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If we have path-connected spaces $A_0 \supseteq A_1 \supseteq A_2 \supseteq \ldots$, is $\bigcap^\infty A_i$ path-connected?

I was thinking that if we take $A_i$ to be a $1/i$-neighborhood of the Koch snowflake $K$, then all the $A_i$ are path-connected and their intersection is $K$...but it seems clear from Is Koch snowflake a continuous curve? that $K$ is a path in the plane and is therefore path-connected. So that doesn't help.

I'm especially interested in subsets of $\mathbb{R}^n$. This question was inspired by Partitions of $\mathbb{R}^2$ into disjoint, connected, dense subsets.

I'm also curious about the same question with "connected" instead of "path-connected".

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2 Answers 2

up vote 4 down vote accepted

No. For $n \in \mathbb N$ let $$ K_n = \left\{x\in \mathbb R^2 \mid x_1 \ge 0, \left|x - \left(\frac 1n, 0\right)\right| = \sqrt{1 + \frac 1n^2} \right\} $$ and $A_i = \bigcup_{n \ge i} K_n$. As we have $K_j \cap K_k = \{(0, \pm 1)\}$ for $j \ne k$, we have $\bigcap_i A_i = \bigcap_n K_n = \{(0, \pm 1)\}$ and so the intersection isn't path connected (it isn't even connected).

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@BrianM.Scott Thanks for spotting my typo. –  martini Nov 7 '12 at 17:00
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It might help the reader to say that $K_n$ is the circular arc in the right half-plane with endpoints $\langle 0,\pm1\rangle$ and centre $\langle\frac1n,0\rangle$; it took me a minute to wade through the definition. Nice example. –  Brian M. Scott Nov 7 '12 at 17:03
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Yes, thank you. We could also define $K_n$ as the broken line defined by the sequence $(0, 1)$, $(1/n, 1)$, $(1/n, -1)$, $(0, -1)$. –  Hew Wolff Nov 7 '12 at 17:24
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For posterity, here is a nice variation on @martini's answer: $A_i = \{(0, 0), (0, 1)\} \cup (0, 1/i) \times (0, 1) \subset \mathbb{R}^2$.

It's easy to see that the $A_i$ are path-connected: they're all homeomorphic to an open disk plus two boundary points. But their intersection is just those two points.

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