# I want to show that $D_{\alpha}$ is a dense open set in $\omega^\omega$

Let $\{f_{\alpha}: \alpha<\kappa \}\subseteq \omega^\omega$.

Asumme that each $f_{\alpha}$ is a stictly increasing function.

Let $D_{\alpha}=\{f \in \omega^\omega: f(n)>f_{\alpha}(n)$ for some $n \in \omega\}$ is a dense open set in $\omega^\omega$ for each $\alpha<\kappa$.

I want to show that $D_{\alpha}$ is a dense open set in $\omega^\omega$.

This may be easy, but I need a suggestion on how to show this.Any suggestion thanks.

• Just verify the definition of dense and open? – Asaf Karagila Sep 10 '15 at 18:50

## 1 Answer

HINT: To show that $D_\alpha$ is open, for each $n\in\omega$ let $D_\alpha(n)=\{f\in D_\alpha:f(n):f_\alpha(n)\}$. Then show that $D_\alpha=\bigcup_{n\in\omega}D_\alpha(n)$, and that each $D_\alpha(n)$ is open; the latter is an immediate consequence of the definition of the product topology.

To show that $D_\alpha$ is dense, let $F\subseteq\omega$ be finite and $\varphi\in{^F\omega}$, and let $B(\varphi)=\{f\in{^\omega\omega}:\varphi\subseteq f\}$. Check that the family of all such sets $B(\varphi)$ is a base for the topology. Show that if $n\in\omega\setminus F$, then $B(\varphi)\cap D_\alpha(n)\ne\varnothing$ by actually constructing an element of $B(\varphi)\cap D_\alpha(n)$.