Can a sequence have infinitely many limits among its subsequences?

Solution 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!).

Solution 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.

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.

Solution 3:

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

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

Solution 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.