12
$\begingroup$

Several existence theorems for partition of unity are known. For example (source),

Proposition 3.1. If $(X,\tau)$ is a paracompact topological space, then for every open cover $\{U_i \subset X\}_{i \in I}$ there is a subordinate partition of unity (def. 2.4).

The axiom of choice is used in the proof of this theorem.

Is there an existence theorem for partition of unity that can be proved without the axiom of choice (including variants such as countable choice, dependent choice, etc)?

For example, what if we strengthen the assumption to separable locally compact metric space?

$\endgroup$
1
  • 4
    $\begingroup$ For separable metric spaces, there's probably a Shoenfield absoluteness argument to remove choice from the proof. More generally, something similar probably works for any metric space with a dense well-ordered subset. (That said, if this does work, a direct proof is probably not that bad in the first place.) $\endgroup$ Commented May 17 at 7:16

1 Answer 1

15
$\begingroup$

The proofs rely, in the background, on Urysohn's Lemma, which follows from the Principle of Dependent Choices but is not provable without some Choice. It is false in the ordered Mostowski model, see Geordnete Läuchli Kontinuen by N. Brunner. More information can be found in Versions of normality and some weak forms of the axiom of choice by Howard, Keremedis, Rubin, and Rubin; the link is to the review on zbMath, the paper is behind a paywall.

Addendum In metric spaces one can define Urysohn functions, but there one runs into a difficulty with paracompactness: in On Stone’s theorem and the Axiom of Choice Good, Tree, and Watson prove that the paracompactness of metric spaces does not follow from the principle of Dependent Choices.

Summary For separable metric spaces the existence of partitions of unity can be proven without Choice, because locally finite refinements can be constructed from countable refinements, and because the continuous functions needed can be defined from the metric. In arbitrary metric spaces the proof of paracompactness needs a certain amount of choice, and in arbitrary paracompact spaces one needs choice twice: first in the proof of Urysohn's Lemma and second in the choice of a set of Urysohn functions that will yield the partition of unity.

$\endgroup$
2
  • 2
    $\begingroup$ Is there no chance to avoid Urysohn's lemma? For metric spaces, the existence of a function separating two given disjoint closed sets can be proved without AC. ( $\frac{d(x, A)}{d(x, A)+d(x, B)}$ is $0$ on $A$ , $1$ on $B$.) $\endgroup$
    – BonBon
    Commented May 17 at 10:31
  • 2
    $\begingroup$ @BonBon The paracompactness of metric spaces needs a certain amount of choice too. I'll add that to the answer. $\endgroup$
    – KP Hart
    Commented May 17 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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