Will the boundary of domain be path-connected, in $\mathbb R^2$?

Question

In the plane, $\mathbb{R}^2$, let $E \subset \mathbb{R}^2$ be a precompact, open and simple connected set.

Then, must the boundary of $E$ be path-connected?

Here, precompact means that its closure is compact.

My feeling is the statement is true. It looks obvious, if we draw a picture. Does anyone know this statement, and where is its proof?

If it's a wrong statement, does anyone know what condition we need adding?

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Tips:

In $\mathbb{R}$, any bounded open set can be a counterexample.

In $\mathbb{R^3}$, donut can be a counterexample.

The reason I assume "simply connected" instead of "connected" is that if I only assume "connected", a punctured set (or say a set with holes) will be a counterexample.

Answer 1

It is false. Let $B$ be the following compact subset of $\mathbb R^2$:

It is the union of the boundary of a square $R$ and a copy of the topologist's sine curve

$$S = \{(x,\sin(1/x) \mid x \in (0,1]\} .$$

This set is connected, but not path-connected. To see this one can adapt the standard proof for the closed topologist's sine curve $S' = S \cup (\{0\} \times [-1,1])$ in math.stackexchange.

The region $E = R \setminus B$ is open in $\mathbb R^2$ and one can show that it is simply connected. But the boundary of $E$ is $B$ which is not path-connected.

Another counterexample is based on the toplogist's whirlpool $W$. This a compact subset of the unit disk $D$ which is connected but not path-connected. The region $E = D \setminus W$ is open in $\mathbb R^2$ and one can show that it is simply connected. Intuitively it is clear that we can find a homeomorphism $h : (0,1) \times (0,\infty) \to E$. The boundary of $E$ is $W$ which is not path-connected.