Any closed subset of $\mathbb C$ is the set of limit points of some sequence

Let $X$ be a closed subset of $\mathbb C$ with the usual norm.

Prove that there exists some complex sequence $(u_n)$ such that $X$ is exactly the set of limit points of $(u_n)$.

For the sake of clarity, let me re-define the set of limit points of a sequence.

$l$ is a limit point of $(u_n)$ iff there a subsequence of $(u_n)$ that converges to $l$.

Equivalenty, $l$ is a limit point of $(u_n)$ iff for every neighbourhood $V$ of $l$, there are infinitely many natural numbers $n$ such that $u_n ∈ V$.

I proved the converse of the assertion above, and I’m stuck with the direct implication. I can’t build such $(u_n)$ with my bare hands, and I tried contradiction, without success.

Solutions Collecting From Web of "Any closed subset of $\mathbb C$ is the set of limit points of some sequence"

A subspace of a normed space is separable, hence $X$ is separable. Let $\{x_n\}_{n \in \mathbb{N}}$ be a countable dense subset.

Now $X$ has a countable number of isolated points (if $x \in X$ is isolated, pick a ball around $x$ not intersecting $X$, then this ball contains a point of $\mathbb{Q}[i]$ and hence we get an injection $\{\text{isolated points}\} \to \mathbb{Q}[i]$ and this last set is countable). Let $I = \{y_m\}_{m \in \mathbb{N}^*}$ be the set of isolated points (maybe there are repetitions, it’s not a problem).

Form the sequence $y_{n,m}$ by setting:
$$y_{n,m} = \begin{cases}
x_n & m = 0 \\
y_m & m > 0
\end{cases}$$
And let $z_k = y_{f(k)}$ where $f : \mathbb{N} \to \mathbb{N} \times \mathbb{N}^*$ is a bijection.

This sequence meets any isolated point of $X$ an infinite number of times by design, so that isolated point will be a limit point of $z_k$.

If $x \in X$ is not an isolated point, let $V$ be a neighborhood of $x$ in $X$; then $V$ is infinite. Since the initial sequence $\{x_n\}$ was dense in $X$, it’s also dense in $V \cap X$, but this set is infinite (because $x$ isn’t an isolated point of $X$), and so $\{z_k\} \cap V \cap X$ will be infinite too (because a finite subset wouldn’t be dense in an infinite one, because $\mathbb{C}$ is a metric space).

And any limit point of $\{z_k\} \subset X$ lies in $\bar X = X$. This concludes.

For a sequence $(a_n)$ the set of its limit points is

$$\cap_{n\ge 1} \overline{\{a_n, a_{n+1}, \ldots \}}$$

From a sequence $(r_n)$ construct the sequence

$$r_1,\ r_1, r_2,\ r_1, r_2 ,r_3,\ \ldots,\ r_1, r_2, \ldots, r_n, \ldots $$

with the set of limit points $\overline{\{r_1, r_2, \ldots \}}$

Obs: Works in general topological spaces.

This is a simpler approach.

To prove the wanted result, it suffices to build a sequence $(v_n)\in X^{\mathbb N}$ such that, $$\forall \lambda \in X, \forall \epsilon>0, \forall N \in \mathbb N, \exists n_0 \geq N, |v_{n_0}-\lambda|\leq \epsilon$$


Since $\mathbb C$ is separable, there exists a dense, countable subset $D$ of $\mathbb C$.

Let’s define $(u_n)$ such that $D=\{u_n | n \in \mathbb N\}$.

Since $X$ is closed, for each $n\in \mathbb N$, one can define the projection $v_n$ of $u_n$ on $X$, that is $v_n \in \{ x\in X \;\;|\;\;|u_n-x|\; \text{is minimal }\}$.


It remains to prove that $(v_n)$ meets our requirements.

Let $\epsilon >0$ and $N\in \mathbb N$.

Let $\displaystyle A=\{n\in \mathbb N \;| \;\; |u_n-\lambda|\leq \frac{\epsilon}3\}$.

It’s easy to check that $A$ is infinite.

Moreover, if $n\in A$, then $|v_n- \lambda|\leq \epsilon$

Otherwise, $$\epsilon \leq |v_n-\lambda|\leq |v_n-u_n|+|u_n-\lambda|\leq |\lambda-u_n|+|u_n-\lambda|\leq\frac{2\epsilon}3$$

Since $A$ is infinite, choose any $n_0 \in A$ with $n_0 \geq N$. We have$|v_{n_0}-\lambda|\leq \epsilon$


PS: One sees that the result still holds if $\mathbb C$ is replaced with any separable metric space that has Bolzano-Weierstrass property.