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I'm reading this book about structure of the real line. One of the things that's proven there is that $\mathfrak{b}\leq non(\mathcal{M}).$ And the proof goes as follows:

For every $f\in\omega^\omega$ let us define $K_f$ as follows: $$K_f=\lbrace g\in\omega^\omega:\ g\preceq f\rbrace.$$ We observe that $$K_f=\bigcup_n \bigcap_{k>n} \lbrace g\in\omega^\omega:\ g(k)\leq f(k)\rbrace.$$ So for every $f$ a set $K_f$ is a union of nowhere-dense sets a thus - first category subset of space $\omega^\omega$.

Suppose that $\kappa<\mathfrak{b}$ and that $\lbrace f_\alpha:\ \alpha<\kappa\rbrace\subseteq\omega^\omega$. Then there exists a function $f\in\omega^\omega$ such that $$(\forall \alpha<\kappa)(f_\alpha \preceq f).$$ But then we have $$\lbrace f_\alpha:\ \alpha<\kappa\rbrace\subseteq K_f,$$ so $non(\mathcal{M})\geq\mathfrak{b}$.

My main question is why are we proving this for $\omega^\omega$ and not for $\mathbb{R}$? Are meager sets in $\omega^\omega$ somehow connected to meager sets of $\mathbb{R}$?

Sarghh
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2 Answers2

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Often $\omega^\omega$ is easier to work with than $\mathbb{R}$, and despite their obvious topological differences for many purposes in logic they're basically equivalent:

  • They're (low-level) Borel isomorphic, so descriptive set theoretic questions above the first few levels of the Borel hierarchy can't distinguish between them.

  • There's a homeomorphism between $\omega^\omega$ and $\mathbb{R}\setminus\mathbb{Q}$; since $\mathbb{Q}$ is "small" in $\mathbb{R}$ (= countable, and so a fortiori meager and null), this means that many of their cardinal invariants are equal.

  • Moreover there's also a continuous bijection from $\omega^\omega$ to $\mathbb{R}$ - see e.g. here (whose inverse of course is badly discontinuous). However, in my experience that's actually less useful than the bulletpoint above.

In particular, for the question in the OP the second bulletpoint above guarantees that we can answer the question for $\mathbb{R}$ by answering it for $\omega^\omega$.

Noah Schweber
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A simple answer: $\omega^\omega$ is just homeomorphic to the set of irrationals in $\Bbb R$. So if we cover it by $\kappa$ many non-meager sets, we can just add, say, one rational to each in the irrationals-version and get a similar cover for the reals and vice versa is similar as well. So the covering numbers for $\omega^\omega$ and $\Bbb R$ are always the same.

Henno Brandsma
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