# How to obtain a countable dense subset disjoint from the closure of a free sequence

I've encountered some dificulties while reading the article, Hereditary normality versus countable tightness in countably compact spaces, from Nyikos (1992). In particular in a passage of the Reduction Theorem where the author makes the assertion that, given a separable, countably compact, $T_5$ topological space $X$ with an uncountable free sequence, we can take $W = \{x_{\alpha} : \alpha < \omega_{1}\}$ free sequence in $X$ and $D \subset X$ countable dense such that $D \cap \overline{W} = \emptyset$.

If i could verify that every $d \in D$ is such that there is an ordinal $\beta < \omega_{1}$ such that $d \in \overline{\{x_{\alpha} : \alpha < \beta\}}$ then we could construct a new the free sequence by shifting the starting point of the old one.

The point is that the case above is not necessarily true. Other than that I can't see how the countable dense interacts with the initial free sequence in a way that helps the problem.

• I guess it's safe to assume that, if I have to ask for the definition of an "uncountable free sequence", then the question is way over my head? – bof Jul 20 '17 at 4:05
• The paper can be downloaded from here – Henno Brandsma Jul 20 '17 at 13:32
• @bof This is a sequence $(x_\alpha), \alpha < \omega_1$ such that for all $\beta < \omega_1$ : $\overline{\{x_\alpha: \alpha < \beta\}} \cap \overline{\{x_\alpha: \alpha \ge \beta\}} = \emptyset$ – Henno Brandsma Jul 20 '17 at 13:34

## 1 Answer

Let $W_∞ = ⋂_{β < ω_1} \overline{\{x_α: α ≥ β\}}$. As you've noticed, the set $D ∩ W_∞$ is the problem. The point is that $W_∞$ is closed and disjoint with $W$, so $D ∩ W_∞ ⊆ \overline{W} ⊆ \overline{D \setminus W_∞}$, so we may put $D' = D \setminus W_∞$. Now $D' ∩ \overline{W} = D ∩ ⋃_{β < ω_1} \overline{\{x_α: α < β\}} ⊆ \overline{\{x_α: α < γ\}}$ for some $γ < ω_1$. Therefore is is enough to put $W' = \{x_{γ + α}: α < ω_1\}$.