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.
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9If the convergence is uniform, then a limit of continuous functions is continuous. – T. Eskin Jun 05 '12 at 19:54
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Is the range of your map going to be the snowflake curve itself, or the area inside the snowflake curve? In the former case, the usual definition of the snowflake curve directly suggests a concrete continuous parameterization of if. In the latter, the various well-known space-filling curves can be adapted to produce a continuous map from $[0,1]$ onto the interior of the snowflake. – hmakholm left over Monica Jun 05 '12 at 19:57
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1A space is a continuous image of $[0,1]$ if and only if it is compact, connected, locally connected, and second-countable. Hahn and Mazurkiewitz, http://en.wikipedia.org/wiki/Space-filling_curve#The_Hahn.E2.80.93Mazurkiewicz_theorem – Jun 05 '12 at 20:03
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@Henning, I mean the snowflake curve itself, not the interior domain. So you suggest that map is the limit of the sequence of continuous parametrizations in the construction of the snowflake? – Sun Jun 05 '12 at 20:20
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So every compact connected manifold is a continuous image of $[0,1]$? Cool. – Jesse Madnick Jun 06 '12 at 00:51
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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. – t.b. Jun 06 '12 at 07:15
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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.
hmakholm left over Monica
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@Leo: obviously not; there are many points in $\mathbb R^2$ that are not on the snowflake curve. – hmakholm left over Monica May 27 '14 at 02:26
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I mean that if we restrict the range to the Koch curve it becomes subjective. Is the restricted function injective? I'm asking in the light of this question. – Leo May 27 '14 at 02:43
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Would the continuity be enough for us to conclude that the topological dimension of the Koch curve is one? – Leo Jun 01 '14 at 00:01
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@Leo: No, we also need that it is injective. (Counterexample: the Peano curve). But since $\gamma$ is a continuous injection from a compact space to a Hausdorff, it is a homeomorphism from $[0,1]$ to its image. The topological dimension of the image is therefore the same as that of $[0,1]$. – hmakholm left over Monica Jun 01 '14 at 00:32
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What exactly do you mean by "counterexample"? Since you brought up the Peano curve, I must mention that it's a source of major confusion for me: see the question What is the topological dimension of the Peano curve?. – Leo Jun 01 '14 at 00:40
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@Leo: I'm assuming you mean the topological dimension of the set of points that belong to the curve. In that case the Peano curve is an example of a curve that is continuous but whose image has topological dimension $\ne1$. – hmakholm left over Monica Jun 01 '14 at 00:45
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That is exactly what I thought it means but people say 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. – Leo Jun 01 '14 at 00:52