I'm studying the proof of the famous "$5B$ covering lemma":

let $\mathcal{G}$ be a family of closed balls in $\mathbb{R}^n$ satisfying: $$ \sup\{ \text{diam}(B) : B \in \mathcal{G} \} < +\infty $$ then there exists a disjoint, at most countable subfamily $\mathcal{F} \subset \mathcal{G}$ such that $$ \bigcup_{B \in \mathcal{G}} B \subset \bigcup_{B \in \mathcal{F}} 5B $$ The proof starts defining the sequence $(\mathcal{G}_i)_i$: $$ \mathcal{G}_i := \{ B \in \mathcal{G}: \frac{S}{2^{i}} < \text{diam}(B)\le \frac{S}{2^{i-1}} \} $$ where $S=\sup\{ \text{diam}(B) : B \in \mathcal{G} \}$. Now, by means of the Haussdorf's maximal principle, we extract a disjoint subfamily $\mathcal{F}_1 \subset \mathcal{G}_1$ (if $\mathcal{G}_1$ is non empty), which is maximal w.r.t inclusion. Haussdorf's maximal principle (equivalent to the axiom of choice) states that:

Given a family of sets or a family of collection of sets $\mathcal{S}$ such that every subfamily $\mathcal{S}_0$ totally ordered w.r.t inclusion has the property $\cup \{ E : E \in \mathcal{S}_0 \} \in \mathcal{S}$, then there exists an element $S^{*} \in \mathcal{S}$ which is maximal w.r.t inclusion order.

Back to the proof, we consider:

$$ \mathcal{S}= \{ \mathcal{E} \subset \mathcal{G} : \mathcal{E} \ \text{contains disjoint balls} \} $$ it is immediate to see that $\mathcal{S}$ satisfies the hypothesis of Haussdorf's maximal principle (in fact, the totally ordered subsets of $\mathcal{S}$ are those formed by "chains" of elements) , hence there is a maximal element $\mathcal{E}^{*} \in \mathcal{S}$, which contains disjoint balls by definition. We then set $\mathcal{F}_1= \mathcal{E}^{*}$.

I've got two questions in mind: is the previous application of Haussdorf's maximal principle correct (the hypothesis seems to be trivially verified, but maybe I'm missing something)? Is the principle even necessary to extract such maximal subfamily?


1 Answer 1


The use seems correct. But it seems that Zorn's lemma is a bit more appropriate. The point here is that the union over a chain of families which are pairwise disjoint is also a pairwise disjoint family. And if the chain was maximal, then the union is a maximal family.

Of course, this needs to be verified, for sake of completeness, because the existence of maximal chains need not imply the existence of maximal elements. Just look at $\Bbb N$ itself as a chain in itself. It is maximal, but there are no maximal elements.

As for the necessity, yes, some choice is necessary. It is consistent that there is a Dedekind-finite set of reals which is dense. Namely, it is a dense subset which has no countably infinite subests. Now suppose that $D$ is such set, and take $\cal G$ to be $\{B_1(x)\mid x\in D\}$. This is in fact a covering of $\Bbb R$. But any countable subfamily of $\cal G$ is finite, and cannot cover the whole line, even if scaled by a factor of $100^{100^{100}}$.

  • $\begingroup$ Thank you for your answer and your deep examples. How it is consistent the existence of a finite-dense set of reals? I'm not an expert in set theory and this seems strange to me.. $\endgroup$
    – GaC
    Sep 25, 2017 at 9:24
  • 1
    $\begingroup$ Dedekind-finite, not finite. Without assuming choice, the two notions are distinct. $\endgroup$
    – Asaf Karagila
    Sep 25, 2017 at 9:25

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