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I know that all subsets $S$ of a manifold $M$ are not a submanifold of $M$, but can it always be given a structure of differencial manifold ?

I feel like yes, taking the induced topology and the restriction to $S$ of charts from $M$. Am I right ?

Johny06
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    Do you know any examples of topological spaces which aren't manifolds? Can any of those be regarded as subsets of a manifold with the induced topology? – Michael Albanese Sep 08 '22 at 13:21
  • For exemple a cone, which is a subset of a manifold. So the answer is finally no. Thank you. – Johny06 Sep 08 '22 at 13:29
  • Welcome to Math.SE! <> Does the first sentence mean "Not every subset $S$ of a manifold $M$ is a submanifold"? (As written it suggests "Every subset $S$ of a manifold is not a submanifold," which is not true.) Second, "can be given a structure of a differentiable manifold" is vague and therefore ambiguous; can you say precisely what you mean? – Andrew D. Hwang Sep 08 '22 at 14:06
  • Thanks. Yes, I meant "All subsets of M are not some submanifolds of M". For the second part, it mean be able to find a maximal atlas and a topology for which it is hausdorff and at countable basis. – Johny06 Sep 08 '22 at 14:35
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    If the Continuum hypothesis is true, then any (infinite) subset of a $n$-dimensional manifold is either countable or in bijection with $\Bbb R$. In the latter case, it follows that we could endow them with the structure of a manifold by pulling back the structure of $\Bbb R$ thanks to the above mentioned bijection. This does not say much about the subset and its inclusion in the original manifold though (e.g, the topology built from the pullback will almost never be the induced topology as a subset) – Didier Sep 08 '22 at 15:35
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    With the induced topology, this is rarely true. For example, the (connected) union of a filled-rectangle and a segment in the plane (like $(0,1)\times(0,1)\cup (1,2)\times{0}$), with the induced topology, cannot be endowed with any manifold structure since "it cannot have a dimension". – Didier Sep 08 '22 at 15:38
  • That is interesting. That mean that for any such subset, we can't prove that there exist no topology such that it is a 1-dimensional manifold ? – Johny06 Sep 08 '22 at 15:57
  • I thought that the cone, for exemple, was not a manifold for any topology, that sound false now. – Johny06 Sep 08 '22 at 16:06

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