TITLE: Koch curve from Cantor sets (paradox)
QUESTION [6 upvotes]: The Koch curve is normally constructed by taking a line segment, replacing the middle third with two copies of itself forming legs in an equilateral triangle, and then repeating this recursively for every subsegment. See image below. At every step, the length of the curve is multiplied by $4/3$ so the final length is infinite.
Notice that every line segment undergoes the construction of the Cantor set:

Therefore one could consider replacing every line segment with Cantor sets already from the beginning. Thus, start with the Cantor set, take two smaller copies of the Cantor set and make a $\wedge$ in the middle opening, and then repeat recursively. In this case the final length will be zero since the Cantor set has length zero.
Question: How is this paradox resolved? Will the limit set of my construction be different than the ordinary Koch curve? If so, what points are missing?

REPLY [4 votes]: Not every point in the Koch curve is in one of your Cantor sets!  Indeed, this follows immediately from the Baire category theorem: each of your Cantor sets is a closed subset of the Koch curve with empty interior, and the union of countably many such sets will still have empty interior and thus cannot be the whole Koch curve.
The proof of the Baire category theorem also gives a recipe for finding points that are not in any of the Cantor sets.  Let's say the $n$th Cantor set is $C_n$ and the Koch curve is $K$.  Start with a point $x_0\in K\setminus C_0$; then there is some closed ball $B_0$ around $x_0$ which does not intersect $C_0$.  Then pick a point $x_1\in B_0\setminus C_1$.  There is again some closed ball $B_1$ around $x_1$ which does not intersect $C_1$.  Then pick a point $x_2\in B_0\cap B_1\setminus C_2$, and so on.
This procedure gives a sequence of points $(x_n)$ in $K$ and balls $B_n$ such that $x_m\in B_n$ for all $m\geq n$ and $B_n\cap C_n=\emptyset$.  Since $K$ is compact, this sequence $(x_n)$ must accumulate at some point $x\in K$ (in fact, we can choose the balls $B_n$ to have radius converging to $0$ and then $(x_n)$ will be Cauchy and thus will converge).  Then $x\in B_n$ for all $n$, so $x$ cannot be in $C_n$ for any $n$.

Another way to think of it is that the Koch curve is parametrized by a continuous function $f:[0,1]\to \mathbb{R}^2$ which is the limit of functions $f_n:[0,1]\to\mathbb{R}^2$ parametrizing the approximations to the Koch curve.  Each of your Cantor sets corresponds to the image of some Cantor set in $[0,1]$ under $f$, and that Cantor set in $[0,1]$ will have Lebesgue measure $0$ if you've chosen your parametrization reasonably.  So, the union of all those Cantor sets in $[0,1]$ is still a null set (and a set of first category), so most points in $[0,1]$ aren't in any of those Cantor sets.
Note that you can also think of a point $x\in[0,1]$ that is not in any of those Cantor sets as a point such that the sequence $(f_n(x))$ never stabilizes but instead keeps moving infinitely often as you build the successive approximations.  The value of $f(x)$ will then be the limit of this sequence, a point of the Koch curve that was not in any of the approximations but instead is a limit of points in the approximations.