Bounded components of complement of bounded planar domain are simply-connected

Let $D \subset \Bbb R^2$ be a bounded open connected set. Let $E$ be a bounded connected component of $D^c$. Must $E$ be simply connected?

The way I imagine it, domains cannot be "too pathological": they consist of the interior some closed curve (not necessarily Jordan) with some holes inside it, and under this imagination the claim sounds plausible.

What I really "need" is the case where $\partial D$ has just two components, but this seems like a reasonable, more general claim.

I know that a connected set in the plane is simply connected iff its complement does not have bounded components, which seems like a useful tool here.

• Any reason you need this? Indeed, each boundary component is simply-connected, but the fundamental group is not a very good invariant of spaces which fail to be locally connected. – Moishe Kohan Sep 3 '17 at 21:52
• @Crazy Ivan: Are you sure you want to see an answer to this question in its present form? The "correct" invariant is the Chech fundamental group (it is still trivial in this case), not the usual fundamental group. – Moishe Kohan Dec 19 '17 at 19:50
• @MoisheCohen I like OP's version of "simply connected": define it as "the complement does not have bounded components". This is of course specific to the planar setting but seems natural here, where the set $E$ we are considering is a planar continuum. – user357151 Dec 19 '17 at 20:12
• @CrazyIvan: OK, as you wish, I wrote a proof. – Moishe Kohan Dec 20 '17 at 19:55
• $E$ need not be path-connected, but usually the definition of simply-connected includes path-connected (so then $E$ need not be simply-connected since it need not be path-connected). On the other hand, suppose $E$ is not simply connected, so there is a Jordan curve $J$ in $E$ the inside of $J$ contains a point $p$ not in $E$. Say $I$ is the inside of $J$, and $O$ the outside of $J$. If $I$ intersects $D$ then $D\subset I$ and $E\cup O$ is connected, but unbounded, a contradiction, as $E\cup O\subset E$ . If $I$ misses $D$ then $E\cup I$ is connected, hence $E\cup I\subset E$, a contradiction. – Mirko Dec 20 '17 at 23:33

For a subset $B\subset \mathbb{R}^2$ I will use the notation $B^c$ to denote its complement, $\mathbb{R}^2 \setminus B$.

Let me start with a basic observation: If $A$ is a closed subset of $\mathbb{R}^2$, then for each connected component $U$ of $A^c$, the boundary of $U$ is contained in $A$.

Now, back to the question. Let $D$ be an open connected bounded subset of $\mathbb{R}^2$. I will denote $\Omega$ the (unique) unbounded component of $\mathbb{R}^2 \setminus \bar{D}$. Let $E$ be a bounded component of $D^c$. Our goal is to show that $E^c$ contains no bounded components.

It is clear that $\Omega\cap E=\emptyset$.

Let $U$ be a component of $E^c$. Then $U$ and $D$ are open subsets of $E^c$. Since $U$ is a component of $E^c$ and $D$ is connected, the open sets $U$ and $D$ are either disjoint or $D\subset U$.

Case 1: $U\cap D=\emptyset$. Since $D$ is open, we also have that $\bar{U}$ is disjoint from $D$.

By the basic observation, the boundary of $U$ is contained in $E$, hence, $\bar{U}\cap E\ne\emptyset$, thus, $\bar{U}\cup E$ is connected and disjoint from $D$. Thus, $\bar{U}\subset E$ (since $E$ was a component of $D^c$). This is a contradiction since $U\cap E=\emptyset$.

Case 2: $D\subset U$, hence, the boundary of $D$ is contained in the closure of $U$. Hence, $U$ contains the boundary of $\Omega$ and, thus, $\Omega\cap U\ne\emptyset$ and, therefore, $\Omega\cup U$ is connected and is disjoint from $E$. Since $U$ was a component of $E^c$, it follows that $\Omega\subset U$ and, therefore, $U$ is unbounded.

Thus, we conclude that all components of $E^c$ are unbounded. qed

Addendum. The notion of simple connectivity you are using is not equivalent to the usual one even for compact subsets of the plane (the Warsaw circle is a standard example). But your notion is equivalent to simple connectivity in Čech sense. Namely, the following are equivalent for compact connected subsets $K\subset R^2$:

• $K^c$ contains no bounded components.

• $\check{H}^1(K, {\mathbb Z})=0$.

• $\check{\pi}_1(K, x)= \{1\}$ for every $x\in K$, i.e. $K$ is Čech-simply-connected.

Here $\check{H}$ means Čech cohomology and $\check{\pi}_1$ means Čech fundamental group. One proves this equivalence using Alexander duality and Hurewicz theorem.

• Also, this notion of simple connectivity is equivalent to the usual one for connected OPEN sets in the plane (but in the above question the sets $E$ are only known to be closed) – Mirko Dec 21 '17 at 2:37
• @Mirko: Yes, of course. For spaces which are not locally connected, one should use Chech notions instead of the standard ones. – Moishe Kohan Dec 21 '17 at 2:39