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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.

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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.

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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 '14 at 19:37

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