Non separable locally compact connected metric space

Solution 1:

Well, no one will be able to provide an example, since it doesn't exist. In the appendix A to chapter $1$ of his book 'A Comprehensive Introduction to Differential Geometry, volume $1$' Spivak proves that a connected, locally compact, paracompact space is $\sigma$-compact. This last term means that you can write your space as a countable union of compact subspaces. It's easy to see that every $\sigma$-compact space is Lindelöf.

Here you are considering a connected, locally compact metric space, so it is $\sigma$-compact and then Lindelöf. But, in metric spaces, it is equivalent to being separable.

Solution 2:

Meanwhile, the long line is an example of a space that is non-separable, locally compact, connected and locally metrizable, but not metrizable.

The long line is the space $[0,1)\times\omega_1$, consisting of $\omega_1$ many copies of the half-open unit interval, using the (inverse lexical) order topology, where $\omega_1$ is the first uncountable ordinal. These intervals fit together neatly, so that every initial segment of the order looks exactly like an interval in the real line. Thus, the long line is locally compact, locally metrizable and connected. But the long line is non-separable, since every countable subset of $\omega_1$ and hence of the long line is bounded. Lastly, it is not metrizable, since otherwise this would contradict Nuno's answer.

Solution 3:

A connected locally compact group is $\sigma$-compact. Here is an argument that you can read even if you don't have Spivak's book on your desk.

Let $G$ be a locally compact group. Choose a compact neighbourhood $S$ of the identity. The subgroup $H$ of $G$ generated by $S$ is an open subgroup of $G$. Since the unit component $G_0$ of $G$ is the intersection of all open subgroups of $G$ (local compactnes is used here), $G_0$ is contained in $H$.

Assume moreover that $G$ is connected. Then $H = G_0$. In particular $G$ is compactly generated. A fortiori $G$ is $\sigma$-compact.