Existence of a sequence. While reading about Weierstrass' Theorem and holomorphic functions, I came across a statement that said: "Let $U$ be any connected open subset of $\mathbb{C}$ and let $\{z_j\} \subset U$ be a sequence of points that has no accumulation point in $U$ but that accumulates at every boundary point of $U$."
I was curious as to why such a sequence exists. How would I be able to construct such a sequence?
 A: Construct your sequence in stages indexed by positive integers $N$.  At stage $N$, enumerate those points $(j+ik)/N^2$ for integers $j,k$ with $|j|+|k|\le N^3$ that are within distance $1/N$ of the boundary of $G$.  
EDIT: Oops, not so clear that this will give you points in $G$.  I'll have to be a bit less specific.  At stage $N$, consider $K_N = \partial G \cap \overline{D_N(0)}$ where $\partial G$ is the boundary of $G$ and $D_r(a) = \{z: |z-a|&ltr\}$ .  Since this is compact, it can be covered by finitely many open disks $D_{1/N}(a_k)$, $k=1,\ldots,m$, centred at points $a_k \in K_N$.  Since $a_k \in \partial G$, $G \cap D_{1/N}(a_k) \ne \emptyset$.  So we take a point $z_j \in G \cap D_{1/N}(a_k)$ for $k=1,\ldots,m$.     
A: Let $\partial U$ the boundary of $U$ in $\mathbb C$, which we assume to be non-empty.
Let $T = \{t_0, t_1, \dotsc \}$ be a countable dense subset of $\partial U$, and let $v$ a sequence a sequence with value in $T$ such that for all $i$ the sequence $v$ takes the value $t_i$ infinitely many times.
Let $n$ be an integer. The ball $B(t_n, 2^{-n})$ cut $U$, since $\partial U$ is a subset of $\overline U$. Let $u_n$ be an element of $B(v_n, 2^{-n})\cap U$.
The sequence $u$ we just defined satisfies your property.


*

*For all $n$, the point $t_n$ is an accumulation point of $u$. Indeed, by definition of $v$, there is an extraction $\phi$ such that $v_{\phi(k)} = t_n$, for all $k$. But then $|t_n - u_{\phi(k)}| &lt 2^{-\phi(k)}$, hence $u_{\phi(k)} \to t_n$.

*The sequence $u$ has no accumulation points in $U$. Since the distance of $u_k$ to $\partial U$ tends to zero, all accumulation are in $\partial U$, which is disjoint from $U$.

*The set of the accumulation points is a closed set which contains T, and which is disjoint from $U$, thus it is $\partial U$.
