A disjoint union of open balls is of course disconnected. Here it is proved that a locally compact, connected, Hausdorff space is not a countable disjoint union of compact subsets, so a countable disjoint union of closed balls in $\mathbb{R}^n$ can't be connected and locally compact (hence cannot be open or closed connected), which rules out the connected sets for $n=1$. But what if we just ask this disjoint union to be connected?
Please forgive me if this question turns out to be trivial. Thank you in advance for any help.
Edit. Actually I should have asked about something more general, like
Is there a connected subset $\mathbb{R}^n$ that can be written as a countable disjoint union of closed sets in $\mathbb{R}^n$? If not, is there a connected subset $\mathbb{R}^n$ that can be written as a countable disjoint union of closed sets in the subspace topology?
And the answer to the second question above is still negative for $n=1$, or for open subsets of $\mathbb{R}^n$ (the link requires also local connectedness which is satisfied for open connected sets). However, I will keep the question as it was given the discussion already presented.
Edit 2. A connected subset of $\mathbb{R}^3$ as a countable disjoint union of closed sets has be constructed in the existing answer. I would still be interested in the closed ball case: Must a countable disjoint union of closed balls in $\mathbb{R}^n$ not be connected?
Best Answer
Example of countable family of pairwise disjoint closed sets on a plane whose union is connected
Here is a relatively simple example. The other section has an older, more complicated example in $ℝ^3$.
Let $D = \{d_n : n=0,1,\dots\}$ be a dense subset of unit sphere with $d_n \neq d_m$. For each $n=1,2\dots$ define the line segment $$ A_n = \{ λd_n : 2^{-n} ≤ λ ≤ 1 \}, $$ and $A_0 = \{ λd_0 : 0 ≤λ≤1 \}$. Let $\text{Sun} = \bigcup_{n=0}^∞ A_n$. It is a union of closed, pairwise disjoint, connected sets. We show that $\text{Sun}$ is connected. Before proof, here is how $A_0,A_1 \dots, A_5$ might look like.
Proof. Let $U,V$ be open disjoint sets with $\text{Sun} = U \cup V$. Without loss of generality assume that $U$ contains $0\in A_0$. Since $A_0$ is connected, it must be contained in $U$. It suffices to show that $A_1, A_2, \dots$ are in $U$ as well.
First, note that there is $N$ for which ball $B = B(0, 2^{-N})\cap A$ is contained in $U$. Consequently, for each $k>N$ an endpoint of $A_k$, namely $2^{-k}d_k$, is close enough to the origin to be contained in $B \subset U$. Hence, all $A_k$ with $k>N$ are in $U$ ($A_n$'s are connected).
It remains to show that $A_1, A_2, \dots, A_N$ are also in $U$. Pick $1≤t≤N$. By construction $d_t \in A_t$. Assume that this point sits in $V$. Then there would be a neighborhood of $d_t$ with no points in common with $U$. This is impossible because $U$ contains $\{d_k : k>N \} \subset D$ which is dense in the unit sphere since it's just $D$ with finitely many points thrown out. Hence $d_t$ belongs to $U$, and so $A_t$ is in $U$.
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Note. Via the natural embedding, this will work in higher dimensions.
Edit. I have just found a construction of set with the same property in the literature: In the 1922 paper, Anna Mullkin presented perhaps the first example of such set. It's called Mullikin Nautilus and it is constructed by taking an infinite union of such polygonal chains:
See: Certain Theorems Relating to Plane Connected Point Sets.
Old example in $ℝ^3$
We construct a family $\{A_n: n =0,1, \dots\}$ of pairwise disjoint, closed subsets of $ℝ^3$ whose union is connected.
First, let $A_0 =\{(0,t,0): t≤0\}$ be ray. For $n≥1$ set $A_n$ is made of a ray $\{(0,2^{-n},0) : t≤ 0\}$ and a certain polygonal chain. Here is a sketch of the projection of $A_0, A_1, \dots A_5$ to XY plane.
Important bits:
This is all we need to know to prove that $A = \bigcup_n A_n$ is connected. At the end of the proof, we define $A_n$ for $n≥1$ at the end in detail. I will be glad for a verification!
Proof. Say $U,V$ are open (in the subspace topology) disjoint sets with $U\cup V=A$. We will show that one of them is empty. WLOG assume that $U$ contains the origin point $(0,0,0)$. Hence $U$ must contain $A_0$. Also, there must be some $N$ such that for $n≥N$ points $(0,2^{-n},0)$ are in $U$. Becasue each $A_n$ is connected, each $A_n$ with $n≥N$ is in $U$.
When $N=1$ set $U$ contains all $A_n$'s, that is $A\subset U$. So must be $V$ is empty as $U\cap V =∅$. Assume that $N>1$. We show that $U$ contains the remaing $A_{N-1}, A_{N-2}, \dots, A_2, A_1$. Pick any $k \in \{1,2, \dots, N-1\}$ and assume $A_k \subset V$. By construction point $x = (k, 2^{-k}, 0)$ sits in $A_k$. On the other hand, the sequence of points from $U$, namely $(x_{k,t})_{t=N}^{∞} = ((k, 2^{-k}+2^{t}, 0))_{t=N}^{∞}$ where $x_{k,t}\in A_t \subset U$, conveges to $x\in A_k \subset V$. Hence $V$ will contain almost all points of this sequence. But all terms of the sequence are members of $U$. So $U$ and $V$ will have infinitely many points in common! A contradiction with $U$ and $V$ being disjoint.
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$A_1$ consist of:
For $n≥2$ let $A_n$ be a union of: