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do you have an example of a non separable locally compact connected metric space?

Thank you

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up vote 14 down vote accepted

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.

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This explain why it was so hard to find an example. Thank you :-) – Guillaume Brunerie Nov 18 '10 at 22:01

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.

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Oh, I see now that Qiaochu already mentioned this space in a comment. Should I delete? – JDH Nov 19 '10 at 0:36
I think you should leave it. The long line is a great example and you added useful information. – Nuno Nov 19 '10 at 1:38

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.

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This isn't so much an answer (so please don't vote this up) so much as a pointer on some things that don't work.

Whatever one cooks up it can't be an infinite-dimensional Banach space since those aren't locally compact. Uncountably many copies of the unit interval each horizontal line segment some fixed distance away from all others doesn't work either since it's not connected.

This slightly more clever construction is also no good: Consider the set $[0,1] \times [0,1]$ with the pseudometric $d'$ given by

$$ d'((x_1,y_1),(x_2,y_2)) = \begin{cases} \vert x_1 - x_2 \vert & \text{ if } y_1 = y_2 \lor x_1 = 0 \lor x_2 = 0 \\\\ 2 & \text{ if } y_1 \neq y_2 \land x_1,x_2 \neq 0, \end{cases} $$

and let $X$ be its Kolmogorov quotient and $d$ its associated metric. It satisfies all the desired properties except local compactness - $\{0\}\times [0,1]$ doesn't have a compact neighbourhood.

The difficulty in conjuring up an example seems to come from the fact that you want a nonseparable connected space where every point has a separable neighbourhood (compact metric implies 2nd-countable).

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You can easily come up with a nonseparable connected space where every point has a separable neighborhood by taking a "big enough" such space, e.g. the long line. The difficulty that I see is that the one-point compactification cannot be metrizable. – Qiaochu Yuan Nov 18 '10 at 21:50

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