Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
$C$ has to be bounded, or there will be a subsequence that converges to whichever of $\pm\infty$ is in the unbounded direction. – Ross Millikan Oct 21 '11 at 18:15
Are you sure of the "including $\pm\infty$" part? If $\infty$ is considered a "point", then one would expect that this point should also be considered when determining whether $C$ is closed or not. And if so, either $C$ is bounded (in which case there can never be a subsequence converging to $\infty$ anyway), or $C$ contains $\infty$ (in which case it should be allowed as a limit point). – Henning Makholm Oct 21 '11 at 18:28
By the way, you also have to require that $C$ is non-empty. Otherwise there are no sequences at all. :-) – Henning Makholm Oct 21 '11 at 18:33
No we don't need that it is nonempty, if it is empty then take the empty sequence and the statement is vacuously true. – nullUser Oct 21 '11 at 18:47
A sequence is a function with domain $\mathbb{N}$. There's no such thing as an empty sequence. – Chris Eagle Oct 21 '11 at 19:23
up vote 2 down vote accepted

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 $C$.

share|cite|improve this answer
I am okay with using any form of the Axiom of Choice if necessary. – nullUser Oct 21 '11 at 18:36
Can't I enumerate all the rational open intervals without choice? Use a pairing function three times to get all ordered fours of $N$ and use them as endpoints if the fractions are in lowest terms and increasing? – Ross Millikan Oct 21 '11 at 21:47
@Ross, you can enumerate the rational intervals without choice, but you need choice to select an element in each $I\cap C$ simultaneously. – Henning Makholm Oct 21 '11 at 21:50
Don't you need to add in any isolated points in $C$ an infinite number of times in the sequence, too? – Ross Millikan Oct 21 '11 at 22:02
@Ross, yes -- my intention was that this construction would be used together with Dave's one. – Henning Makholm Oct 21 '11 at 22:08

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.

share|cite|improve this answer
Exactly my question, it is not clear at all that an arbitrary closed set has a countable dense subset. – nullUser Oct 21 '11 at 18:25
@Kb100: You may want roughly indicate the mathematical level that is being assumed, as I took it for granted that we could use the fact that $\mathbb R$ is separable. In fact, I'm using the fact that $\mathbb R$ is hereditarily separable, which is equivalent to separability in metric spaces. – Dave L. Renfro Oct 21 '11 at 18:31
I have no idea what that even means =\. This is introductory real analysis. – nullUser Oct 21 '11 at 18:35
@Kb100: Unless I'm overlooking something, I think you have enough if you use Henning Makholm's answer along with mine. However, it would probably be a good idea to try to prove the result for some specific closed sets, such as $[0,1]$ and $\{\frac{1}{n}:\;n=1,2,3,...\} \cup \{0\},$ to get a better feel for what's going on. – Dave L. Renfro Oct 21 '11 at 20:12
Yes his idea was sufficient. I made some preferential changes to the idea but it was rather easy to prove with this bit of help. The general proof was less than a page. Thanks so much everyone! – nullUser Oct 21 '11 at 21:20

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 $C\cup D$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.