Can a sequence have infinitely many limits among its subsequences?

Question

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.

Answer 1

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!).

Answer 2

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.

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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.

Answer 3

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

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

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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!

Answer 8

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 $\frac{1}{n}$.