Covering the Circle with closed connected sets. This is a follow up to this previous post . The answer and the question got me to thinking about the case of $S^{1}$.
The conditions are again the same .
Can $S^{1}$ be covered by finite number of connected closed sets $A_{1},...,A_{n}$ such that

*

*Each has diameter less than a given $\epsilon$ .


*$\cup A_{n}= S^{1}$


*$\displaystyle A_{k}\cap\bigg(\bigcup_{i=1}^{k-1}A_{i}\bigg)$ is connected for all $k\leq n$ ?
As the comments under the answer by Paul Frost suggests , a formal proof will not be easy. However I can see how a particular attempt fails. Suppose I start at some point(say (1,0)) and divide the circumference into a large number of pieces with each of arc length $\frac{2\pi}{m}$ where $m$ is large enough to satisfy condition $1$. Now I order them as we would do with hours on a clock and proceed with the "numbering" anticlockwise.
When we reach $A_{n}$ then we have the $A_{n}\cap\bigg(\bigcup_{i=1}^{n-1}A_{i}\bigg)$ is a doubleton set and hence it is not connected.
Any help or a sketch of how to try and prove this will be highly appreciated.
 A: It is impossible. Let us give a negative answer to the following more general question:
Can $S^{1}$ be covered by a finite number of connected closed proper subsets $A_{1},...,A_n$  of $S^1$ such that $\displaystyle A_{k}\cap\bigg(\bigcup_{i=1}^{k-1}A_{i}\bigg)$ is connected for all $k\leq n$ ?
So let $A_1,\ldots,A_n$ be connected closed proper subsets of $S^1$ such that $\displaystyle A_{k}\cap\bigg(\bigcup_{i=1}^{k-1}A_{i}\bigg)$ is connected for all $k\leq n$.
The connected closed proper subsets of $S^1$ are closed arcs of the form $A(t_0,t_1) = \{ e^{2\pi it} \mid t_0 \le t \le t_1 \} $ with $0 \le t_1 - t_0 < 2\pi$ which connect the two points $z_j = e^{2\pi it_j}$. If $t_0 = t_1$, it is a single-point space. An open arc is a set of the form  $O(t_0,t_1) = \{ e^{2\pi it} \mid t_0 < t < t_1 \} $ with $0 < t_1 - t_0 \le 2\pi$. All arcs $\Gamma$ have a natural ordering which allows to decide whether or not a point $z \in  \Gamma$ lies strictly between two distinct point $z_0, z_1 \in \Gamma$.
We shall now show by induction on $n$ that $\bigcup_{i=1}^n A_i \ne S^1$.
The base case $n=1$ is trivial.
$n \mapsto n+1$:
Assume that we have $A_1,\ldots A_{n+1}$ as above with the property $\bigcup_{i=1}^{n+1} A_i = S^1$. All $A_i$ are closed arcs. We know that $B = \bigcup_{i=1}^n A_i \ne S^1$. It is imposible that $B$ is empty or a single point space because then $B \cup A_{n+1}$ can never be $S^1$. Let $z \in S^1 \setminus B$. Since $S^1 \setminus B$ is open in $S^1$ and $\ne S^1$, there is a maximal open arc $O(\tau_0,\tau_1) \subset S^1 \setminus B$ with  $z \in O(\tau_0,\tau_1)$. Its endpoints $z_j = e^{2\pi i\tau_j}$ are contained in $B$. Clearly $O(\tau_0,\tau_1) \subset A_{n+1} = A(t_0,t_1)$. But this shows that $D = A_{n+1} \cap B$ cannot be connected: If it were, it would be a closed subarc of $A_{n+1}$ containig the points $z_j$, but $z \notin D$ although it lies strictly between $z_0$ and $z_1$.
