Suppose we have an extended real (countably) infinite sequence $(x_n)$. Then consider all of its possible subsequences $(x_{n_k})$. We could then consider the set $$A = \{a\in \overline{\mathbb{R}}:x_{n_k}\rightarrow a \text{ for some subsequence}\}.$$

Must this set necessarily be finite?

Otherwise we have countably many numbers that the original sequence approaches arbitrarily closely countably many times in its tail. Is there some kind of argument that this would require an uncountable sequence?

My thought is we could take the supposed countable set, and then choose some $\epsilon>0$ so that no two limits live in the same $\epsilon$-ball. Far enough down each subsequence, no element in the tail of one subsequence can also be in the tail of another. From here, is the argument like Cantor's diagonalization?

Limits: subseq

$a_1 : x_{n_1}, x_{n_2}, x_{n_3}, \dots$

$a_2 : x_{m_1}, x_{m_2}, x_{m_3}, \dots$

$a_3 : x_{o_1}, x_{o_2}, x_{o_3}, \dots$

We would need uncountably many limits for these subsequences to be unique, and this contradicts the original sequence being countable.

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    $\begingroup$ I would say yes. (SUPPOSE IT IS BOUNDED, then any sequence has a convergent subsequence) If it took (any form of) infinite (distinct) values between some bounds, then they would be arbitrarily close to each other and be indistinct from the same sequence. If however it is finite there must be an $\epsilon$ sufficiently small to "distinguish" between them. As you're using extended reals there are bounds for both. $\endgroup$
    – Alec Teal
    Aug 13, 2015 at 15:35
  • $\begingroup$ @AlecTeal You can easily adapt the arguments given in the answers below to rule out the bounded case too. Enumerate for example all the rationals in $[0,1]$ (or for a classical exercise, look at $x_n = \sin n$). $\endgroup$ Aug 13, 2015 at 15:38
  • $\begingroup$ @NajibIdrissi I posted that before there were answers. I'm just happy I was on the right lines! $\endgroup$
    – Alec Teal
    Aug 13, 2015 at 15:39
  • $\begingroup$ I say more: for any sequence $(a_n)$ you can obtain sequence $(x_n)$ such that for any $a_i$ exists subsequence $x_{n_k}\to a_i$ $\endgroup$ Aug 13, 2015 at 15:42
  • $\begingroup$ Just a side note: by some definitions (e.g., "a sequence is a function with a domain of $\mathbb{N}$"), all sequences are countable. Wikipedia lists this definition: "Formally, a sequence can be defined as a function whose domain is a countable totally ordered set, such as the natural numbers." (emphasis mine) $\endgroup$ Aug 14, 2015 at 18:26

8 Answers 8


Consider the sequence

$$ \{1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,...\} $$

That is, the sequence where you count till $n$, then start over and count till $n+1$. You can easily show that for every $n\in\mathbb{N}$ you can find a subsequence that converges to $n$ (it is indeed a constant subsequence!).

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    $\begingroup$ This is THE minimalist universal solution. Introduce infinitely many limit points, by giving names to each one: 1,2,3,... Construct the simplest sequence converging to each one : the constant sequence. Pack the sequences into one sequence using the greedy algorithm. QED. $\endgroup$ Aug 14, 2015 at 16:13
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    $\begingroup$ Yeah, its pretty minimalist. It may be worth noticing that this counterexample works only for limit sets that are countable or that admit a countable dense subset. $\endgroup$
    – bartgol
    Aug 15, 2015 at 3:40
  • $\begingroup$ No, this works fine without a countable dense subset. Take the reals with the discrete metric. This sequence still has an infinite number of accumulation points. It does not have an accumulation point at every real, however. $\endgroup$ Aug 15, 2015 at 19:24
  • $\begingroup$ What I meant is that if the limit set is uncountable, you cannot build constant sequence converging to all of them and patch them together in a single sequence. Otherwise uncountable would be countable. But you can have a single sequence whose subsequences have limit points that are uncountable. They just can't be all eventually constant. ASCII Advocate's example $\{0.0,0.1,0.2,...0.9,0.01,...0.99,0.001,...0.999\}$ has $[0,1]$ as lmit points. But only for countably many $q$ (the rationals) you can extract an eventually constant subsequence converging to $q$. The others are only "approached".. $\endgroup$
    – bartgol
    Aug 15, 2015 at 20:01

The rational numbers are countable. This means that there is an enumeration of the rationals by $\Bbb N$. In other words, there is a sequence $q_n$, such that each rational number appears exactly once in that sequence.

