The proof from Van Mill, et al to prove $\omega_1$ is dually discrete

63 Views Asked by At

Here is the proof from

J. van Mill, V.V. Tkachuk, R.G. Wilson; Classes defined by stars and neighbourhood assignments, Top. Appl. 154 (2007), pp.2127–2134, MR2324924, link

to prove $\omega_1$ is dually discrete. (A topological space $X$ is dually discrete if for any neighbourhood assignment $\{ O_x : x \in X \}$ there is a discrete subspace $Y \subseteq X$ such that $\bigcup_{x \in Y} O_x = X$.)

enter image description here

I have small questions on the proof.

  1. Why for any non-isolated point $\alpha \in \omega$ there is $f(\alpha)<\alpha$ such that $(f(\alpha), \alpha]\subset O_\alpha$?

  2. How to use the pressing down lemma to prove that there is a uncontable $A\subset \omega_1$...

  3. Why does $\bigcup\{O_\alpha: \alpha \in B\} $ contain $\omega_1\setminus (\beta+1)$?

Thanks for your help.

1

There are 1 best solutions below

0
On BEST ANSWER
  1. For each non-isolated $\alpha \in \omega_1$ (these are the countable limit ordinals), we have an (open) neighbourhood $O_\alpha$ of $\alpha$. But the family $$\{ ( \gamma , \alpha ] = \{ \xi : \gamma < \xi \leq \alpha \} : \gamma < \alpha \}$$ is a neighbourhood base for $\alpha$, and so there must be a $\gamma < \alpha$ such that $( \gamma , \alpha ] \subseteq O_\alpha$. So for each limit $\alpha < \omega_1$ let $f(\alpha)$ denote such a $\gamma$.

  2. The function $f$ is clearly regressive ($f(\alpha) < \alpha$) on the stationary set (even closed unbounded set) of limit ordinals, and so the Pressing Down Lemma immediately concludes that it is constant on a stationary (hence uncountable) subset.

  3. Given any $\beta < \xi < \omega_1$ since $B$ is uncountable there is an $\alpha \in B$ such that $\xi \leq \alpha$. But then $\xi \in ( \beta , \alpha ] = ( f(\alpha) , \alpha ] \subseteq O_\alpha$.