# Every $f\in\omega^\omega$ is bounded by the “increasing enumeration” of the intersection of a countable dense set and a dense open set in $\mathbb{R}$

I am studying the theorem 2.2.6 of "On the structure of the real line" of book Bartosznky-Judah.

In the proof of theorem 2.2.6 the part $(4) \to (5)$

$(4)$ for every family of dense open subsets of $\mathbb{R}$, $\{D_\alpha: \alpha < \kappa \}$ and a countable dense $X\subseteq \mathbb{R}$, there exists a countable dense subset $Y \subseteq X$ such that $|Y\setminus D_\alpha|<\aleph_o$ for $\alpha<\kappa$.

$(5)$ $\mathfrak{b}>\kappa$

Fix a countable dense subset $X=\{q_n:n \in \omega \}$ of $\mathbb{R}$. For an subset $Y\subseteq X$ let $f_Y \in \omega^\omega$ be the increasing function such that $Y=\{q_{f_{Y}(n)}:n \in \omega\}$.

I do not know why:

For every function $f\in \omega$ prove that there exists an open dense $D\subseteq \mathbb{R}$ such that $f(n) \leq f_Y(n)$ for all but finitely many $n$, where $Y=D \cap X$.

Any contribution. I would be very grateful.

• meta.math.stackexchange.com/a/9960/822 – Nate Eldredge Sep 4 '15 at 16:01
• I'm confused. Isn't it true that if $Y$ and $Z$ are subsets of $X$ and $Y\subseteq Z$, then $f_Z \le f_Y$? Then letting $U_n$ ($n < \omega$) be a countable base for $\mathbb{R}$, we have for every open $V$ that there exists an $n$ such that $f_{V\cap X} \le f_{U_n\cap X}$. Then letting $g$ be a $<^*$-bound on the functions $f_{U_n\cap X}$, we have $f_{V\cap X} <^* g$ for all open $V$. – Paul McKenney Sep 5 '15 at 13:18