5
$\begingroup$

(The following question arose in a joint research with Adam Przeździecki and Boaz Tsaban.)

For a $\sigma$-ideal $\mathcal{I}$ of subsets of the unit interval $[0,1]$, define
$$\newcommand{\card}[1]{\left|#1\right|}\newcommand{\cov}{\operatorname{cov}}\newcommand{\sub}{\subset} \cov(\mathcal{I}):=\min\{\card{\mathcal{A}}: \mathcal{A}\sub \mathcal{I}, \bigcup\mathcal{A}=[0,1]\}. $$ Let $\mathcal{E}$ be the family of all $F_\sigma$ Lebesgue null subsets of the unit interval, and $\mathcal{N}$ be the family of all Lebesgue null subsets of the unit interval.

  1. Let $\kappa_\mathcal{E}$ be the minimal cardinal number such that some measure one set is covered by $\kappa_\mathcal{E}$ elements of $\mathcal{E}$.
  2. Similarly, let $\kappa_\mathcal{N}$ be the minimal cardinal number such that some measure one set is covered by $\kappa_\mathcal{N}$ elements of $\mathcal{N}$.

We have $\kappa_\mathcal{E}\leq \cov(\mathcal{E})$, and $\kappa_\mathcal{N}=\cov(\mathcal{N})$.

Question 1: Is it provable that $\kappa_\mathcal{E}=\cov(\mathcal{E})$?

Question 2: Does the cardinal number $\cov(\mathcal{E})$ change if we work in a closed positive subset of the unit interval instead of the whole interval?

A negative answer for Question 2 implies a positive answer for Question 1.

$\endgroup$
10
  • $\begingroup$ The ideal generated by closed measure zero sets is discussed in Bartoszynski-Judah, section 2.6. $\endgroup$
    – Goldstern
    Dec 13, 2016 at 16:46
  • $\begingroup$ @Goldstern: Thanks, we already consulted this reference, and failed to find an answer there. $\endgroup$ Dec 13, 2016 at 17:13
  • $\begingroup$ For Q1, use the fact if a family of sets can cover a set of positive measure, then their rational translates can cover everything. $\endgroup$
    – Ashutosh
    Dec 13, 2016 at 17:22
  • $\begingroup$ @PiotrSzewczak I assumed that you have checked it. I thought it is worth mentioning, because other people might use the results or references there - either to find an answer, or perhaps to ask interesting related questions. $\endgroup$
    – Goldstern
    Dec 13, 2016 at 17:31
  • 1
    $\begingroup$ @Ashutosh: Could you explain it? For a meager set of positive measure, its rational translates cannot cover everything. $\endgroup$ Dec 13, 2016 at 18:14

1 Answer 1

3
$\begingroup$

The answer to the first question is yes: it is always true that $\kappa_{\mathcal E} = \mathrm{cov}(\mathcal E)$. The answer to the second question is no.

As Piotr points out, a negative answer to the second question implies a positive answer to the first. [This is because if $A \subseteq [0,1]$ is a set of measure one, then it contains a closed $K \subseteq [0,1]$ of positive measure (indeed, we can get the measure of $K$ as close to $1$ as we like). A negative answer to the second question means that it takes at least $\mathrm{cov}(\mathcal E)$ sets in $\mathcal E$ to cover $K$; since $K \subseteq A$, it takes at least $\mathrm{cov}(\mathcal E)$ sets in $\mathcal E$ to cover $A$ too.]

So it suffices to show that the answer to the second question is no: the $\mathcal E$-covering number is the same for any positive-measure closed subsets of $[0,1]$.

To see this, fix a closed $K \subseteq [0,1]$ of positive measure $\alpha$. Let $\mathcal F$ be a family of sets in $\mathcal E$ that covers $K$. Our goal is to cover $[0,1]$ with a family of sets in $\mathcal E$ of size $|\mathcal F|$. This shows that the $\mathcal E$-covering number for $[0,1]$ is no bigger than the analogous number for $K$. It is obvious that the $\mathcal E$-covering number for $[0,1]$ is no smaller than the analogous number for $K$, since any cover of $[0,1]$ by sets in $\mathcal E$ is also a cover of $K$.

Define a map $\phi: K \rightarrow [0,1]$ by $$\phi(x) = \frac{1}{\alpha} \cdot m(K \cap [0,x])$$ for all $x \in K$, where $m$ denotes the Lebesgue measure. Using the fact that $K$ is closed, one can show that $\phi$ is surjective. In particular, $$\phi(\mathcal F) = \{\phi[A \cap K] : A \in \mathcal F\}$$ is a family of sets that covers $[0,1]$. To finish the proof, it is enough to show that the map $A \mapsto \phi[A \cap K]$ sends sets in $\mathcal E$ to sets in $\mathcal E$: then $\phi(\mathcal F)$ is the desired $\mathcal E$-cover of $[0,1]$.

From the definition of $\phi$, we see $$m(\phi[(0,b)]) = \frac{1}{\alpha} \cdot m(K \cap (0,b))$$ and taking complements we get $$m(\phi[(a,1)]) = \frac{1}{\alpha} \cdot m(K \cap (a,1)).$$ and taking intersections, we see that for any interval $(a,b) \subseteq [0,1]$, $$m(\phi[(a,b) \cap K]) = \frac{1}{\alpha}(m((a,b) \cap K)).$$ That is, $\phi$ can increase the measure of an interval by at most a factor of $\frac{1}{\alpha}$. With a little more (routine) work, this shows that the map $A \mapsto \phi[A \cap K]$ sends null sets to null sets.

Observe that $\phi$ is continuous (the inverse image of an open interval in $[0,1]$ is an open interval in $K$). Therefore $\phi$ maps compact sets to compact sets. Therefore the map $A \mapsto \phi[A \cap K]$ sends compact sets to compact sets. Since $[0,1]$ is compact, this means that it also sends $F_\sigma$ sets to $F_\sigma$ sets. So we have showed that the map $A \mapsto \phi[A \cap K]$ sends null $F_\sigma$ sets to null $F_\sigma$ sets, and we are done.

$\endgroup$
0

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.