Now suppose that $r$ is any real number, then we can define by induction $q_{n_k}$ to be such that $n_k>n_j$ for all $j<k$, and $|r-q_{n_k}|<\frac1k$. So $q_{n_0}$ is the least indexed rational of distance $<1$ from $r$, and $q_{n_1}$ is the least indexed rational, whose index appears after $n_0$, and is of distance $<\frac12$ from $r$, and so on.

We go along the sequence, and use the least possible index whenever we get close enough. The fact that each index is preceded only by finitely many indices ensures that such index will be found eventually.

So it means that every real number is a limit of a subsequence (and $\pm\infty$ if you want to include them here).

So you have a sequence which has $2^{\aleph_0}$ limit points. One strange remark is that while we do not know if $2^{\aleph_0}=\aleph_1$ or not, we can say with confidence that if a sequence [of real numbers] has infinitely many limit points, then it either has $\aleph_0$ or $2^{\aleph_0}$ of them.

If you want to talk about larger topological spaces, then $\beta\Bbb N$ has a sequence which has $2^{2^{\aleph_0}}$ limit points, although there you need to talk about nets and not sequences. Simple cardinal arithmetics show that in a Hausdorff space, this is also the maximal number of limit points a sequence (or rather, a countable net) can have.

There is also non-Hausdorff spaces, but then you can just construct example where a singleton is dense, so a constant sequence has every point as a limit point, and then you have virtually no limitations.

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    $\begingroup$ But do these subsequences have their elements ordered as they were in the original sequence? I am assuming a subsequence is the original sequence with some elements deleted and the original order maintained otherwise. $\endgroup$
    – Pburg
    Aug 13, 2015 at 15:38
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    $\begingroup$ I was expecting that comment, actually. I guess I should have included this in the original answer, let me add this. $\endgroup$
    – Asaf Karagila
    Aug 13, 2015 at 15:39
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    $\begingroup$ @user43687: Half the weird examples from measure theory class began with "Pick an enumeration of the rationals ..." $\endgroup$
    – Asaf Karagila
    Aug 13, 2015 at 15:47
  • $\begingroup$ What topological space does $\beta \mathbb{N}$ denote? $\endgroup$ Aug 14, 2015 at 3:28
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    $\begingroup$ @DavidZhang The Stone–Čech compactification of the natural numbers. $\endgroup$
    – Dennis
    Aug 14, 2015 at 4:34

$\sin(n) _{n=1}^{\infty}$

(originally entered $\sin(1/n)$, because I was thinking of $\sin(1/x)$ as $x \to 0$.)

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    $\begingroup$ Nice. But a bit of explanation might be welcome. $\endgroup$
    – vonbrand
    Aug 13, 2015 at 18:40
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    $\begingroup$ This is called the "Topologists Sine Curve" (for purposes of googling) in the ->0 limit you can find many sources explaining that it has all of [-1,1] as its limit. You can adapt the same argument to get the same thing for the -> infinity limit. $\endgroup$ Aug 14, 2015 at 16:23

The example that convinced me that it was the case was the sequence

$$ \left(0, 1, 0, \frac{1}{2}, 1, 0, \frac{1}{3}, \frac{2}{3}, 1, 0, \frac{1}{4},…\right)$$

which eventually reach all the rationals in $[0,1]$, and thus whose adherence is the whole $[0,1]$ segment. So it is a rational sequence which uncountably many adherence values.


Let $\{x\}$ denote the fractional part of a real number $x$, i.e., $$\{x\} = x - \lfloor x \rfloor.$$

Since $\pi$ is an irrational number, the sequence $$\{a_n\} = \{\{n\pi\}: n = 1, 2, 3, \ldots\}$$ is dense in $[0, 1]$. (for the proof, see here.) Therefore, every point of $[0, 1]$ is a limit of some subsequence of $\{a_n\}$.

This example provides an example that the set of limits of subsequences can be made uncountably infinite.


