# How Nwd captures Meager

Let Nwd denote the collection of closed, nowhere dense subsets of $2^{\mathbb N}$, and Meager the $\sigma$-ideal generated by Nwd. Then all of the cardinal invariants of Meager can be expressed in terms of the inclusion partial ordering on Nwd.

To begin, every partial order P carries two very important invariants:

• cf(P) = the least cardinality of a cofinal subset of P
• add(P) = the least cardinality of an unbounded subset of P

These invariants are preserved under an appropriate notion of morphism. Namely, P is said to be Tukey reducible to Q if there is a function $f\colon P\to Q$ such that $f^{-1}$ carries bounded sets to bounded sets.

Theorem. If P and Q are Tukey bireducible, then cf(P)=cf(Q) and add(P)=add(Q).

(The proof is not difficult.)

Now, I had always thought that Nwd and Meager were so closely related that they should have the same invariants, and in fact, that they would be Tukey bireducible. This intuition is essentially correct, but the real story is just a little bit more technical than that. In the remainder of this post, I will explain this in detail.

My intuition turns out to be half-correct: in fact cf(Nwd)=cf(Meager), but this is not the case for add(). In fact since Meager is a $\sigma$-ideal, add(Meager) is always uncountable, whereas add(Nwd) is countable! (Just think of an enumeration of the rationals….) We can take care of this using an ad-hoc crutch, namely, let:

• $\sigma$-add(P) = the least cardinality of a subset of P which cannot be dominated by a countable subset of P

Then it is clear that $\sigma$-add(Nwd) is exactly add(Meager). What is more, $\sigma$-add() is again preserved under Tukey bireducibility and even the weaker (and more suitable) notion of $\sigma$-Tukey reducibility: P is said to be $\sigma$-Tukey reducible to Q if there is a function $f\colon P\to Q$ such that $f^{-1}$ carries bounded sets to sets which are dominated by a countable family.

Theorem. If P and Q are $\sigma$-Tukey bireducible, then cf(P)=cf(Q) (if they are infinite) and $\sigma$-add(P) = $\sigma$-add(Q).

Again, the proof is very direct. This tells us that $\sigma$-Tukey reducibility is just what we need to properly state relationship between Nwd and Meager.

The Main Theorem. Nwd is $\omega$-Tukey equivalent to Meager.

This is the sense in which Nwd captures all of the cardinal invariants of Meager. In the remainder of the post, let me outline a proof of the theorem.

First let’s find a $\sigma$-Tukey map from Meager to Nwd; in fact in this direction we will find a Tukey map. To proceed, we’ll need some notation. For any sequence $\tau\in2^{<\NN}$ and any subset $F\subset 2^\NN$, let $\tau\concat F$ denote the set $\set{\tau\concat x:x\in F}$ and let $F^{-\tau}$ denote the set $\set{x:\tau\concat x\in F}$.

We can now define the map. Begin by fixing a sequence $\tau_n$ of pairwise incomparable elements of $2^{<\NN}$. Then given a meager set $M$, we first write $M\subset\bigcup A_n$ where $A_n$ are in Nwd. Then, define $f(M)=\text{the closure of }\bigcup\tau_n\concat A_n\;.$ (Actually to ensure that $f(M)$ is still nowhere dense, we should select the $\tau_n$ sparse enough.) To see that $f$ is Tukey, suppose that $F$ is in Nwd and that $f(M)\subset F$. Then for all $n$, we have that $A_n\subset F^{-\tau_n}$. It follows that $M\subset\bigcup F^{-\tau_n}$, as desired.

Next, let's attack the problem of finding a $\sigma$-Tukey map from Nwd to Meager. (And no, the identity map does not work!) In similarity with our first construction, we will let $f(F)=\text{the closure of }\bigcup\tau_n\concat F\;.$ But now we have to be a bit more careful about how we select the $\tau_n$. Instead of fixing the $\tau_n$ in advance, this time they will depend on $F$. As before, we will ensure they are sparse enough that $f(F)$ is meager. But additionally, we can inductively choose them dense enough that the following condition holds:

(*) Whenever $N_\sigma\cap f(F)\neq\emptyset$ there exists $n$ with $\sigma\subset\tau_n$.

Now, to see that $F$ is $\sigma$-Tukey, suppose that $M$ is a meager set such that $f(F)\subset M$. Since $M$ is meager, we can write $M\subset\bigcup A_m$ where $A_m$ are in Nwd. Since $f(F)$ is closed, Baire’s category theorem (!) implies that some $A_m$ must have nonempty interior in the sense of the subspace topology on $f(F)$. This means that there is $\sigma$ such that $N_\sigma\cap f(F)\neq\emptyset$ and $N_\sigma\cap f(F)\subset A_m$. Now appealing to (*) there exists $n$ such that $\sigma\subset\tau_n$. But this implies that $\tau_n\concat F\subset N_\sigma\cap f(F)\subset A_m$, and so $F\subset A_m^{-\tau_n}$.

Putting this together, we conclude that if $f(F)\subset M$ then $F$ is dominated by one of the countable family of $A_m^{-\tau}$ for $m\in\NN$ and $\tau\in2^{<\NN}$, as desired.