Is it possible to fill an $n-$dimensional unit cube with countable number of non-overlapping $n-$dimensional balls?
By $n-$dimensional unit cube, I am thinking of the set $$ C:=\{\ x\in\Bbb R^n\ |\ 0\le x_i\le 1\ \text{for all}\ i=1,2,\dots,n\ \} $$ where $x=(x_1,\dots,x_n)$. An $n-$dimensional ball centered at $x_0$ is defined to be $$ B(x_0;r):=\{\ x\in\Bbb R^n\ |\ d(x_0,x)<r\ \} $$ where $d$ is the Euclidean distance function.
A sequence of balls $(B_1,B_2,\dots)$ is said to fill the unit cube if they are pairwise-disjoint, each $B_i\subset C$ and $$ \sum_{i=1}^{\infty}\mu(B_i)=1 $$ where $\mu$ is the Lebesgue measure on $\Bbb R^n$.
Intuitively, I feel that it should be possible to fill the unit cube with many balls. However, I can't think of a way to prove it.
Since $C$ is separable, I thought about enumerating its rational points into a sequence $\{c_1,c_2,c_3,\dots \}$ and form balls around them such that $B_i:=B(c_i,r_i)$ where $$ r_i:=\sup\{\ r\in\Bbb R^+\ |\ B(c_i;r)\ \text{is contained in $C$ and doesn't intersect $B_1,B_2,\dots,B_{i-1}$} \}. $$ I don't know if this method works or not. I would really appreciate if anyone can suggest a way to finish the prove, suggest a new method, or give a negative answer to my question. An answer that only works in dimension $2$ or $3$ is also very welcomed.
Yes, this is possible. First, let me set up a bit of machinery. Define a dyadic cube to be an open cube of the form $\prod_{i=1}^n \left(\frac{a_i}{2^k},\frac{a_i+1}{2^k}\right)$ where $a_1,\dots,a_n\in\mathbb{Z}$ and $k\in\mathbb{N}$. Write $N\subset\mathbb{R}^n$ for the set of points which have at least one coordinate that is a dyadic rational; note that $N$ has measure $0$. Note that if $U\subseteq\mathbb{R}^n$ is open and $x\in U\setminus N$, there is a dyadic cube $D$ such that $x\in D\subseteq U$ (since if $x\not\in N$ then there are dyadic cubes of arbitrarily small diameter containing $x$). Moreover, there is a maximal such cube. It follows that $U\setminus N$ is contained in the union of all the maximal dyadic cubes contained in $U$. Moreover, these maximal dyadic cubes are disjoint, since if two dyadic cubes intersect then one must be contained in the other.
To sum up, we have shown that if $U\subseteq\mathbb{R}^n$ is open, there is a (necessarily countable) disjoint union of dyadic cubes contained in $U$ that has full measure in $U$.
Now let us consider your question. Start by picking any ball $B\subset C$. Let $U=C\setminus \overline{B}$. As above, we can partition almost all of $U$ into dyadic cubes. We can then choose scaled down versions of $B$ inside each of these dyadic cubes, and repeat: take the complement of the closures of these balls inside each dyadic cube, and partition almost all of those sets into dyadic cubes. We then take scaled down versions of $B$ inside all these new dyadic cubes, and so on.
At each step of this process, we add new balls which take up $\mu(B)/\mu(C)$ of the measure we have not yet covered with balls. Thus the measure not covered by balls after $N$ steps is $(1-\mu(B)/\mu(C))^N$. This converges to $0$, so after iterating infinitely many times we have "filled" $C$.
More generally, a similar argument shows that if $B\subset C$ is any set of positive measure such that $\mu(B)=\mu(\overline{B})$, then $C$ can be "filled" by scaled copies of $B$.