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Let X = compact connected set in $R^2$. Let $X^c$ be its complement.

Am I right to say this:

The number of connected components of $X^c$ roughly refers to the number of "holes" in $X$. So, I can make $X^c$ contain any arbitrary number of connected components by starting with a closed and bounded set, then deleting the same number of open sets from its interior and defining the resulting set as $X$?

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You can definitely make connected components that way. – Karolis Juodelė Jul 27 '12 at 15:44
Closed and bounded is the same as compact. You should start with a simply connected compact set, and then remove $n-1$ simply connected open subsets (the "outside" will also be a connected component). – Andrew Jul 27 '12 at 15:44
Sorry, I meant to say start with close + bounded + simply connected set (i.e. compact + simply connected). Thanks! I'll answer this question myself later. – Legendre Jul 27 '12 at 15:56
@Andrew I didn't really get it. So you mean, the remaining part after taking out $n-1$ simply connected open subsets is also connected, right? In that sense, we can have as many connected component in $X^c$ as we wish, right? Please correct me if I got it wrong thank you! – MatheMagic Oct 21 '14 at 18:45
up vote 1 down vote accepted

The answer is yes. The explanation is in the comments.

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