Let $X$ be a paracompact topological space, let $C$ be an open cover, and let $\mathscr B$ be a basis for the topology. Does there always exist a locally finite refinement that consists of basis sets?

The reason this is trickier than it sounds is that, while every set in the refinement is a union of basis sets, this is definitely not guaranteed to be a locally-finite sort of union, and taking a locally finite refinement of this lands you outside the world of basis sets...

Now, let me add that I don't quite remember the context that led me to consider this question, but I'm still interested in finding out the answer.

  • 4
    $\begingroup$ What if $X$ is an infinite-dimensional Hilbert space, and $\mathscr B$ is the set of all open balls, and $C=\{X\}$? Does $X$ have a locally finite cover by open balls? $\endgroup$ – bof Jun 24 '19 at 11:47
  • $\begingroup$ @bof Interesting. Is $X$ paracompact? $\endgroup$ – silvascientist Jun 24 '19 at 13:11
  • 1
    $\begingroup$ Metric spaces are paracompact. $\endgroup$ – bof Jun 24 '19 at 13:31

The property in question was introduced by Ralph Ford in his dissertation, 1963. A space is totally paracompact if every open basis has a locally finite subcover.

The property may be further differentiated, depending on whether you start your definition with "every open basis $\mathcal B$" (which I assume you mean, and which coincides with total paracompactness), or you say "there is an open basis $\mathcal B$, etc", which requires consideration of some other conditions to make the latter property different from paracompactness, given that the topology itself is a basis. One variation is base-paracompactness, when you say there is a base $\mathcal B$ of cardinality the weight of the space such that every open cover has a localy-finite refinement with elements of this base. Another variation is if one says there is a base $\mathcal B$ such that every base $\mathcal B'\subseteq\mathcal B$ has a locally-finite subcover, this is called base-base paracompactness.

The comment by bof above provides an example of a paracompact space that is not totally paracompact and answers your question. I will put together some links to the literature where you can read more.

H. H. Corson, T. J. McMinn, E. A. Michael, J. Nagata,
Bases and local finiteness.
Notices Amer. Math. Soc. 6 (1959), 814 (abstract).
One result announced there:
The base of all bounded, open, convex sets in a reflexive, infinite-dimensional Banach space is coarse (i.e. it has no locally-finite subcover). Also the irrationals have a coarse base.

H. H. Corson, Collections of convex sets which cover a Banach space.
Fund. Math. 49 (1960/1961), 143–145.
This paper has a proof of the result announced earlier.

R. Ford, Basic properties in dimension theory,
Dissertation, Auburn Univ., (1963).
He proved that small and large inductive dimension coincide for totally paracompact metric spaces.
http://topology.auburn.edu/tp/reprints/v25/tp25101.pdf (the above is not a link to the dissertation itself, and the following is a link that I don't know if it works)

A.Lelek, On totally paracompact metric spaces.
Proc. Amer. Math. Soc. 19 (1968), 168–170
A. Lelek, Some cover properties of spaces
Fundamenta Mathematicae (1969)
Volume: 64, Issue: 2, page 209-218
One of the properties considered is total paracompactness.

John O’Farrell,
Some methods of determining total paracompactness,
Dissertation, Auburn University (1982).
John O’Farrell, The Sorgenfrey line is not totally metacompact,
Houston J. Math 9 (1983), no.2, 271-273.
The irrationals with the usual topology are not totally paracompact. The Sorgenfrey line is not totally paracompact.

Zoltan Balogh and Harold Bennett
Total Paracompactness of Real GO-Spaces
Proceedings of the American Mathematical Society
Vol. 101, No. 4 (Dec., 1987), pp. 753-760
Real GO-spaces here means take the reals as a set, with a generalized-ordered topology: Any topology stronger than the usual, that has a basis of order-convex sets. Some examples are the Sorgenfrey line and the Michael line.

Francisco Gallego Lupiañez
Total paracompactness and Banach spaces
Proc. Amer. Math. Soc. 103 (1988), 210-214
(Seems very relevant, though I just discovered it with google.)

John E. Porter, Base-paracompact spaces
Topology and its Applications 128 (2003) 145–156
Base-paracompactness is a less restrictive property, there is a base $\mathcal B$ of cardinality of the weight, such that every open cover has a locally-finite refinement with elements of $\mathcal B$. At present 6/26/2019, it is an open question if every paracompact space is base-paracompact. The introduction lists some literature about total paracompactness.

Strashimir G. Popvassilev, Base-family paracompactness
Houston Journal of Mathematics Volume 32, No. 2, 2006
Base-family paracompactness is a related property, and the introductions lists some of the results and literature relevant to total paracompactness.

Strashimir G. Popvassilev, Base-cover paracompactness,
Proc. Amer. Math. Soc. 132 (2004), 3121-3130
Another related property and discussion of literature.

Strashimir G. Popvassilev
Base-base paracompactness and subsets of the Sorgenfrey line
Mathematica Bohemica (2012) Volume: 137, Issue: 4, page 395-401
Another related property and an open question.

Gary Gruenhage, Generalized metrizable spaces
In Recent progress of general topology III,
see section 20. Base paracompact

  • $\begingroup$ Hi Mirko. Thanks for this comprehensive answer! It certainly piques my interest. $\endgroup$ – YuiTo Cheng Jun 26 '19 at 12:28
  • $\begingroup$ @YuiToCheng You are very welcome, my pleasure $\endgroup$ – Mirko Jun 26 '19 at 17:48

Interesting. Going back to my reading of Lee's Introduction to Topological Manifolds, I found the following theorem which I now realize is what prompted me to go here and ask this question which had been lingering in the back of my brain for a while. Note that this is not what made me originally consider the question.

Theorem 1.15 (Manifolds are Paracompact). Every topological manifold is paracompact. In fact, given a topological manifold $M$, an open cover $\mathcal X$ of $M$, and any basis $\mathscr B$ for the topology of $M$, there exists a countable, locally finite open refinement of $\mathcal X$ consisting of elements of $\mathscr B$.

Here is the proof (as given by Lee):

Given $M$, $\mathcal X$, and $\mathscr B$ as in the hypothesis of the theorem, let $\left(K_j\right)_{j=1}^\infty$ be an exhaustion of $M$ by compact sets (Proposition A.60). For each $j$, let $V_j = K_{j+1}/\operatorname{Int}(K_j)$ and $W_j = \operatorname{Int}(K_{j+2}) / K_{j-1}$ (where we interpret $K_j$ as $\emptyset$ if $j < 1$). Then $V_j$ is a compact set contained in the open subset $W_j$. For each $x \in V_j$, there is some $X_x \in \mathcal X$ containing $x$, and because $\mathscr B$ is a basis, there exists $B_x \in \mathscr B$ such that $x \in B_x \subseteq X_x \cap W_j$. The collection of all such sets $B_x$ as $x$ ranges over $V_j$ is an open cover of $V_j$, and thus has a finite subcover. The union of all such finite subcovers as $j$ ranges over the positive integers is a countable open cover of $M$ that refines $\mathcal X$. Because the finite subcover of $V_j$ consists of sets contained in $W_j$, and $W_j \cap W_{j'} = \emptyset$ except when $j-2 \le j' \le j+2$, the resulting cover is locally finite $\blacksquare$

This seems to apply not only to manifolds, but also to second-countable locally-compact Hausdorff spaces (for which the Proposition A.60 holds, namely that they admit an exhaustion by compact sets). However, local compactness here seems key to making a locally finite refinement out of basis elements, so it seems like the statement may very well not hold in general. I have no idea how one might prove this one way or the other, though.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.