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Obviously a Weierstrass curve is not a smooth manifold, but it seems like a Weierstrass curve should be a topological manifold (which I now see is a suspicion supported by this post), since it is a continuous image of the real line.

That said, I am a particularly concerned about the locally Euclidean requirement. I know that a Weiestrass curve generally has a non-integer Hausdorff dimension (see here)--does that impact this requirement?


marked as duplicate by Moishe Kohan, José Carlos Santos general-topology Apr 30 at 8:51

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A Weierstraß curve $W \subset \mathbb R ^2$ is the graph $G(w) = \{ (x,w(x) \mid \in J \}$ of a Weierstraß function $w : J \to \mathbb R$ (which is continuous but nowhere differentiable). Here $J$ is an interval. The map $i : J \to G(w), i(x) = (x,w(x))$, is a homeomorphism. This is true for any graph of any continuous function $f : X \to Y$ between toplogical spaces.

Therefore $G(w)$ is a topological manifold. It is also a topological submanifold of $\mathbb R ^2$. See my answer to The graph of $f:\mathbb R^2\to \mathbb R$ can be embbeded in $\mathbb R^2$ or only in $\mathbb R^3$?.

Moreover, it can be given the structure of a differentiable manifold: Simply take the single chart $i^{-1} : G(w) \to J$. However, it is not a differentiable submanifold of $\mathbb R ^2$.

Hausdorff dimension is not related to your question. It is not invariant under homeomorphisms. See Is the Hausdorff dimension invariant under homeomorphisms?.

  • $\begingroup$ Ah, very good. I especially appreciate your comments about differentiability and the Hausdorff dimension. $\endgroup$ – AegisCruiser Apr 29 at 18:27

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