13
$\begingroup$

On the one hand, $\mathbb{Q}$ is neither complete (as a metric space) nor locally compact (as a topological space). On the other hand, $\mathbb{R}$ is both complete and locally compact.

My question is, what is the relationship between completeness and local compactness? e.g. Does one imply the other?

$\endgroup$

2 Answers 2

11
$\begingroup$

Note that the Baire space $\mathbb{N}^\mathbb{N}$ (which is homeomorphic to the space of irrationals) is a completely metrizable space which is not locally compact (in fact all compact subsets of $\mathbb{N}^\mathbb{N}$ have empty interior).

While locally compact metric spaces may not be complete (for example, as in Brad's answer, the open unit interval $(0,1)$ under the usual metric), these spaces will always be completely metrizable. There are at least two ways to see this result:

  1. Every locally compact metric space will be an open subset of its completion, and every G$_\delta$ subset of a complete metric space is completely metrizable.
  2. Every locally compact completely regular space is Čech-complete (i.e., is a G$_\delta$ subset of its Stone-Čech (or any other) compactification), and a metric space is completely metrizable iff it is Čech-complete.
$\endgroup$
2
  • $\begingroup$ So if a topological space is both locally compact, and also metrizable, then it is also completely metrizable? Or is there a subtlety I am missing? $\endgroup$ Commented Mar 29, 2013 at 9:29
  • $\begingroup$ @user18921: Yes, every locally compact metrizable space is completely metrizable. $\endgroup$
    – user642796
    Commented Mar 29, 2013 at 9:41
4
$\begingroup$

Neither implies the other. In addition to the examples you give, there are metric spaces which are complete and not locally compact (e.g. an infinite-dimensional Hilbert space) and metric spaces which are locally compact and not complete (e.g. the open interval $(0,1)$).

$\endgroup$
10
  • 3
    $\begingroup$ But locally compact metric spaces should be completely metrizable, if I am not mistaken. $\endgroup$
    – Asaf Karagila
    Commented Mar 28, 2013 at 12:21
  • $\begingroup$ @AsafKaragila wouldn't one need second countability or something similar? $\endgroup$ Commented Mar 28, 2013 at 12:34
  • $\begingroup$ @Alexander: I don't see why. Completeness is a local property. $\endgroup$
    – Asaf Karagila
    Commented Mar 28, 2013 at 12:36
  • 2
    $\begingroup$ A locally compact metric space will be open in its completion, and will therefore be itself completely metrizable (being a G$_\delta$ subset of a complete metric space). $\endgroup$
    – user642796
    Commented Mar 28, 2013 at 12:45
  • 2
    $\begingroup$ By the way, I would like to remind all users that completeness is not a topological concept at all, since a complete topological space may be homeomorphic to a non-complete one: think of $\mathbb R$ and $(0,1)$. Completeness makes sense for a metric space (and I don't mean metrizable space: the metric must be part of the structure) or more generally for a uniform space. Uniform spaces form a rather passé structure, which can still be found in Bourbaki (but probably not in many other other places) and due to André Weil. $\endgroup$ Commented Mar 28, 2013 at 19:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .