Is Koch snowflake a continuous curve? For Koch snowflake, does there exits a continuous map from $[0,1]$ to it?
The actural construction of the map may be impossible, but how to claim the existence of such a continuous map? Or can we conside the limit of a sequence of continuous map, but this sequence of continuous maps may not have continuous limit.
 A: Consider the snowflake curve as the limit of the curves $(\gamma_n)_{n\in \mathbb N}$, in the usual way, starting with $\gamma_0$ which is just a equilateral triangle of side length 1. Then each $\gamma_n$ is piecewise linear, consisting of $3\cdot 4^n$ pieces of length $3^{-n}$ each; for definiteness let us imagine that we parameterize it such that $|\gamma_n'(t)| = 3(\frac 43)^n$ whenever it exists.
Now, it always holds that $|\gamma_{n+1}(t)-\gamma_n(t)|\le 3^{-n}$ for every $t$ (because each step of the iteration just changes the curve between two corners in the existing curve, but keeps each corner and its corresponding parameter value unchanged). This means that the $\gamma_n$'s converge uniformly towards their pointwise limit: At every $t$ the distance between $\gamma_n(t)$  and $\lim_{i\to\infty}\gamma_i(t)$ is at most $\sum_{i=n}^\infty (1/3)^i$ which is independent of $t$ and goes to $0$ as $n\to\infty$.
Because uniform  convergence preserves continuity, the limiting curve is a continuous function from $[0,1]$ to the plane.
