Balogh continuum topology

Assume AC (The Axiom of Choice) and put $\kappa =2^{{\mathrm{\aleph }}_0}$ and $C={}^{\kappa }2$. Fix a map $D:C\to C$, $D(c)=d_c$, supplied by Balogh combinatorial map. For $\alpha <\kappa$, $s\in [C{]}^{<\omega }$ and $a\in [\kappa {]}^{<\omega }$ set

$F(\alpha ,s,a)=\{\beta <\kappa :(\mathrm{\forall }c\in s)d_c(\beta )=c(\alpha )\}\setminus a,$

and let ${\mathcal{F}}_{\alpha }$ be the family of all these sets. It may contain the empty set; it is not assumed to be a proper filter. When $s=\mathrm{\varnothing }$, the condition on $c$ is vacuous and $F(\alpha ,\mathrm{\varnothing },a)=\kappa \setminus a$.

On $X=\kappa \times \omega$ declare $O\subseteq X$ open when, for every $(\alpha ,n+1)\in O$, there is $F\in {\mathcal{F}}_{\alpha }$ with $F\times \{n\}\subseteq O$. No extra condition is placed on points at height zero. The next lemma proves that this rule defines a topology and describes its neighborhood bases.

Write $L_n=\kappa \times \{n\}$ and $U_n=\kappa \times \{0,\dots ,n\}$ for $n<\omega$, and use $U_{-1}=\mathrm{\varnothing }$ in formulas involving the bottom level. This occurrence of $-1$ is only an indexing convention; every point of $X$ has a natural-number height.