the building of the wall was complete. In fact, there are said to be gaps which have never been built in at all, although that’s merely an assertion which probably belongs among the many legends which have arisen about the structure and which, for individual people at least, are impossible to prove with their own eyes and according to their own standards, because the structure is so immense.”
Imagine then, how this would look from space (you can see the Wall from there, or so the fraudsters at NASA would have us believe). In the early days of construction, you wouldn't be able to see it at all--it would be scattered, barely-visible segments, much like the Cantor set. In fact, it's possible to build the Cantor set in this way, via the following process. Define two functions \(F_0\) and \(F_1\) on the unit interval \([0,1]\) by \[ F_0(x) = \frac{1}{3}x\] \[F_1(x) = \frac{1}{3}x + \frac{2}{3}.\] Now, start with a number \(x_0\) in the open interval \((0,1)\) and iteratively apply one of the functions \(F_0\) or \(F_1\) randomly. The map \(F_0\) takes a point two-thirds of the way toward \(0\) and \(F_1\) takes a point two-thirds of the way toward \(1\). So, if a point is in, say \((1/3,2/3)\), then both \(F_0\) and \(F_1\) take it into one of the complementary intervals. If a point is in \((1/9,2/9)\) then it maps to either \((1/27,2/27)\) or \((19/27,20/27)\), and so on. No matter what we do, by iterating these maps indefinitely we end up at a point in \(C\).
Now, this isn't really the right metaphor since the Wall is getting filled in, while the Cantor set is built by chipping away, but it sure feels like the same idea. The workers getting shipped from one location to another, seemingly at random, to build this thing that no one individual can verify the existence of; points moving around the interval until they settle at points in \(C\). Beautiful and unimaginable all at once.