Monday, 22 September 2008

gn.general topology - A question about unbounded connected subsets of the plane.

The following gives a partial answer: no such unbounded connected set may exist with the further assumption that it is closed. Actually, the argument generalizes for any locally compact metric space. I'm not completely sure that a much simpler or even trivial proof may exist, though.



Let Gamma be a closed unbounded connected subset of the plane. Let xinGamma and let B:=B(x,r) be an open ball around x. I claim that the connected component of x in GammacapbarB meets partialB, which shows that Gamma does contain non-trivial bounded connected subsets.



For any epsilon>0, consider the epsilon-neighborhood of Gamma, that is Gammaepsilon:=cupyinGammaB(y,epsilon). It is an open unbounded connected subset of the plane.
Let Uepsilon be the connected component of x in GammaepsiloncapB. Since the latter is locally connected, Uepsilon is both an open and closed subset of it in the relative topology. It is therefore an open subset of Gammaepsilon; however it is not closed in it, because Gammaepsilon is connected. Therefore barUepsilon is a closed connected set that meets partialB, and of course contains x. Since the set of all connected closed subsets of a compact metric space is compact in the Hausdorff distance, taking a limit as epsilonto0 we get a bounded connected subset of Gamma connecting x with partialB (this also passes to the limit).



Rmk One could state the above in terms of the one-point compactification of Gamma, and more generally for compact connected metric spaces. The trick of approximating a metric space with a locally connected metric space is made possible via the Kuratowski embedding (one defines Xepsilon as an epsilon nbd of X in the embedding).



PS: Of course the same affirmative conclusion holds, even more directely, if Gamma is assumed to be open, which is another case included in the original assumption of completely metrizable.

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