Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recently, I have read Brendle's article Between p-points and nowhere dense ultrafilters [Isr. J. Math. 113, 205-230]. In this paper, he noted that $\mathrm{cof}(\mathcal{C},\mathcal{M}) = \mathrm{cov}(\mathcal{M})$, where $\mathcal{C}$ represents the set of all closed countable subsets of the real line. But I don't know how to prove this; a proof sketch would be appreciated.

I also want to know about $\mathrm{cof}(\mathcal{C}',\mathcal{M})$, where $\mathcal{C}'$ represents the set of all countable subsets of the real line.

share|improve this question
What is cof(C,M)? –  Stefan Geschke Mar 15 '12 at 19:05
@Stefan: cof(C,M) is the least cardinality of a subset F of M such that any element of C is contained in some element of F. –  Ramiro de la Vega Mar 15 '12 at 23:42
Jialiang: Brendle mentions that the easiest way to show cof(C,M)=cov(M) is to use Bartoszynski´s characterization of cov(M). Have you checked the Bartoszynski-Judah´s paper that Brendle refers to? –  Ramiro de la Vega Mar 15 '12 at 23:44
It does make sense: $M$ is the collection of meager sets, and countable sets are meager. –  Justin Palumbo Mar 16 '12 at 0:28
add comment

1 Answer

up vote 3 down vote accepted

Let's work on Baire space $\omega^\omega$. $\mathrm{cof}(\mathcal{C},\mathcal{M})$ is the least size of a collection of meager sets so that any countable closed set is a subset of the member of the collection. $\mathrm{cof}(\mathcal{C}',\mathcal{M})$ is the same, but where $\mathcal{C}'$ is all possible countable sets. $\mathrm{cov}(\mathcal{M})$ is the least size of a collection of meager sets whose union equals $\omega^\omega$.

I claim that all three numbers are equal. In other words, in terms of cardinality it doesn't matter whether we ask for a collection of meager sets cofinal over the countable sets, the closed countable sets, or even just the one element sets.

$\mathrm{cof}(\mathcal{C},\mathcal{M})\leq \mathrm{cof}(\mathcal{C}',\mathcal{M})$ is straight from the definition.

$\mathrm{cov}(\mathcal{M})\leq\mathrm{cof}(\mathcal{C},\mathcal{M})$ is almost as straightforward; since sets of size 1 are closed any collection of meager sets cofinal in the countable closed sets covers the whole space.

$\mathrm{cof}(\mathcal{C}',\mathcal{M})\leq \mathrm{cov}(\mathcal{M})$: We use a characterization of $\mathrm{cov}(\mathcal{M})$ due to Bartoszynski (this is Theorem 2.4.1 in his book with Judah): it is the least size of a family $\mathcal{F}\subseteq\omega^\omega$ so that any $g\in\omega^\omega$ is eventually different from some $f\in\mathcal{F}$, meaning $(\forall^\infty n)f(n)\not=g(n)$. So fix such a family. Also fix a partition of $\omega$ into countably many infinite pieces $\{A_k:k\in\omega\}$. For each $f\in\mathcal{F}$ let $M_f$ be the set of all $x\in\omega^\omega$ for which there exists a $k$ so that $(\forall^\infty n\in A_k)f(n)\not=x(n)$. Then $M_f$ is meager. We claim the collection of $M_f$ is cofinal over the countable sets.

Fix $C=\{h_k:k\in\omega\}$ a countable set. Let $g_C\in\omega^\omega$ be given by $g_C(n)=h_k(n)$ if $n\in A_k$. There is some $f\in\mathcal{F}$ so that $g_C$ is eventually different from $f$. And this implies that $C\subseteq M_f$. QED.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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