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Before posting this question,I just asked a similar question:a question about connected open sets in $R^2$. I got several nice answers.Now I want to ask:

Let $U$ be a nonempty connected open set in $\mathbb{R}^2$ and $U\not=\mathbb{R}^2$.I want to ask if there must exist an open ball $B\subset \mathbb{R}^2$ such that $B\not\subset U$ and $B\cap U$ is a nonempty connected open set.

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  • $\begingroup$ The answer provided by @S. Carnahan to your original question seems to come close to answering the present modified version. At least it seems you need that $U$ is fractal (or has fractal boundary). $\endgroup$
    – Mirko
    Commented Nov 27, 2014 at 15:56
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    $\begingroup$ I do not think S. Carnahan's answer can solve my question! $\endgroup$
    – user173856
    Commented Nov 27, 2014 at 16:02
  • $\begingroup$ You know better than me what your question is, yet, the way I think of it is different: Does there exist $x$ in the boundary of $U$ such that for every $r>0$ the ball $B(x,r)$ intersects $U$ in a connected set? The answer to your question would be trivially yes if $U$ is bounded (just take a large enough ball containing $U$), but on the other hand it doesn't feel like boundedness should have anything to do with this question. Or, you may ask if there are $x\in Bd(U)$ such that for every $\epsilon$ there is $\delta$ with $U\cap B(x,\delta)$ connected. I don't know but might be edited some way. $\endgroup$
    – Mirko
    Commented Nov 27, 2014 at 16:21

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The answer is no. It is sufficient to find a simple curve from the origin to infinity with the property that every circle intersects it at least twice. The region will be the complement of this curve.

Let us begin with the graph of $y=\sqrt{x}\sin(1/x), x\geq 0$. This is a simple curve, and its complement is connected. The complement almost has the desired property: most circles intersect this graph at least twice, so the intersection of the corresponding discs with our region are not connected.

It is true that there are circles that intersect our graph only once: these are circles of sufficiently large radius which look almost line vertical lines near the places where they intersect our graph. But it is easy to modify our curve so that all circles (and all straight lines) with intersect it at least twice. For this we arrange appropriate zig-zag's on our graph near the $x$-axis.

If this is not clear enough I will scan a picture after the holidays.

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    $\begingroup$ Squashing the spiral $r=e^\theta$ in the $y$-direction yields even more elegant realization of the same idea. $\endgroup$
    – fedja
    Commented Nov 27, 2014 at 18:37
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Yet worse example: take $U$ to be the complement of a pseudo-arc $P$ in the unit sphere. For any open disc $D$, such that $P \not\subset \bar D$ and $D\not\subset U$ we have $D\cap U$ is not connected.

Take a point in $P$ as the north pole and consider the stereographic projection of $U$ to the plane. For the obtained set $U'$ and any open topological disc $D$ such that $D\not\subset U'$ the set $D\cap U'$ is not connected.

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