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I have heard of the idea of a sequence converging to more than one limit, but I cannot imagine how it would work.

Could someone give me an example of such a case, and explain how it works?

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Sequences cannot converge to more than one limit, since the definition of convergence allows one to prove that a limit is unique if one exists. However, a sequence may have multiple convergent subsequences with distinct limits. –  David H Apr 16 '14 at 20:31
This is not possible. Convergence implies the limit is unique. It is possible to have a sequence with two subsequences which converge independently to different values. E.g. the sequence $-1,0,-1,0,-1,0,\ldots$ has two subsequences which converge to different things but the sequence itself does not converge. –  Cameron Williams Apr 16 '14 at 20:32
So the idea is that some sequences have different limits (i.e. different converging subsequences), but the sequence (in its entirety) can only converge to one limit? –  ellya Apr 16 '14 at 20:33
You don't want to say that the sequence has a limit if it has two subsequences that converge to different values but that's the general idea. –  Cameron Williams Apr 16 '14 at 20:35

2 Answers 2

up vote 26 down vote accepted

I have heard of the idea of a sequence converging to more than one limit, but I cannot imagine how it would work.

You need a space that isn't Hausdorff for that.

If you don't know what a topological space is, come back to it later when you do. If you know, the simplest example is a space with at least two points and the indiscrete topology, then every sequence converges to every point.

All metric spaces (hence all subsets of an $\mathbb{R}^n$ in the usual topology) are Hausdorff, in those spaces, a sequence can have at most one limit.

A somewhat interesting example is the line with a doubled origin. We take two distinct symbols $0_1,0_2 \notin \mathbb{R}$, and let $X = (\mathbb{R}\setminus\{0\}) \cup \{0_1,0_2\}$. As a basis of open sets, we take the intervals $(a,b) \subset \mathbb{R}\setminus\{0\}$, and the sets of the form $(-\varepsilon,0) \cup \{0_k\} \cup (0,\varepsilon)$ for $\varepsilon > 0$ and $k = 1,2$. The open sets are then unions of such sets. The space is not Hausdorff, because every neighbourhood of $0_1$ intersects every neighbourhood of $0_2$ - the intersection contains a set of the form $(-\delta,0)\cup (0,\delta)$ for a $\delta > 0$, and the sequence $(2^{-n})_{n\in\mathbb{N}}$ for example converges to both, $0_1$ and $0_2$. The line with the doubled origin serves as an example of a space that is locally homeomorphic to $\mathbb{R}$ - every point has an open neighbourhood that is homeomorphic to $\mathbb{R}$ - but not Hausdorff, illustrating that the Hausdorff requirement in the definition of a Manifold is not redundant.

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You could also mention the possibility to have two distinct limits but for two distinct topologies. –  gniourf_gniourf Apr 16 '14 at 20:37
(BTW this is the only really valid answer, the other answers all assume Hausdorff). –  gniourf_gniourf Apr 16 '14 at 20:39
Can you expand on the structure of the sequence, I would like to see a concrete example, it helps to get my head around such things, cheers! –  ellya Apr 16 '14 at 20:40
@ellya In an indiscrete space, every sequence converges to every point. So that example is not particularly enlightening. But the problem is that interesting examples of non-Hausdorff spaces that arise in nature are rare and usually not easy to explain. Let me think a little whether I can come up with something that's not too artificial and simple enough. –  Daniel Fischer Apr 16 '14 at 20:47
In the indescrete argument, i.e $\tau=\{\emptyset,X\}$, is it that the only plausible sequence is $x_n=X$, in which case the limit is every point in $X$? Or have I missed something? –  ellya Apr 16 '14 at 20:52

As the other answers point out, if a sequence converges, it can only have one limit. However, a sequence that doesn't converge may have infinitely many convergent subsequences, all converging to different limits.

For instance, choose an enumeration of the rationals, $\{q_n\}$. Then for every real number $r$ there is a subsequence of $\{q_n\}$ converging to $r$.

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