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I believe the following to be true, but am worried my intuition does not account for fractally things:

Let $K\subset\bar{\mathbb{C}}$ ($\bar{\mathbb{C}}$ being the Riemann sphere) be closed (thus compact) and suppose that $K$ and $K^c$ are both connected (thus $K$ is the complement of an open simply connected sub-set of $\bar{\mathbb{C}}$), and let $U\subset\mathbb{R}^2$ be any open set containing $K$. Then there is a list of balls $B_1,\ldots,B_N\subset U$ such that $\displaystyle K\subset\bigcup_{i=1}^NB_i$ and $\displaystyle\bigcup_{i=1}^NB_i$ is simply connected.

I thought this would be straight forward but am not so sure now.

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    $\begingroup$ From your title, I assume you mean $K$ is simply connected and compact? $\endgroup$ – manthanomen Nov 26 '13 at 2:19
  • $\begingroup$ @manthanomen Thanks, of course you are right. $\endgroup$ – Trevor J Richards Nov 26 '13 at 2:39
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    $\begingroup$ Thanks to John for pointing out my mistake about simple-connectedness. I have replaced "simply-connected" with "$K$ and $K^c$ are both connected". $\endgroup$ – Trevor J Richards Nov 26 '13 at 5:47
  • $\begingroup$ May be it can be proved that $K$ is contractible and there is a list of balls $B_1,\ldots,B_N\subset U$ such that $\displaystyle K\subset\bigcup_{i=1}^NB_i$ and $\displaystyle\bigcup_{i=1}^NB_i$ admits a deformation retraction onto $K$. $\endgroup$ – Alex Ravsky Nov 26 '13 at 6:55
  • $\begingroup$ @Alex Ravsky, Perhaps so. I noticed that the restriction on $K$ is equivalent to $K$ being the complement of an open simply connected subset of the Riemann sphere. Is it the case that the complement of an simply connected open set in $S^2$ is contractible? $\endgroup$ – Trevor J Richards Nov 27 '13 at 2:04
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I figured it out; we do it in two steps.

Assume that $K$ is bounded. First cover $K$ in finitely many balls $B_1,B_2,\ldots,B_N$, all contained in $U$ (easy application of compactness and openness). It is easy to see that a union of finitely many balls has at most finitely many bounded components of its complement. Let $D$ be some bounded component of

$$\left(\displaystyle\bigcup_{i=1}^NB_i\right)^c.$$

Draw a path from some point in $D$ to $\infty$ which does not intersect $K$. In the Riemann sphere, this path is compact, so we may cover this path with finitely many balls $C_1,\ldots,C_M$. Define

$$K'=\left(\displaystyle\bigcup_{i=1}^NB_i\right)\setminus\left(\bigcup_{i=1}^MC_i\right).$$

Do this for each of the finitely many bounded components of $$\left(\displaystyle\bigcup_{i=1}^NB_i\right)^c.$$

Then we obtain a new set $K'\subset U$ which is the union of finitely many balls minus finitely many balls which has the same properties as $K$, namely that $K'$ is compact and both $K'$ and ${K'}^c$ are connected. Note also that the boundary of $K'$ consists of finitely many smooth pieces.

Now that $\partial K'$ is piecewise smooth, with finitely many "pieces", it is easy to show that this $K'$ has the desired property, and whichever cover works for $K'$ will also work for $K$.

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