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This question seems quite classical, but I don't quite know what subarea of topology it falls into.

Suppose that removing the set $S$ disconnects the 2-torus $\mathbb{T}^2 = \mathbb{R}^2\diagup\mathbb{Z}^2$, i.e. $\mathbb{T}^2$ can be written as the pairwise disjoint union of $S$ and two open sets $U, V$. Must $S$ be `at least one-dimensional' in some sense?

As a specific question, must $S$ contain a nontrivial (not a singleton) continuous curve?

I phrased the question for manifolds because I imagine any answer probably holds in greater generality, but I would be happy even with any info for this specific case.

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    $\begingroup$ Are there any background assumptions on the set $S$? Is it locally closed, for instance, or just anything? $\endgroup$ Commented Dec 5 at 23:00
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    $\begingroup$ The type of argument I have in mind is that if $Z \subseteq X$ is closed inside a connected $n$-dimensional manifold $X$ such that $U = X \setminus Z$ is not connected, then the long exact sequence for (sheaf) cohomology with compact support gives $\ldots \to H^{n-1}_c(Z,\mathbf Z) \to H^n_c(U,\mathbf Z) \to H^n_c(X,\mathbf Z)$, forcing $H^{n-1}_c(Z,\mathbf Z) \neq 0$. So $\mathrm{covdim} Z \geq n-1$. $\endgroup$ Commented Dec 5 at 23:13
  • $\begingroup$ So in the case I care about, it is actually closed and nowhere dense. If that helps, I'm interested in a solution in this case. But somehow I thought the fact should be true more generally! $\endgroup$ Commented Dec 5 at 23:14
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    $\begingroup$ After looking it up in google, it seems indeed there is this well-known construction, the pseudo-arc, which can disconnect $\mathbb{R}^2$ (or $\mathbb{T}^2$) but is totally path-disconnected $\endgroup$
    – Saúl RM
    Commented Dec 8 at 18:46
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    $\begingroup$ Thanks! You might have missed my comment below (on the solution itself) where i mentioned that the pseudoarc was a possible counterexample. But I didn't have an explicit reference that it could disconnect R^2, thank you for that! $\endgroup$ Commented Dec 9 at 1:16

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Let my write up my comment as an answer. The easiest argument I know uses sheaf cohomology. It requires working through a bit of theory, but ultimately gives a very flexible tool for proving all sorts of things about topological spaces.

Half of this argument could be carried out in singular or de Rham cohomology as well, at least if $Z$ is a CW complex or a submanifold, respectively. But I don't know how to do the comparison between cohomological dimension and covering dimension that way — this ultimately uses Čech cohomology. Also, sheaf cohomology (resp. compactly supported sheaf cohomology) is well-behaved on arbitrary paracompact (resp. locally compact) Hausdorff spaces, so it's exactly suited for this situation where we don't know that much about $Z$.

Lemma. If $X$ is a connected $n$-manifold and $Z \subseteq X$ is closed such that $U = X \setminus Z$ is disconnected, then $\operatorname{covdim} Z \geq n-1$.

Since $X$ and therefore $Z$ is metrisable, I think the covering dimension coincides with the inductive dimension (which I know little about).

Proof. The sequence $0 \to \mathbf Z_U \to \mathbf Z_X \to \mathbf Z_Z \to 0$ of sheaves on $X$ gives a long exact sequence $$\ldots \to H^{n-1}_c(Z,\mathbf Z) \to H^n_c(U,\mathbf Z) \to H^n_c(X,\mathbf Z);$$ see for instance [Iversen, Ch. III, 7.6]. By Poincaré duality, the latter two are free $\mathbf Z$-modules whose ranks are given by the number of connected components of $U$ and $X$ respectively. The hypothesis $\operatorname{rk} H^n_c(U,\mathbf Z) > \operatorname{rk} H^n_c(X,\mathbf Z)$ therefore forces $H^{n-1}_c(Z,\mathbf Z) \neq 0$.

Since $Z$ is paracompact, it is explained in [Bredon, Ch. II, discussion on p. 122 after Cor. 16.34] that $\operatorname{covdim}(Z)$ agrees with $\dim_{\mathbf Z}(Z)$ (see definition [Bredon, Def. 16.6]). Since $Z$ is paracompact, the preceding proposition 16.5 shows that $\dim_{\mathbf Z}(Z) = \dim_{\mathrm{cld},\mathbf Z}(Z)$ is the 'usual' sheaf-theoretic dimension, i.e. the minimal $k$ such that $H^{k+1}(Z,\mathscr F) = 0$ for all sheaves of $\mathbf Z$-modules $\mathscr F$ on $Z$. Moreover, [Bredon, Cor. 16.33(a)$\Leftrightarrow$(b)] shows that then also $H^{k+1}_c(Z,\mathscr F) = 0$ for all sheaves of $\mathbf Z$-modules $\mathscr F$ on $Z$. Since $H^{n-1}_c(X,\mathbf Z) \neq 0$, we conclude that $\dim_{\mathbf Z}(Z) \geq n-1$. $\square$


References.

[Bredon] G. E. Bredon, Sheaf theory. Graduate Texts in Mathematics 170. Springer, New York, 1997. ZBL0874.55001.

[Iversen] B. Iversen, Cohomology of sheaves. Universitext. Springer, Berlin, 1986. ZBL1272.55001.

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  • $\begingroup$ I can't possibly vet all of the cohomology, but it seems reasonable, thank you! Can I ask: do you know if closed + covdim 1 implies containing a nontrivial continuous curve? I tried to read up a bit but haven't found answer. $\endgroup$ Commented Dec 6 at 19:04
  • $\begingroup$ @RonniePavlov Although the final statement you're after sounds plausible, I'm not sure I believe that this should follow just by knowing the covering dimension. Point set topology is not at all my expertise, and topological spaces can be pretty weird. $\endgroup$ Commented Dec 6 at 21:28
  • $\begingroup$ No prob, thanks for your help thus far! I read a bit of Engleking's "Dimension Theory" long ago and recall it being helpful, I guess it's back to that to read a bit. $\endgroup$ Commented Dec 6 at 21:33
  • $\begingroup$ As a nonexpert I could be misreading, but it looks like (a) a space of positive inductive dimension must contain a nontrivial continuum, i.e. a connected compact metric space. (b) However, a continuum need not contain a nontrivial continuous image of [0,1] (which I guess is called an arc); something called the pseudoarc is a counterexample. The psuedoarc sounds a little like what Saul RM was suggesting a construction for in a comment above. General point-set topology is indeed weird! $\endgroup$ Commented Dec 6 at 22:33

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