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Related question: Are manifold subsets submanifolds?

Assume all manifolds, topological or smooth discussed here have dimensions and do not have boundary.

Let $A'$ and $B'$ be sets with $A' \subseteq B'$.

Question A: Are these correct?

  1. As far as I know, all sets can be made into topological spaces.

  2. By (1), make the sets $A'$ and $B'$ into, respectively, the topological spaces $A$ and $B$.

  3. I haven't thought about whether there are some sets that cannot be given topological spaces that enable them to become smooth or topological manifolds, but as far as I know, some topological spaces cannot be made into smooth manifolds or even topological manifolds...such as ones that aren't Hausdorff I guess.

  4. By (3), assume $A$ and $B$ from (2) can be made into smooth manifolds $(A,\mathscr A)$ and $(B,\mathscr B)$ where $\mathscr A$ and $\mathscr B$ are smooth atlases.

  5. By (4) and the above related question, $(A,\mathscr A)$ is not necessarily a (regular/an embedded) smooth submanifold of $(B,\mathscr B)$ or even an immersed smooth submanifold.

Question B: Does there exist a smooth atlas $\mathcal A$ where $(A,\mathcal A)$ becomes a smooth submanifold of $(B,\mathscr B)$?

  • I hope Question $B$ is equivalent to both

  • (C) "If a topological subspace can become a smooth manifold, then can it become a smooth submanifold?"

  • (D) "Can smooth manifold subsets $(N,\mathscr N)$ of smooth manifolds $(M, \mathscr M)$ always be made into smooth submanifolds $(N,\mathcal N)$ of $(M, \mathscr M)$?"

  • If one of (1)-(5) is wrong, then $B$, $C$ or $D$ could be meaningless or, if meaningful, not equivalent to the other meaningful ones. Please answer the meaningful ones among (B),(C) and (D), and please point out which are equivalent or not.

  • Not concerned about uniqueness at this point. You can say something about uniqueness if you want.

BCLC
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1 Answers1

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Your statements in Question A are all essentially correct. Whether or not you can put a manifold topology on a set depends only on the cardinality of the set. For sets with cardinality greater than that of $\mathbb R,$ the answer depends on your axioms:

  • If your definition of manifold includes second-countability, then any set strictly larger than $\mathbb R$ cannot be made a manifold.
  • If not, every set can be made a 0-dimensional manifold. (If you insist on positive dimension, I think by assuming the axiom of choice you can prove that all sets with cardinality at least that of $\mathbb R$ can be made manifolds.)

Regardless, I think this point is a tangent from your main concern:

Question B. You can answer this without needing to think about smooth structure at all: If $A$ has an atlas making it a smooth submanifold of $B,$ then $A$ is a topological submanifold of $B.$ This makes things much easier, since there is no arbitrary choice to be made when talking about topological submanifolds: a subset $A \subset B$ is a topological submanifold if and only the induced topology makes $A$ a topological manifold.

So, if $A$ has some atlas making it a smooth submanifold, then $A$ is a topological manifold in the induced topology. Thus any subset $A \subset B$ which is not a topological manifold in the induced topology (e.g. one that is not locally Euclidean) provides a counterexample.

Anthony Carapetis
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    First-countability is not relevant (it's immediate from being locally Euclidean), only second-countability is. The continuum hypothesis is also not relevant--it would be relevant only for cardinalities less than that of $\mathbb{R}$, not greater. – Eric Wofsey Jul 23 '19 at 15:05
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    @EricWofsey: thanks for the corrections. The CH remark was a hangover from before I remembered 0-dimensional manifolds exist and reframed that point with the >|R| condition. – Anthony Carapetis Jul 24 '19 at 00:16
  • Thanks Anthony Carapetis. Actually, I think I realized a way (or another way) my question is kind of wrong. I actually think I implicitly assume "From the atlases $\mathscr A$ and $\mathscr B$, we can already say whether or not $(A, \mathscr A)$ is or isn't a regular/an embedded submanifold of $(B, \mathscr B)$." Is this statement wrong (whether or not I actually implicitly assumed it)? –  Aug 01 '19 at 03:14
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    @SeleneAuckland: So long as $A$ is a subset of $B$, this statement is not wrong - it's just asking whether the inclusion map $A \to B$ is a smooth embedding when considered as a function between manifolds $(A,\mathscr A) \to (B,\mathscr B).$ – Anthony Carapetis Aug 01 '19 at 03:26
  • Anthony Carapetis thank you so much. Will analyze your answer later. –  Aug 01 '19 at 03:27
  • @EricWofsey and Anthony Carapetis, I analyzed. Not sure I understand. Counterexample to what? It sounds like you're answering the question "Does every $\mathcal A$ make $(A,\mathcal A)$ a submanifold of $(B,\mathscr B)$?" I know the answer is no. My question is if there necessarily exists some $\mathcal A$ that makes $(A,\mathcal A)$ a submanifold of $(B,\mathscr B)$. Edit: Oh I forgot to say topological or smooth. I'll edit the question now. –  Sep 11 '19 at 03:27
  • @SeleneAuckland: I mean a counterexample $A$ to the claim "for all $A$, there exists some $\mathcal A$ s.t. ...". If $A$ is not a topological submanifold, then every $\mathcal A$ will fail to make it a smooth submanifold. – Anthony Carapetis Sep 11 '19 at 03:33
  • @AnthonyCarapetis That should be part of answering question (A), I think. Anyway, question (B) is supposed to be as follows (I might have been unclear): Question (B) assumes there exists some $\mathscr A$ for $A$ and then asks if there is a $\mathcal A$ (again, given the existence of a $\mathscr A$). So some but not all topological subspaces $X$ can become topological manifolds or topological submanifolds, and thus, I ask about that $X$'s that can become. –  Sep 11 '19 at 03:34