Is the "Runge hull" of an open set open?

Question

Let $X\subset\mathbb{C}$. The ''Runge hull'' $h(X)$ of $X$ is defined to be the the union of $X$ and all the precompact connected components of $\mathbb{C}\setminus X$.

If $X$ is open, is $h(X)$ necessarily open?

Alternatively, this can be stated as follows. Is the union of all unbounded connected components of a closed set in $\mathbb{C}$ closed? The set $\mathbb{C}$ may also be replaced by an arbitrary manifold.

The terminology comes from Lectures in Riemann Surfaces by O. Forster, although he did not call it explicitly call it the ''Runge hull''. (Also, I noticed that this is not the standard definition, hence the quotation marks.) In the book, it was shown that if $X$ is closed or compact, so is $h(X)$. However, the author did not say anything about the openness of $h(X)$. After thinking for a while, I could netiher give a proof nor construct a counterexample.

Thanks in advance!

Answer 1

Yes; this is true more generally for any locally compact Hausdorff space. It follows easily from the following lemma.

Lemma: Let $Y$ be a locally compact Hausdorff space, let $C\subseteq Y$ be a connected component which is compact, and let $U$ be an open neighborhood of $C$. Then there is a clopen set $V\subseteq Y$ such that $C\subseteq V\subseteq U$.

Proof: Since $C$ is compact and $Y$ is locally compact, $U$ contains a precompact neighborhood of $C$, so we may assume $U$ is precompact. Now since $\overline{U}$ is compact Hausdorff, $C$ is not just a component of $\overline{U}$ but a quasicomponent. Thus $C$ can be separated from each point of $\overline{U}\setminus U$ by clopen sets of $\overline{U}$, and by compactness of $\overline{U}\setminus U$ there is then a single clopen set $V$ of $\overline{U}$ such that $C\subseteq V\subseteq U$. Then $V$ is closed in $Y$ since $\overline{U}$ is closed in $Y$, and $V$ is open in $Y$ since it is open in $U$ and $U$ is open in $Y$. Thus $V$ is clopen in $Y$ and is the desired set.

Now if $Z$ is a locally compact Hausdorff space and $X\subseteq Z$ is an open subset, let $Y=Z\setminus X$. If $C$ is a compact connected component of $Y$, then the lemma gives a precompact clopen set $V\subseteq Y$ containing $C$. Then $V$ is a union of compact connected components of $Y$, and it follows that the union $W$ of all compact connected components of $Y$ is open in $Y$. It follows that $W\cup X$ is open in $Z$, and $W\cup X$ is exactly the Runge hull of $X$.

Answer 2

Edit: The proof below is incorrect; see the comments below.

I claim that the union of all unbounded connected components of a closed set $C\subset\mathbb{C}$ is closed. Consider the one-point compactification of $\mathbb{C}\cup\{\infty\}$. Then $C\cup\{\infty\}$ is closed in the topology of $\mathbb{C}\cup\{\infty\}$. The unbounded components of $C$ are precisely those which lie in the connected component of $\{\infty\}$ in $C\cup\{\infty\}$. Since connected components are always closed, we conclude that the union of unbounded connected components of $C$ together with $\{\infty\}$ is closed in $C\cup\{\infty\}$. Hence the union of unbounded connected components of $C$ is closed in $\mathbb{C}$.

The exact same argument works with $\mathbb{C}$ replaced by any manifold (or indeed any locally compact space, although I'm not so sure about this).