A simple and explicit example with rational numbers is to list the fractions $\frac{a}{b}$ in $[0,1)$, first with denominator $b=1$, then the ones with denominator $b=10$, then the ones with $b=100$, then (et cetera), listing the fractions for each value of $b$ in increasing order.

Given the decimal representation of a number $r = 0.d_1 d_2 d_3...$ its $n$ digit approximation appears in list at position $(111...111 + (d_1 d_2 \cdots d_n)_{10} + 1)$ where there are $n$ ones repeated, and $(\dots)_{10}$ is the integer whose base $10$ digits are $d_1 d_2 \cdots d_n$.

This is a formula for extracting from the list a subsequence converging to any given value of $r$ in $[0,1]$ that is specified by a sequence of base 10 digits.


One approach - similar in spirit to some others - is to begin with an ordered list of infinitely many sequences, each of which converges to something different. Let us write them row by row.

So: I have in mind a first row whose elements, read left to right, converge to $1$; a second row whose elements, read left to right, converge to $1/2$; and, more generally, a $k$th row whose elements, read left to right, converge to $1/k$.

Next, let us produce our desired sequence by going through this table (a table that extends indefinitely to the right, and indefinitely downwards) by using the same sort of "dove-tailing" approach often used to provide an enumeration of the (positive) rational numbers.

And we won't even erase numbers that have already appeared!

In this way, we produce a single sequence consisting of all numbers in the original table, which means, in particular, there are infinitely many subsequences converging to different limits (by construction, we have, at least, a subsequence converging to $1/k$ for each $k \in \mathbb{Z}^{+}$).

Bonus: The construction described here could be imitated for any countably infinite set of limits!

  • $\begingroup$ If you take the convergent sequences to be constant, this is exactly the "minimalist universal" solution (the dovetailing a la Cantor lists the constants in the same order as that solution, and the difference between 1 2 3 ... and 1, 1/2, 1/3, ... is one of naming). The deviation from minimalism in this answer is that the constant subsequences can be loosened slightly to convergent subsequences with the same limits. $\endgroup$ Aug 15, 2015 at 3:27
  • $\begingroup$ @ASCIIAdvocate Yes, and the re-naming could be accomplished by the bonus listed at the end. My response is mostly to indicate yet another way of thinking about the question and its answer; similarly, though dove-tailing is often used to enumerate the positive fractions, one could accomplish the same by listing $a/b$ ordered first by $a+b<n$ as $n$ increases, and, if $a+b = c+d$, listing $a/b$ before $c/d$ provided $a<c$. The enumeration begins: $1/1,1/2, 2/1, 1/3, 2/2, 3/1, \ldots$ (I believe that multiple ways of thinking about the problem and an answer produce a deeper understanding.) $\endgroup$ Aug 15, 2015 at 6:17

My thought is we could take the supposed countable set, and then choose some ϵ>0 so that no two limits live in the same ϵ-ball.

I think your argument here is correct, if you can find such an $\epsilon$. You give no justification that such $\epsilon$ must exist, and indeed, it in general doesn't (Asaf Karagila's answer for the sequence of rationals shows this).

Or for a simpler example, take a sequence similar to bartgol's:

$$\left\{\frac{1}{1}, \frac{1}{1}, \frac{1}{2}, \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \cdots\right\}.$$

Clearly each $\frac{1}{n}$ is a limit of a subsequence of this sequence, and you can't find an $\epsilon > 0$ that separates every $\frac{1}{n}$.

  • $\begingroup$ The answer by bartgol has such an $\epsilon$, but the argument still fails... $\endgroup$ Aug 13, 2015 at 23:23
  • $\begingroup$ Yes, you make a good point about the $\epsilon$. Still, even with the $\epsilon$, I think I see that my diagonalization application must be incorrect even with correct premises. If I figure out how to articulate it, I will post an edit. $\endgroup$
    – Pburg
    Aug 14, 2015 at 13:02
  • $\begingroup$ I see. Yes, the $\epsilon$ argument only gives you a finite number of points of the set of points is bounded (as there can only be a finite number of $\epsilon$ balls in a bounded subset of $\mathbb{R}$). If it isn't, the set can be countable, but not uncountable. I'm not sure about using diagonalization as an argument. $\endgroup$
    – asmeurer
    Aug 14, 2015 at 19:03

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