So, pretty self-explanatory title: I was wondering wether if every point that the Koch curve has does necesarilly belong to one of its approximating polygons. Just in case, on the picture below, the first approximating polygons are shown.
My intuition says "yes" because the Koch curve is not space-filling, but I do not really trust my intuition with this kind of questions.
Thank you in advance!
The $n$th approximating polygon (as you have drawn it) can be described in the obvious way by a function $f_n:[0,1]\to\Bbb R^2$. And then we can define the Koch curve by $$f(x)=\lim_{n\to\infty}f_n(x)$$ The behaviour of $f_n(x)$ as $n\to\infty$ depends on the base-$4$ expansion of $x$. Specifically, $x$ belongs to an approximating polygon if and only if its base-$4$ expansion terminates.
So the set of points on the Koch curve which belong to an approximating polygon is countable.