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The question: Can a closed simple 2D curve (by 2D curve, I mean a 1-dimensional continuum on a 2-dimensional plane) be a less-than-1-dimensional fractal (by dimension, I mean Hausdorff dimension) ?

I've searched through some famous fractals, none of them satisfy. Maybe there is a theorem that prevents the answers, but I'm too new to topology to notice.

Thanks in advance.

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    No. The topological dimension is one by assumption and the Hausdorff dimension always exceeds the topological dimension. This is proved for compact sets as theorem 6.3.11 in [Measure, Topology, and Fractal Geometry](http://www.springer.com/us/book/9780387747484) and more generally in section 3.1 of [Integral, Probability, and Fractal Measures](http://www.springer.com/us/book/9780387982052) – 2017-02-02
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    @MarkMcClure: By "exceeds" you mean $\ge$, of course. Just for the record, the theorem itself is due to L. Sznirelman. – 2017-02-02
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    @MoisheCohen Correct. – 2017-02-02

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