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Find an example of a sequence $V_n$ of closed and connected subsets of the Euclidean plane satisfying $V_n \supseteq V_{n+1}$ such that $\cap_{i=1}^{\infty}V_i$ is not connected.

Normally I would show how far I got on my own but I can't figure this one out, I have been thinking of different pictures in my head to come up with an idea but I can't. If anyone could give me a tip or an example that would be great. Thanks in advance.

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Consider two disjoint half-planes connected by a "bridge" that goes each time farther towards infinity, something like $$V_n = \{ (x,y) \in \mathbb{R}^2 : |x| \ge 1 \} \cup \mathbb{R} \times [n, + \infty).$$

Then $V_n$ is closed (union of two closed sets) and connected (it's even path connected), but $\bigcap_n V_n$ is the disjoint union of two half planes and is disconnected.

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    $\begingroup$ Or a stair that goes infinitely high. Go removing one step at a time upwards... :-) $\endgroup$ – André Caldas Apr 2 '14 at 14:23
  • $\begingroup$ @nik Thanks for your answer. So we would have something like $V_n = \{ (x,y)\in \mathbb R^n: |x| \geq 1 \}\cup \{ (x,y) \in \mathbb R^2 : |x| \leq 1, y\geq n \}$? The first set being the two closed sides of the plane and the last set being the bridge? $\endgroup$ – Slugger Apr 2 '14 at 14:30
  • $\begingroup$ @Slugger: Yes, that's exactly what I had in mind. Notice by the way that the sets in question aren't compact. $\endgroup$ – Najib Idrissi Apr 2 '14 at 14:31
  • $\begingroup$ Thanks a lot, that's a nice example. @Andre Caldas, If I understand correctly, wouldn't you end up with an the empty set if you take the infinite intersection? $\endgroup$ – Slugger Apr 2 '14 at 14:37
  • $\begingroup$ @Slugger: Nope, you would get this: gartic.uol.com.br/yelle/desenho-jogo/perna-de-pau in Portuguese we call it "perna de pau". Literally, "pegleg". But I guess this is not called "pegleg" in English... $\endgroup$ – André Caldas Apr 2 '14 at 14:54

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