Paracompact space

In mathematics, a paracompact space is a topological space in which every open cover admits a locally finite open refinement. Paracompact spaces are sometimes also required to be Hausdorff. Paracompact spaces were introduced by Dieudonné (1944).

Contents

Definitions of relevant terms

\bigcup_{\alpha \in A}U_{\alpha} = X.
\left\{ \alpha \in A�: U_{\alpha} \cap V(x) \neq \varnothing \right\}
is finite.

Note the similarity between the definitions of compact and paracompact: for paracompact, we replace "subcover" by "open refinement" and "finite" by "locally finite". Both of these changes are significant: if we take the above definition of paracompact and change "open refinement" back to "subcover", or "locally finite" back to "finite", we end up with the compact spaces in both cases.

A hereditarily paracompact space is a space such that every subspace of it is paracompact. This is equivalent to requiring that every open subspace be paracompact.

Examples and counterexamples

Some examples of spaces that are not paracompact include:

Properties

Partitions of unity

Paracompactness has little to do with the notion of compactness, but rather more to do with breaking up topological space entities into manageable pieces. The most important feature of paracompact Hausdorff spaces is that they are normal and admit partitions of unity subordinate to any open cover. This means the following: if X is a paracompact Hausdorff space with a given open cover, then there exists a collection of continuous functions on X with values in the unit interval [0, 1] such that:

In fact, a Hausdorff space is paracompact if and only if it admits partitions of unity subordinate to any open cover (see below). This property is sometimes used to define paracompact spaces (at least in the Hausdorff case).

Partitions of unity are useful because they often allow one to extend local constructions to the whole space. For instance, the integral of differential forms on paracompact manifolds is first defined locally (where the manifold looks like Euclidean space and the integral is well known), and this definition is then extended to the whole space via a partition of unity.

Proof that paracompact hausdorff spaces admit partitions of unity

An hausdorff space X\, is paracompact if and only if it every open cover admits a subordinate partition of unity. The if direction is straightforward. Now for the only if direction, we do this in a few stages.

Lemma 1: If \mathcal{O}\, is a locally finite open cover, then there exists open sets W_{U}\, for each U\in\mathcal{O}\,, such that each \bar{W_{U}}\subseteq U\, and \{W_{U}:U\in\mathcal{O}\}\, is a locally finite refinement.
Lemma 2: If \mathcal{O}\, is a locally finite open cover, then there are continuous functions f_{U}:X\to[0,1]\, such that \operatorname{supp}~f_{U}\subseteq U\, and such that f:=\sum_{U\in\mathcal{O}}f_{U}\, is a continuous function which is always non-zero and finite.
Theorem: In a paracompact hausdorff space X\,, if \mathcal{O}\, is an open cover, then there exists a partition of unity subordinate to it.
Proof (Lemma 1): Let \mathcal{V}\, be the collection of open sets meeting only finitely many sets in \mathcal{O}\,, and whose closure is contained in a set in \mathcal{O}. One can check as an exercise that this provides an open refinement, since paracompact hausdorff spaces are regular, and since \mathcal{O}\, is locally finite. Now replace \mathcal{V}\, by a locally finite open refinitement. One can easily check that each set in this refinement has the same property as that which characterised the original cover.
Now we define W_{U}=\bigcup\{A\in\mathcal{V}:\bar{A}\subseteq U\}\,. We have that each \bar{W_{U}}\subseteq U\,; for otherwise letting x\in U\setminus\bar{W_{U}}\,, we take V\in\mathcal{V},\ni x\, with closure contained in U\,; but then (x\in )V\subseteq W_{U}(\subseteq\bar{W_{U}}\not\ni x)\, a contradiction. And it easy to see that \{W_{U}:U\in\mathcal{O}\}\, is an open refinement of \mathcal{O}\,.
Finally, to verify that this cover is locally finite, fix x\in X\,; let N\, a neighbourhood of x\, meeting only finitely many sets in \mathcal{V}\,. We will show that N meets only finitely many of the W_{U}\,. If W_{U}\, meets N\,, then some A\in\mathcal{V}\, with \bar{A}\subseteq U\, meets N\,. Thus \{U\in\mathcal{O}:U\text{ meets }N\}\, is the same as \bigcup_{A\in\mathcal{V}:A\text{ meets }N}\{U\in\mathcal{O}:\bar{A}\subseteq U\}\, which is contained in \bigcup_{A\in\mathcal{V}:A\text{ meets }N}\{U\in\mathcal{O}:A\text{ meets }U\}\,. By the setup of \mathcal{V}\,, each A\in\mathcal{V}\, meets only finitely many sets in \mathcal{O}\,. Hence the right-hand collection is a finite union of finite sets. Thus \{W_{U}:U\in\mathcal{O},\text{ meets }N\}\, is finite. Hence the cover is locally finite.

