For any closed subset of $\mathbb{R}$ there is a sequence in $\mathbb{R}$ whose sequential limits is equal to the that subset Question: Let $A$ be a closed subset in $\mathbb{R}$. Prove that there exists a sequence $x_n$ in $\mathbb{R}$ whose set of subsequential limits is exactly equal to $A$. 
My approach:
I think this sort of makes sense me if we our space was $\mathbb{C}$, however, I'm not even sure if this statement is true or not. I was thinking if it's possible to prove it by starting to say that the set of all subsequential limits of a sequence in a metric space is closed. However, I'm not sure if for any closed subset in $\mathbb{R}$, we can construct such sequence.
 A: I'm going to assume that you know basic things about compactness. And give you two steps to begin with.


*

*If $A$ is a closed and bounded set, then it is compact. Prove, even in general, that a compact metric space is always the closure of a countable set (or in other words, a compact metric space is always countable). You can do that by finding a particular sequence of open covers, refine each to a finite subcover and use that to define the countable set.

*If $A$ is closed and unbounded, then it can be written as the countable union of closed and bounded sets. Namely, if $A$ is closed, then it is the union of countably many compact sets. What do you know about the countable unions of countable sets? Now apply the previous case.
Or, prove directly,


*If $A\subseteq\Bbb R$ is covered by open sets, then there is a countable subcover. Use similar tricks to the previous two steps and find a countable dense subset. (This property of $\Bbb R$ is called hereditarily Lindelöf.)

A: First suppose $K\subseteq M$ is a compact subset of $\textit {any}$ metric space $M$.
For $n\in \mathbb N$, we have the open cover 
$\mathcal A'_{n}=\left \{ B_{1/n}(x) \right \}_{x\in K}$ 
from which we can extract a finite subcover $\mathcal A_{n}=\left \{ B_{1/n}(x_{ni}) \right \}_{1\leq i\leq J_{n}}$ where $J_{n}\in \mathbb N$. 
Then $\left \{ x_{ni} \right \}_{n\in \mathbb N,1\leq i\leq J_{n}}$ is the desired sequence, expressed as a matrix and enumerated in the usual way. 
Let $x\in K$ and $n\in \mathbb N $. Since $\mathcal A_{n}$ is a cover of $K$, there is an $1\leq i\leq J_{n}$, a $x_{ni}\in \left \{ x_{ni} \right \}_{n\in \mathbb N,1\leq i\leq J_{n}}$ and a  $B_{1/n}(x_{ni})$ that contains $x$ so that $d(x_{ni},x)<1/n$ and this gives us a subsequence of $\left \{ x_{ni} \right \}_{n\in \mathbb N,1\leq i\leq J_{n}}$ which converges to $x$. 
Notice that this argument will work even if $\mathcal A'_{n}$ has a properly countable (not finite) subcover. 
Now, let $C$ be closed and unbounded in $\mathbb R$ and take $A_{n}=C\cap \overline B_{n}(0)$. Then the $A_{n}$ are compact and $\bigcup _{n\in \mathbb N}A_{n}=C$. Therefore, this case reduces to the previous one, because for each $A_{n}$ we can construct a sequence $\left \{ x_{k}^{n} \right \}_{k}$, which hs a convergent subsequence to every $x\in A_{n}$. Then $\left \{ x_{k}^{n} \right \}_{k,n}$ is a sequence for which for every $x\in C$ there is a subsequence converging to $x$. 
A: Let $\mathcal I_1$ be the set of unit length intervals with endpoints at the integers.
Let $\mathcal I_2$ be the set of half unit length intervals with endpoints at the half-integers.
Let $\mathcal I_3$ be the set of quarter unit length intervals with endpoints at the quarter-integers.
…
Let $\mathcal I_n$ be the set of $2^{-n}$ length intervals with endpoints at the numbers $k / 2^{-n}$, $k \in \mathbb{Z}$.
…
Now start defining the terms of the sequence. 
The first batch of $0$ to $2$ terms are chosen in order, one in each of the intervals of $\mathcal I_1$ contained in $[-1,+1]$ that intersect $A$.
The second batch of $0$ to $8$ terms are chosen in order, one in each of the intervals of $\mathcal{I}_2$ contained in $[-2,+2]$ that intersect $A$.
The third batch of $0$ to $48$ terms are chosen in order, one in each of the intervals of $\mathcal I_3$ contained in $[-3,+3]$  that intersect $A$.
…
The $n$th batch of $0$ to $n 2^{n+1}$ terms are chosen in order, one in each of the intervals of $\mathcal I_n$ contained in $[-n,+n]$ that intersect $A$.
…
A: Hint: $A$ contains a countable dense subset.
