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I want to show that not every locally compact topological space is compact.

I have one example which I am not sure if it is a correct example which is simply the whole real line.

Is it a correct example? Can you provide more examples of such topological spaces?

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$\Bbb R$ is indeed locally compact but not compact. –  Adam Hughes Aug 31 at 20:37
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Probably the simplest is infinite discrete space. There are loads of examples, for example noncompact locally compact topological groups. And you might find this useful for future counterexample hunting. –  Marcin Łoś Aug 31 at 20:41

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There are TONS of examples of locally compact spaces which are not compact. I'll start with the easy ones and build some more complicated ones for those interested in alternate examples.

As you suspect, $\Bbb R$ is one such. Take any real number $x\in\Bbb R$, and any $\epsilon >0$, then the set $[x-\epsilon, x+\epsilon]$ is a compact neighborhood of $x$--it contains the open set $(x-\epsilon, x+\epsilon)$--and clearly the open covering

$$\left\{(n, n+1):n\in\Bbb Z\right\}\cup\left\{\left(n-{1\over 2}, n+{1\over 2}\right):n\in\Bbb Z\right\}$$

has no finite subcover, in fact it has no proper subset which is still a cover! To see this, note that if you omit some of the sets of the form $\left(n-{1\over 2},n+{1\over 2}\right)$ then you cannot get the integer $n$, and if you omit a set of the form $(n,n+1)$ you miss the real number $n+{1\over 2}$.

You can easily extrapolate this to higher dimensional Euclidean spaces.

In fact, if you want to kill this without any explicit examples at all, you can just use the Heini-Borel theorem which forces all compact sets to be bounded in Euclidean spaces.


For other options, there is always taking any collection of compact sets and taking a disjoint union of infinitely many of them. By definition of the topology on a union (it's a direct limit topology) the set is locally compact, but it is very much not compact.


Take any locally compact field which is not compact, such as the $p$-adic numbers, $\Bbb Q_p$. Then by some abstract theory such as can be found in the first chapter of André Weil's Basic Number Theory, the field has an absolute value on it. And the absolute value is continuous, hence the image of a compact set is compact.

However, the image of the whole field is the set $$\{p^n: n\in\Bbb Z\}\cup\{0\}\subseteq \Bbb R$$ which is not even bounded, and we know that in a metric space, compact sets are bounded! In fact, since the multiplicative group of the field is also locally compact, if it were compact we'd have to map to a compact subgroup of $\Bbb R^+$, for which there is only the trivial subgroup! I.e. it's only ever possible to be compact if you have the trivial absolute value!


In the same vein as the previous two examples combined: you can take a space like the integers, $\Bbb Z$--a disjoint union of one-point sets in their usual topology--which are naturally endowed with the discrete topology, then noting that compact sets which are discrete are automatically finite you get that they (and any other infinite discrete space) are not compact. Furthermore, since discrete sets are closed and closed subsets of compact sets are compact, this gives you another proof that $\Bbb R$ is not compact, since $\Bbb Z$ is a closed subgroup in $\Bbb R$.

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The real line is indeed such an example, as is $\mathbb{R}^n$ for all $n > 0$. Take any compact Hausdorff space $X$, pick a non-isolated point $x$ in it, then $X \setminus \{x\}$ is another example. In fact any proper open, non-closed subset of any compact Hausdorff space.

The last follows as open subsets of compact Hausdorff spaces are locally compact, and if the subspace is not closed, it cannot be compact (in a Hausdorff space).

(It turns out that every example is essentially this: a locally compact non-compact Hausdorff space has a one-point compactification, so is of the form $K\setminus\{x\}$ for some compact Hausdorff $K$ and $x \in K$.)

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