Example of a sequence with more than one limit.

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?

Solution 1:

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.