 

 

 

 

\blacksquare\, (Lem 1)

Proof (Lemma 2): Applying Lemma 1, let f_{U}:X\to[0,1]\, be coninuous maps with f_{U}\upharpoonright\bar{W}_{U}=1\, and \operatorname{supp}~f_{U}\subseteq U\, (by Urysohn's lemma for disjoint closed sets in normal spaces, which a paracompact hausdorff space is). Note by the support of a function, we here mean the points not mapping to zero (and not the closure of this set). To show that f=\sum_{U\in\mathcal{O}}f_{U}\, is always finite and non-zero, take x\in X\,, and let N\, a neighbourhood of x\, meeting only finitely many sets in \mathcal{O}\,; thus x\, belongs to only finitely many sets in \mathcal{O}\,; thus f_{U}(x)=0\, for all but finitely many U\,; moreover x\in W_{U}\, for some U\,, thus f_{U}(x)=1\,; so f(x)\, is finite and \geq 1\,. To establish continuity, take x,N\, as before, and let S=\{U\in\mathcal{O}:N\text{ meets }U\}\,, which is finite; then f\upharpoonright N=\sum_{U\in S}f_{U}\upharpoonright N\,, which is a continuous function; hence the preimage under f\, of a neighbourhood of f(x)\, will be a neighbourhood of x\,.

 

 

 

 

\blacksquare\, (Lem 2)

Proof (Theorem): Take \mathcal{O}*\, a locally finite subcover of the refinement cover: \{V\text{ open }:(\exists{U\in\mathcal{O}})\bar{V}\subseteq U\}\,. Applying Lemma 2, we obtain continuous functions f_{W}:X\to[0,1]\, with \operatorname{supp}~f_{W}\subseteq W\, (thus the usual closed version of the support is contained in some U\in\mathcal{O}\,, for each W\in\mathcal{O}*\,; for which their sum constitutes a continuous function which is always finite non-zero (hence 1/f\, is continuous positive, finite-valued). So replacing each f_{W}\, by f_{W}/f\,, we have now — all things remaining the same — that their sum is everywhere 1\,. Finally for x\in X\,, letting N\, be a neighbourhood of x\, meeting only finitely many sets in \mathcal{O}*\,, we have f_{W}\upharpoonright N=0\, for all but finitely many W\in\mathcal{O}*\, since each \operatorname{supp}~f_{W}\subseteq W\,. Thus we have a partition of unity subordinate to the original open cover.

 

 

 

 

\blacksquare\, (Thm)

Comparison with compactness

Paracompactness is similar to compactness in the following respects:

It is different in these respects:

Product related properties

Although a product of paracompact spaces need not be paracompact, the following are true:

Both these results can be proved by the tube lemma which is used in the proof that a product of finitely many compact spaces is compact.

Variations

There are several variations of the notion of paracompactness. To define them, we first need to extend the list of terms above:

\mathbf{U}^{*}(x)�:= \bigcup_{U_{\alpha} \ni x}U_{\alpha}.
The notation for the star is not standardised in the literature, and this is just one possibility.
\left\{ \alpha \in A�: x \in U_{\alpha} \right\}
is finite.

A topological space is:

The adverb "countably" can be added to any of the adjectives "paracompact", "metacompact", and "fully normal" to make the requirement apply only to countable open covers.

Every paracompact space is metacompact, and every metacompact space is orthocompact.

As the name implies, a fully normal space is normal. Every fully T4 space is paracompact. In fact, for Hausdorff spaces, paracompactness and full normality are equivalent. Thus, a fully T4 space is the same thing as a paracompact Hausdorff space.

As an historical note: fully normal spaces were defined before paracompact spaces. The proof that all metrizable spaces are fully normal is easy. When it was proved by A.H. Stone that for Hausdorff spaces fully normal and paracompact are equivalent, he implicitly proved that all metrizable spaces are paracompact. Later M.E. Rudin gave a direct proof of the latter fact.

See also

Notes

  1. ^ Rudin, Mary Ellen. A new proof that metric spaces are paracompact. Proceedings of the American Mathematical Society, Vol. 20, No. 2. (Feb., 1969), p. 603.
  2. ^ C. Good, I. J. Tree, and W. S. Watson. On Stone's Theorem and the Axiom of Choice. Proceedings of the American Mathematical Society, Vol. 126, No. 4. (April, 1998), pp. 1211–1218.
  3. ^ Hatcher, Allen, Vector bundles and K-theory, preliminary version available on the authors homepage

References