Given closed $C \subseteq \mathbb{R}$ find a sequence with subsequences convergent to every point in $C$ and nowhere else Given $C \subseteq \mathbb{R}$ closed, find a sequence such for every point in $C$ there is a subsequence of your sequence which converges to that point, and that there is no subsequence of your sequence which converges to a point outside of $C$ (excluding $\pm \infty$).
If I take an enumeration of $\mathbb{Q}$ that will have subsequences which converge everywhere in $\mathbb{R}$. I want to somehow cut out the portions of $\mathbb{Q}$ which are not near $C$. Initially I thought to just take $\mathbb{Q} \cap C$ but this didn't work.
Perhaps I could modify this trick a little to get it to work?
Or is there a better way to do it?
 A: I think this requires (countable) choice in general. Enumerate all open intervals with rational endpoints, and for each such interval $I$ choose a point in $I\cap C$ if one exists (and throw away the interval otherwise). The set of all the chosen points will be dense in $C$.
A: This might be easier: Let $\{x_{1},\;x_{2},\; x_{3},\;...\}$ be a countable dense subset of $C$ and let the sequence be
$$x_{1},\; x_{1},\; x_{2},\;x_{1},\;x_{2},\;x_{3},\;x_{1},\;x_{2},\;x_{3},\;x_{4},\;...$$
There are some details to fill in (what if the set is finite, how do we know that no subsequences converge to a point not in $C$, etc.), which I'll leave to you.
A: Here is an explicit construction. 
For every $n\geqslant0$ call $I(n)$ the integer interval $I(n)=[2\cdot4^{n},8\cdot4^{n}-1]$. For every $k$ in $I(n)$, let $a(k)=2^{-n}(k-5\cdot4^{n})$, and $x(k)$ any point in $C$ such that $|x(k)-a(k)|=\min\{|x-a(k)|\mid x\in C\}$. Then $\mathfrak X=(x(k))_{k\geqslant2}$ is a sequence of elements of $C$ whose limit set is $C\cup D$ where $D\subseteq\{-\infty,+\infty\}$ is such that $D$ contains $-\infty$ if and only if $\inf C=-\infty$ and $D$ contains $+\infty$ if and only if $\sup C=+\infty$. 
To see this, first note that every $x$ in $C$ is such that $|x|\leqslant3\cdot2^n$ for $n$ large enough. For every such $n$, there exists $k$ in $I(n)$ such that $|x-a(k)|\leqslant2^{-n-1}$. Since $|x(k)-a(k)|\leqslant2^{-n-1}$, $|x-x(k)|\leqslant2^{-n}$ hence $x$ is a limit point of $\mathfrak X$. 
Finally, $\mathfrak X\subseteq C$ hence every limit point of $\mathfrak X$ is in the closure of $C$ in $\overline{\mathbb R}$, that is, in $C\cup D$.
