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MathematicsHow many components does the complement of a connected compact set in $\mathbb{R^2}$ have?
[+1] [2] Jill Clover
[2012-10-29 03:49:10]
[ general-topology ]
[ https://math.stackexchange.com/questions/223249/how-many-components-does-the-complement-of-a-connected-compact-set-in-mathbbr ]

How many components of complement of a compact and connected set in $\mathbb{R^2}$has?

Are there any tricks to solve the problem?

[+5] [2012-10-29 03:54:17] Brian M. Scott [ACCEPTED]

The complement of a compact, connected set in $\Bbb R^2$ can have any finite number of components, and it can also have countably infinitely many components. The complement of a line segment has one component. The complement of a circle has two. The complement of a figure $8$ has three. By adjoining more and more circles, you can get any finite number. Finally, if you take the union of circles of radius $\frac1n$ and centre $\left\langle\frac1n,0\right\rangle$, you get a compact, connected set whose complement has infinitely many components.


This Q appeared on my feed today. You didn't mention that the complement, being open in $\Bbb R^2$, cannot have uncountably many components. - DanielWainfleet
@DanielWainfleet: I enumerated the possible values. - Brian M. Scott
The reason why only at most countably many components can happen is that since $\mathbb{R}^2$ is locally connected, all components of open subsets of $\mathbb{R}^2$ are open, and since $\mathbb{R}^2$ is separable, the disjoint family of open sets admits an injection into a countable set. - Jakobian
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[0] [2012-10-29 03:52:33] MJD

It could be anything. Consider a closed disc with $n$ separated open discs deleted from it. Such a set is closed and bounded, hence compact; it is clearly connected. But its complement has $n+1$ connected components.


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