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So it is weekend! and I am reading a nice book, "The Poincaré conjecture", written by a mathematician (Donal O'Shea, topologist). The book introduces step by step basic concepts of Topology, and talks about the mathematical advances since Poincaré stated the conjecture up to the moment it was demonstrated, providing a glimpse of the solution that Perelman gave to the conjecture. It is not technical, narrative, a very light reading.

My question is about the following: when the author defines a manifold, states that "if a manifold has a boundary, its boundary will be one dimension lower".

Then I thought about fractals, for instance: can a fractal be a manifold? (I understand that manifold means the same as "surface" in this context) and then, if the answer is yes, then if it has boundaries, are the boundaries strictly one dimension lower?

My beginner doubt is basically if the statement of the author would be still valid in that case. Maybe I am confusing the "fractal dimension" concept with the generic "dimension" concept.

Thank you!

UPDATE 2015/08/28

There is a very nice follow-up on this question, with an interesting explanation as well by another user here (link).

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iadvd
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If you mean a set whose fractional Hausdorff measure is finite and non-zero, then this will not be a manifold. But it will have a topological boundary. That won't be a manifold boundary because it isn't a manifold. Not in the usual sense of the word "manifold" anyway.

Alan U. Kennington
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    It would be nice if you could clarify why a set with fractional Hausdorff measure cannot be a manifold, e.g. why it cannot be locally homeomorphic to $\Bbb{R}^n$. – A.P. Jun 27 '15 at 09:47
  • @A.P. it also seems unclear what the correct analog of $\mathbb R^n$ might be for $n\notin\mathbb N_0$, if there is one. – Dejan Govc Jun 27 '15 at 10:36
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    @DejanGovc That's exactly why Kennington is saying that a fractal isn't a manifold in the usual sense. It may be possible to properly generalise the notion of manifold to cover fractals, although I doubt that this is the case. – A.P. Jun 27 '15 at 10:43
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    The concept of topological dimension is what you would look for if you wanted to know why a fractional dimensional set can't be a manifold. First, it is known that all manifolds are of integer dimension. Second, it is known that there is one and only one Hausdorff dimension for which the set's measure is non-zero and non-infinite. (This is a kind of a deep analysis kind of thing in geometric measure theory.) Consequently a manifold must have integral dimension, not fractional. It can't be both. – Alan U. Kennington Jun 27 '15 at 10:51
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    @AlanU.Kennington thank you for the explanation, I did not know the restriction of manifolds being of integer dimension. Please just one more beginner question: can a fractal contain a manifold? for instance I was thinking about a fractal that is a normal surface in the XY-plane but fractalic only in Z. – iadvd Jun 27 '15 at 11:59
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    @iadvd I guess you mean a smooth manifold, which is certainly a common assumption. Otherwise, the graph of a nowhere differentiable function can be both a manifold and a fractal. The boundary of the Mandelbrot set contains a line segment, a circle, and a cardioid. So I guess a fractal can contain a smooth manifold. – Mark McClure Jun 27 '15 at 12:09
  • @MarkMcClure I see!! thank you for the explanation. – iadvd Jun 27 '15 at 12:14
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    @MarkMcClure Why do you say that the boundary $B$ of the Mandelbrot set contains, e.g., a cardioid $C$? I see that you can (somewhat roughly) approximate part of $B$ with $C$, but why must $C$ be actually contained in $B$? It sounds to me like saying that the Koch curve contains a six-pointed star. – A.P. Jun 27 '15 at 12:53
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    @A.P. It can, in fact, be proved that every complex parameter on the cardioid $c=e^{i t} \left(2-e^{i t}\right)/4$ is on the boundary of the Mandelbrot set. I can't find a convenient pointer to that fact at the moment and proving it would take us a bit too far astray from this question. Perhaps you could raise the question on the site? I'd answer it this evening, if no one else has by then. – Mark McClure Jun 27 '15 at 13:20
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    @MarkMcClure Thanks for your offer. I've asked the question. – A.P. Jun 27 '15 at 14:09
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    @A.P. I am enjoying and learning from you guys, thanks again. – iadvd Jun 27 '15 at 14:15
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    The way I understood the question was that the phrase "be a manifold" means that the set is a regular submanifold of some Euclidean space, and that could be a $C^k$ embedding with $k\ge0$, not necessarily $k>0$. Then locally the set must be the graph of a continuous function. The embedding map must also be injective. Even if you forget regularity of the embedding, you get preservation of the topological dimension, which you don't get for the Peano curve, for example, which is not injective. So these technical details need to be made more precise. – Alan U. Kennington Jun 27 '15 at 14:24
  • can a fractal be a manifold locally! – Erik Kaplun Sep 06 '20 at 08:36