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

Question

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.

Maybe you can build such a sequence using an enumeration of $X \cap \mathbb Q + i\mathbb Q$, not sure though. — AlexR, Oct 13 '14 at 15:30
@NajibIdrissi Yeah, thought it wouldn't be that easy, though given this specific set you can of course take an enumeration of $\mathbb Q + \sqrt2 i$... Looks like one may be able to generalise that. — AlexR, Oct 13 '14 at 15:33
@NajibIdrissi Though quite relevant, I don't think it's a duplicate. — Daniel Fischer, Oct 13 '14 at 15:34
@NajibIdrissi could you elaborate on how this is a duplicate ? — Gabriel Romon, Oct 13 '14 at 15:34
@G.T.R Though I don't think it's a duplicate, if you take $X$ to be the subspace the question says that $X$ is seperable so it has a countable dense subset, wich you can enumerate for your sequence $u_m$ — AlexR, Oct 13 '14 at 15:36
Very similar to Given closed $C \subseteq \mathbb{R}$ find a sequence with subsequences convergent to every point in $C$ and nowhere else. — Dave L. Renfro, Oct 14 '14 at 17:37
For a more interesting result, prove that such a sequence can always be chosen so that for each point in the closed set, there exists a subsequence convering to that point whose indices have upper asymptotic density $1$. — Dave L. Renfro, Oct 14 '14 at 17:46

score 4 · Accepted Answer · edited Apr 13 '17 at 12:19

4

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.

edited Apr 13 '17 at 12:19

Community

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answered Oct 13 '14 at 15:35

Najib Idrissi

56,269

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Close, but you need to treat isolated points of $X$ specially. – Daniel Fischer Oct 13 '14 at 15:37
@NajibIdrissi I don't understand why $X$ has to be separable. Can you elaborate ? – Gabriel Romon Oct 13 '14 at 15:46
@G.T.R $\mathbb{C}$ is separable ($\mathbb{Q}[i]$ is dense), now use the answer I linked. – Najib Idrissi Oct 13 '14 at 15:48
@Daniel I updated my answer (you were right, there was some work to do; I don't believe it's a duplicate anymore). – Najib Idrissi Oct 13 '14 at 15:51
I think you meant $I = {y_m}_{m \in \mathbb{N}^*}$ – Gabriel Romon Oct 13 '14 at 16:09
@G.T.R Good catch. – Najib Idrissi Oct 13 '14 at 16:25
@NajibIdrissi It took me time to understand the gist of your proof, but I concur. Great answer. – Gabriel Romon Oct 13 '14 at 17:08
@Stefan Yes, of course. – Najib Idrissi Oct 13 '14 at 18:52
@NajibIdrissi I added a simpler proof. – Gabriel Romon Oct 14 '14 at 17:10

orangeskid · Answer 2 · 2025-02-04T18:52:29.297

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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.

$\bf{Added:}$

Let's adress why $X$ has a dense subset $\{r_n\}$. Start with $\mathbb{C}$ a separable metric space, and $X$ a non-void subspace of $\mathbb{C}$. Take $(q_n)_n$ a sequence with $\overline{\{q_n\}}= \mathbb{C}$. Now, for every $n$, choose $r_n \in X$, such that $d(q_n, r_n) \le 2 d(q_n, X)$. Let us show that $\{r_n\}$ is dense in $X$. Indeed, take $x \in X$, $\epsilon>0$. Take $q_n$ such that $d(x, q_n)< \epsilon$. Then $d(q_n, X) < \epsilon$, and so $d(q_n, r_n) < 2 \epsilon$. We get $d(x,r_n)< 3 \epsilon$.

Note: a subspace of a separable topological space might not be separable, but it is so if the space is metrizable.

edited Feb 04 '25 at 18:52

answered Oct 13 '14 at 17:01

orangeskid

56,630

Sorry, I'm not understanding. How does this address the question? – Andrés E. Caicedo Oct 14 '14 at 17:23
@Andres Caicedo: Perhaps $(r_{n})$ is intended to be a sequence of elements in the closed set that is dense in the closed set? – Dave L. Renfro Oct 14 '14 at 17:31
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Hi @Dave. Perhaps. It seems puzzling that the existence of such a sequence is treated as a blackbox, while the small detail of how to ensure that such a sequence works is elaborated. – Andrés E. Caicedo Oct 14 '14 at 17:33
@Andrés E. Caicedo: addressed the existence of the sequence. – orangeskid Feb 04 '25 at 18:54

Gabriel Romon · Answer 3 · 2014-10-14T18:11:53.127

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.

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

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