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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 $(\forall^infty (\forall^\infty n)f(n)\not=g(n)$. 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.

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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 $f\in\mathcal{F}$. 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$. QED. 2 added 47 characters in body; deleted 71 characters in body; added 3 characters in body Let's work on Baire space$\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}'$is all possible countable sets. (The definition makes sense, although$\mathcal{C}'$is not an ideal).$\mathrm{cov}(\mathcal{M})$is the least size of a collection of meager sets whose union equals$\omega^\omega$. I claim that all three numbers are equal.$\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})$is almost as straightforward; since sets of size 1 are closed any collection of meager sets cofinal in the countable closed sets covers all the setswhole space.$\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}$. So fix such a family. Also fix a partition of$\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.

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