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Basic statements
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A topological space is called locally compact if every point has a compact neighbourhood.
Or rather, if one does not at the same time assume that the space is Hausdorff topological space, then one needs to require that these compact neighbourhoods exist in a controlled way, e.g. such that one may find them inside every prescribed open neighbourhood (def. below) and possibily such that they are topological closures of smaller open neighbourhoods (def. below).
There are various definitions in use, which all coincide if the space is also Hausdorff (prop. below).
A locally compact Hausdorff space may also be called a local compactum; compare at compactum.
Local compactness is one of the conditions that are often required by default for working with topological spaces: locally compact spaces are a class of “nice topological spaces”.
(local compactness via compact neighbourhood base)
A topological space is locally compact if every point has a neighborhood base consisting of compact subspaces. This means that for every point $x \in X$ every open neighbourhood $U_x \supset \{x\}$ contains a compact neighbourhood $K_x \subset U_x$.
Alternatively:
(local compactness via compact closures inside neighbourhoods)
A topological space $X$ is called locally compact if for every point $x \in X$ and every open neighbourhood $U_x \supset \{x\}$ there exists a smaller open neighbourhood $V_x \subset U_x$ whose topological closure is compact and still contained in $U$:
If $X$ is a Hausdorff topological space then definition is equivalent to definition .
Generally definition implies definition . We need to show that Hausdorffness implies the converse.
Hence assume that for every point $x \in X$ then every open neighbourhood $U_x \supset \{x\}$ contains a compact neighbourhood. We need to show that it then also contains the closure $Cl(V_x)$ of a smaller open neighbourhood and such that this closure is compact.
So let $K_x \subset U_x$ be a compact neighbourhood. Being a neighbourhood, it has a non-trivial interior which is an open neighbouhood
Since compact subspaces of Hausdorff spaces are closed, it follows that $K_x \subset X$ is a closed subset. This implies that the topological closure of its interior as a subset of $X$ is still contained in $K_x$ (since the topological closure is the smallest closed subset containing the given subset): $Cl(Int(K_x)) \subset K_x$. Since subsets are closed in a closed subspace precisely if they are closed in the ambient space, $Cl(Int(K_x))$ is also closed as a subset of the compact subspace $K_x$. Now since closed subsets of compact spaces are compact, it follows that this closure is also compact as a subspace of $K_x$, and since continuous images of compact spaces are compact, it finally follows that it is also compact as a subspace of $X$:
(remark on terminology)
As for compact spaces (this remark), some authors choose to include the Hausdorff condition as a matter of course, calling locally compact not-necessarily-Hausdorff spaces ‘locally quasi-compact’. We will not follow that convention here, but the reader should be warned that without the Hausdorff hypothesis, there are several inequivalent notions of local compactness in the literature; see the English Wikipedia for a survey and counterexamples.
Note, however, that a topological space $X$ satisfying Definition is regular, because, as is immediate from the definition, closed neighbourhoods then form a neighbourhood basis of $X$, which is equivalent to regularity. Thus we only need that $X$ be $T_{0}$ for it to in fact be Hausdorff.
Every discrete space is locally compact.
(open subspaces of compact Hausdorff spaces are locally compact)
Every open topological subspace $X \underset{\text{open}}{\subset} K$ of a compact Hausdorff space $K$ is a locally compact topological space.
In particular every compact Hausdorff space itself is locally compact.
Conversely, every locally compact Haudorff space $X$ arises in this way, since it can be considered an open subspace in its one-point compactification $X \sqcup \{\infty\}$. See there this example.
The real numbers, complex numbers, and $\mathfrak{p}$-adic completions of algebraic number fields (with respect to a prime ideal $\mathfrak{p}$ in the ring of integers) are locally compact. In positive characteristic $p$, the field of Laurent series $\mathbb{F}_q((t))$ over a finite field with $q$ elements, topologized with respect to a discrete valuation, is locally compact. In fact, any non-discrete locally compact field must be of one of these types; they are called local fields
Finite product topological spaces of locally compact spaces are locally compact.
Closed subspaces of locally compact spaces are locally compact. (Hence locally compact spaces form a finitely complete category.)
(topological manifolds are locally compact topological spaces)
topological manifolds, being locally homeomorphic to the Euclidean metric spaces $\mathbb{R}^n$, are locally compact, via examples and .
