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.
Is Koch snowflake a continuous curve?
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3@Jesse: That's much easier than Hahn-Mazurkiewicz since you're locally Euclidean, not only locally connected. Cover your manifold with finitely many images of closed balls in $\mathbb{R}^n$. For each closed ball take a Peano curve. Link the Peano curves together. – 2012-06-06
1 Answers
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.
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0That is exactly what I thought it means but [people say](http://math.stackexchange.com/a/816466/51387) that the topological dimension of the Peano curve is one (and I've seen a few papers mentioning the same). Which can only imply that they mean the graph of the curve (and not the image). But then continuity would be enough to conclude that the topological dimension of the Koch curve is one. – 2014-06-01