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Must the complement of the countable union of disjoint arcs (in the plane) be connected?

A (Jordan) arc is the injective image of a continuous function from $[0,1]$ to $\Bbb R^2$. It is known that, if I have finitely many pairwise disjoint arcs, the complement of their union must be connected, and even path-connected. However, this no longer is true when we allow ourselves infinitely many arcs (take the union of line segments of the form $\{x\}\times[0,1]$ for $x\in\Bbb R$).

In fact, the "path-connected" part is not even true for countably many arcs!

I know three counterexamples (special thanks to Balarka and Alessandro):

  • Let $U_r$ be a closed circular arc of $270$ degrees, with the $90^\circ$ opening facing up, of radius $r$. Let $D_r$ be the same but with the opening facing down. Then the complement of: $$U_{1/2}\cup D_{2/3}\cup U_{3/4}\cup D_{4/5}\cup\dotsb$$ is not path-connected.
  • The complement of: $$\bigcup_{n=1}^\infty(\{1/n\}\times[-n,n])$$ is not path-connected. This gives an example made of only straight lines, though it's unbounded.
  • Let $S_1=[-1,1]\times\{1\}$ and $S_{-1}=[-1,1]\times\{-1\}$ be the top and bottom sides of a square. Let $T_{-1,n}=\{-1-\frac1n\}\times[-1,1]$ and $T_{1,n}=\{1+\frac1n\}\times[-1,1]$ be sequences of segments approaching the left and right sides of the same square. Then the complement of: $$S_{-1}\cup S_1\cup\bigcup_{n=1}^\infty T_{-1,n}\cup\bigcup_{n=1}^\infty T_{1,n}$$ is not path-connected. This gives an example that is bounded, is made of straight lines, and has all of its lines the same length.

However, while these complements are all not path-connected, they all are connected. Hence the question: Must the complement of countably many disjoint arcs be connected?

I've thought about this all day and I haven't been able to solve it. Homology seems useless, since it can only measure path components, not connected components. On the other hand, if we replace $\Bbb R^2$ with other spaces, it can become false (take for example the plane minus open disks of radius $\frac13$ centered at the integers on the $x$-axis), so the proof must use some special property of $\Bbb R^2$.

Akiva Weinberger
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  • http://math.stackexchange.com/questions/2103193/if-we-remove-a-countable-family-of-simple-non-closed-curves-which-are-mutually?noredirect=1&lq=1 – Moishe Kohan Feb 24 '17 at 07:03
  • And another one, essentially equivalent question: http://math.stackexchange.com/questions/1130572/connectedness-of-the-complement-of-the-countable-union-of-closed-jordan-regions – Moishe Kohan Feb 24 '17 at 07:09

2 Answers2

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Connectedness should follow from Sitnikov Duality theorem if you consider subsets $A$ (union of arcs) of $S^2$ rather than $R^2$: $$ 0=H^c_1(A)\cong \check{H}^0(S^2 -A). $$ It follows that $S^2-A$ is connected. I am still thinking about subsets of $R^2$.

Moishe Kohan
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  • OK, but homology measures path components, not connected components: http://math.stackexchange.com/questions/317882/is-there-a-homology-theory-that-counts-connected-components-of-a-space. – Mike F Feb 24 '17 at 21:58
  • That link says that Čech homology measures quasicomponents, which I think is the same as connected components for spaces like $\Bbb R^n$. So that might be what's used here. – Akiva Weinberger Feb 24 '17 at 22:19
  • @AkivaWeinberger can you explain what you mean by "measures quasicomponents" or provide a source? thanks! – Forever Mozart Feb 25 '17 at 01:18
  • @ForeverMozart: The business with quasicomponents is not really relevant. The relevant thing is that is a space is not connected then 0-th reduced Chech cohomology group is nonzero, which pretty much follows from the definition. – Moishe Kohan Feb 25 '17 at 02:33
  • @MikeF: What homology? In any case, it is Sitnikov compactly supported homology we are talking about, not singular homology. – Moishe Kohan Feb 25 '17 at 02:35
  • @AkivaWeinberger: The business with quasicomponents is not really relevant. The relevant thing is that is a space is not connected then 0-th reduced Chech cohomology group is nonzero, which pretty much follows from the definition. – Moishe Kohan Feb 25 '17 at 02:36
  • @ForeverMozart I was quoting the link that Mike F posted – Akiva Weinberger Feb 26 '17 at 00:23
  • Could you point me to some resources to help me understand this answer? I only know singular and simplicial homology. – Akiva Weinberger Feb 27 '17 at 15:38
  • @AkivaWeinberger: Massey's book, not the standard one but the one mentioned in the link. But this is a hard read. – Moishe Kohan Feb 27 '17 at 16:02
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I think if we take $\gamma_{n}:[-1,1]\rightarrow\mathbb{R}^2,\ \gamma_{n}(t)=(1/n,nt)$ we have the desired set.

Veridian Dynamics
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