TITLE: Finding simply connected open sets between compact ones and general open ones in $\mathbb R^2$.
QUESTION [18 upvotes]: In a paper  I am reading (not a published one), the following is considered obvious:
Let $K$ be a compact and connected subset of $\,\mathbb R^2$, with $\mathbb R^2\smallsetminus K$ connected, and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open  $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open  $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.
I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries. Any ideas?

REPLY [2 votes]: I want to go at it with complex analysis. Probably overkill, but potentially quite short. I'll use two facts:

Any open subset $X \subset \overline{\mathbb{C}}$ is simply connected if and only if both $X$ and its complement are connected.
Let $X \subset \overline{\mathbb{C}}$ be open and simply connected in $\overline{\mathbb{C}}$. If its complement has at least two points, then there is a biholomorphism $g : X \to B(0,1)$ (Riemann mapping theorem).

If $K$ is a point, then the result is obvious (take a small ball centered around $K$). Otherwise, we can see $K$ as a connected compact subset of the Riemann sphere $\overline{\mathbb{C}}$. Then $\overline{\mathbb C} \setminus K$ is a connected open subset of the Riemann sphere. By the first fact, $\overline{\mathbb C} \setminus K$ is simply connected. By the second fact, and since $K$ has at least two points, there is a biholomorphism $g : \overline{\mathbb C} \setminus K \mapsto B(0,1)$.
Since $\overline{\mathbb{C}} \setminus U$ is compact in $\overline{\mathbb C} \setminus K$, its image by $g$ is compact in $B(0,1)$. Hence, there exists $\varepsilon > 0$ such that $g(\overline{\mathbb{C}} \setminus U) \subset \overline{B}(0,1-\varepsilon)$.
Take $V := g^{-1} \bigl( \overline{B}(0,1-\varepsilon)^c\bigr) \cup K$, and delete the infinity if it happens to be in this set. Then $\overline{\mathbb C} \setminus V$ is biholomorphic (with $g$) to the disk $\overline B(0,1-\varepsilon)$, which ensures that is is connected (and stays connected if you have to delete the infinity).
The motivation of this proof is that, with the conformal mapping theorem, I can make the complementary of $K$ look like a disk, which allows me to shrink it a little without having to worry about potential pathologies.