# Is a Weierstrass curve a topological manifold? [duplicate]

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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?

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## 1 Answer

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?.

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