Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $\kappa$ is infinite cardinal and $X$ is a topological space with $\operatorname{w}(X)\leq\kappa$, then, for every base $\mathcal{B}$ of $X$, there exists an $\mathcal{B}_0\subseteq \mathcal{B}$ such that $\mathcal{B}_0$ is a base for $X$ and $|\mathcal{B}_0|\leq\kappa$. The proof of this result is quite similar for network. I would like to know if the result holds for $\pi$-base

share|cite|improve this question
up vote 6 down vote accepted

Paul’s given you the answer; here’s a proof. Let $\kappa=\pi w(X)$, and let $\mathscr{P}$ be a $\pi$-base for $X$ such that $|\mathscr{P}|\le\kappa$. Let $\mathscr{B}$ be any $\pi$-base for $X$. For each $P\in\mathscr{P}$ there is a $B_P\in\mathscr{B}$ such that $B_P\subseteq P$. Let $\mathscr{B}_0=\{B_P:P\in\mathscr{B}\}$; clearly $|\mathscr{B}_0|\le\kappa$. Moreover, if $U$ is any non-empty open set in $X$, there is some $P\in\mathscr{P}$ such that $P\subseteq U$ and hence $B_P\subseteq P\subseteq U$, so $\mathscr{B}_0$ is a $\pi$-base for $X$.

share|cite|improve this answer

The answer is Yes. A $\pi$-base for $X$ is a collection $\mathscr V$ of non-empty open sets in $X$ such that if $R$ is any non-empty open set in $X$, then $V \subseteq R$ for some $V \in \mathscr V$.

Example: $\{\{n\}: n\in \omega\}$ is a $\pi$-base for $\beta \omega$.

The $\pi$-weight of $X$ is defined as follows: $$\pi w(X)=\min\{|\mathscr V|: \mathscr V \text{ a } \pi-base \text{ for } X\}+\omega .$$

Note that $d(X)\le\pi w(X)\le w(X)$. The cardinal function $\pi$-weight is not monotone; e.g., $\pi w(\beta \omega)=\omega$ but $\pi w(\beta \omega \setminus \omega)=2^\omega$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.