In fact this applies also to locally Euclidean spaces which are not necessarily paracompact Hausdorff topological spaces, such as the long line.
The only Hausdorff topological vector spaces that are locally compact are finite-dimensional Euclidean spaces. More generally, a TVS is locally compact if and only if its Hausdorff reflection has finite dimension.
(countably infinite products of non-compact spaces are not locally compact)
Let $X$ be a topological space which is not compact. Then the product topological space of a countably infinite set of copies of $X$
is not locally compact.
Since the continuous image of a compact space is compact, and since the projection maps $p_i \;\colon\; \underset{\mathbb{N}}{\prod} X \longrightarrow X$ are continuous, it follows that every compact subspace of the product space is contained in one of the form
for $K_i \subset X$ compact.
But by the nature of the Tychonoff topology, a base for the topology on $\underset{\mathbb{N}}{\prod} X$ is given by subsets of the form
with $U_i \subset X$ open. Hence every compact neighbourhood in $\underset{\mathbb{N}}{\prod} X$ contains a subset of this kind, but if $X$ itself is non-compact, then none of these is contained in a product of compact subsets.
The space of rational numbers as a subspace of the real numbers with the Euclidean topology is not locally compact since its compact subsets all have empty interior.
(proper maps to locally compact spaces are closed)
Let
$(X,\tau_X)$ be a topological space,
$(Y,\tau_Y)$ a locally compact topological space according to def. ,
$f \colon X \to Y$ a continuous function.
Then:
If $f$ is a proper map, then it is a closed map.
Perhaps the most important consequence of local compactness (as defined above) for categorical topology is that locally compact spaces are exponentiable, i.e., if $Y$ is locally compact, then $Y \times -: Top \to Top$ has a right adjoint $(-)^Y: Top \to Top$. In fact, this is almost an abstract definition of local compactness: for sober spaces, local compactness is equivalent to being exponentiable. Cf. the situation for locales: a result of Hyland is that locale is locally compact if and only if it is exponentiable. (See exponential law for spaces and compact-open topology for more details.)
As noted above, locally compact spaces form a finitely complete full subcategory of $Top$. It is not true that arbitrary products of locally compact spaces are locally compact. However, some important examples of locally compact spaces are constructed as restricted direct products, as follows.
Let $(X_p, K_p)_{p \in P}$ be a collection of pairs of spaces where each $X_p$ is locally compact and $K_p \subseteq X_p$ is a compact open subspace. The restricted direct product of the collection is the colimit of the filtered diagram consisting of spaces
where $F$ ranges over all finite subsets of $P$, together with inclusions $D_F \subseteq D_{F'}$ where $F \subseteq F'$. We observe that each of the $D_F$ is locally compact, and that a filtered colimit or union of a system of open inclusions of locally compact spaces is again locally compact. Therefore, restricted direct products are locally compact, under the hypotheses stated above.
These hypotheses are of course pretty severe; important examples of such restricted direct products include topologized adele rings and idele groups. In the case of adele rings, the collection of pairs is $(K_{\mathfrak{p}}, O_{\mathfrak{p}})$ where $K_{\mathfrak{p}}$ is the $\mathfrak{p}$-adic completion of a number field $K$ and $O_{\mathfrak{p}}$ is the $\mathfrak{p}$-adic completion of the ring of integers $O \subseteq K$.
In any event, the category of locally compact spaces does not admit general infinite products. If it did, then so would the category of locally compact Hausdorff spaces, and so would the category of locally compact Hausdorff abelian groups. However, there is no product of countably many copies of the real numbers in $LCHAb$, for if there were, then by utilizing the universal property of the product, it would become a Hausdorff TVS over the real numbers, in contradiction to the fact that the only locally compact Hausdorff TVS are finite-dimensional.
Locally compact spaces are closed under coproducts in $Top$. They do not admit many types of colimits generally; in some sense this is a raison d'être for compactly generated spaces: they are precisely the colimits in $Top$ of diagrams of locally compact spaces.
Under Gelfand duality the category of compact Hausdorff topological spaces is equivalent to the opposite category of commutative C-star algebras. With some care there are generalizations of this also to locally compact topological spaces. See at Gelfand duality for more.
Locally compact Hausdorff spaces are paracompact whenever they are also second-countable.
Last revised on January 19, 2018 at 06:48:39. See the history of this page for a list of all contributions to it.