A topology question on circle

Question

$F_{1}$ and $F_{2}$ are two disjoint compact sets in $\mathbb{S}^{1}$, does there exist an arc $\overset{\frown}{AB}$ adjoining $F_{1}$ and $F_{2}$?

Here ''adjoin'' means $A\in F_{1}$ $B\in F_{2}$ and the (open) arc $\overset{\frown}{AB}$ contains no point of $F_{1}\cup F_{2}$.

Could anyone give a proof or a counterexample? I am working on analysis, this is a question related to my research. I would like to know whether it is true or not.

Answer 1

Yes it's true. Consider the function

$$\Delta\colon F_1 \times F_2 \to \mathbb{R}$$

defined by restricting the distance function of $S^1$. Since $F_1$ and $F_2$ are compact so is the product, and therefore by the Extreme Value Theorem $\Delta$ has a minimum value $\Delta(x_0, y_0) = d_0$. Since $F_1 \cap F_2 = \emptyset$, $x_0 \neq y_0$. Then you can find your arc from $x_0$ to $y_0$ avoiding $F_1 \cup F_2$, otherwise there would be another pair with a smaller distance.

Answer 2

I think there does exist such an arc.

Consider the map $\theta: S^1\to [0,2\pi)$. Both the compact sets $F_1$ and $F_2$ will have closed, bijective images in $[0,2\pi)$. Choose an interval in $[0,2\pi)\setminus \theta(F_1\cup F_2)$. The pre-image of such an interval is the desired arc.