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

I'm studying some applications of small cardinals related to the Michael's Problem. Recall that we say that a space $X$ is a Michael space if X is a regular Lindelöf space such that $X\times \omega^\omega$ is not Lindelöf, and the question of whether such a space can be constructed without additional axioms is what we call the Michael's Problem.

Now, in this paper, the authors prove a theorem (cf. Theorem 2.4 of the paper) that gives sufficient conditions for a space $X$ to be non-productively Lindelöf. Then, they use the Theorem 2.4 to obtain Michael spaces under some hypotheses related to small cardinals.

Given a family $\mathscr{F}\subseteq\omega^\omega$, we call the $\mathscr{F}$-topology over $\omega^\omega$ the smallest topology that contains the usual one and is such that each $K_f:=\{g\in\omega^\omega:g\leq f\}$ for $f\in\mathscr{F}$ is open ($g\leq f$ iff $g(n)\leq f(n)$ for all $n\in\omega$).

The idea is to assure that the $\mathscr{F}$-topology satisfies the hypotheses of Theorem 2.4, from which the existence of a Michael space follows as a corollary. Since my questions are about some technical conditions, I won't explicit the Theorem 2.4.

Question 1 (related to Propositions 3.2 and 3.3 of the paper). How to prove that "If $\mathscr{F}$ is a dominating family, then $\{K_f:f\in\mathscr{F}\}$ is an open cover for $\omega^\omega$ with the $\mathscr{F}$-topology"?

This is obvious if we consider $\mathscr{F}$ dominating with respect to $\leq$, but I don't see how to prove it if $\mathscr{F}$ is dominating with respect to $\leq^*$ ($f\leq^* g$ iff $f(n)\leq g(n)$ for all but finitely many $n\in\omega$). The problem is that when they assume $\frak{d}=\omega_1$ (which it yields $\frak{b}=\frak{d}$), they use a scale as the family $\mathscr{F}$, and such a scale is dominating with respect to $\leq^*$. So, I suppose that "should be possible" to prove that $\{K_f:f\in\mathscr{F}\}$ is an open cover when $\mathscr{F}$ is dominating in $(\omega^\omega,\leq^*$), or, given a scale $\mathscr{F}$ with respect to $\leq^*$ one can construct a scale $\mathscr{F}$ dominating in $(\omega^\omega,\leq)$.

Question 2 (related to Proposition 3.6 of the paper). How to prove that "if $\frak{b}=\frak{d}=$ $cov (\mathcal{M})$, then there exists a strong scale, i.e., there exists a well ordered family $\{f_{\alpha}:\alpha<\frak{d}\}$ such that for every $f\in\omega^\omega$ there is an $\alpha<\frak{d}$ such that $f\leq f_\alpha$"?

Since $\omega_1\leq\frak{b}\leq\frak{d}$ and $\omega_1\leq cov(\mathcal{M})\leq\frak{d}$, an answer for this question also answers Question 1.

Thank you.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

It is not true that if $\mathscr{F}$ is dominating then $\{K_f:f\in\mathscr{F}\}$ is a cover of $\omega^\omega$. For example it is possible to have a dominating $\mathscr{F}$ such that $f(4)=6$ for every $f \in \mathscr{F}$ (the value of the functions at a single integer won´t change the fact that the family is or is not dominating); then the constant function $c_7\in\omega^\omega$ won´t be covered.

To construct a strong scale under $\frak{b}=\frak{d}$:

Fix a dominating family $\{g_\alpha : \alpha \in \frak{d} \}$; then do the usual at limit stages (i.e. for limit $\lambda$ let $f_\lambda$ dominate both $\{g_\alpha : \alpha \in \lambda\}$ and $\{f_\alpha : \alpha \in \lambda\}$), and for ordinals of the form $\alpha=\lambda+n$ with $\lambda$ limit and $n \in \omega$ define $f_\alpha=f_\lambda+ c_n$, where $c_n$ is the constant function with value $n$. Then $\left< f_\alpha : \alpha \in \frak{d}\right>$ is a strong scale.

share|improve this answer
Instead of building a strong scale from scratch, you could modify an existing scale, which is well-known to exist if $\mathfrak b=\mathfrak d$. If $\{g_\alpha:\alpha<\kappa\}$ is a scale, then a strong scale (increasing with respect to $\leq^*$ but dominating with respect to $\leq$) is given by setting, for $\alpha=\lambda+n$ as in Ramiro's answer, $f_\alpha(x)=xg_\lambda(x)+n$. –  Andreas Blass Feb 13 at 19:37

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.