3
$\begingroup$

Given a topological space $(X,\tau)$, we say that a matching is a collection of non-empty open sets that are pairwise disjoint.

Given an infinite cardinal $\kappa$, is there a $T_2$-space with $|X|\geq\kappa$ and a cardinal $\alpha<|X|$ with the following properties?

  1. there is a dense subset $D\subseteq X$ with $|D| = \alpha$, and
  2. there is a matching $M\subseteq (\tau\setminus\{\emptyset\})$ with $|M| = \alpha$.
$\endgroup$

1 Answer 1

2
$\begingroup$

The answer is yes. Given a cardinal $\kappa$, let $\lambda = 2^\kappa$ (this denotes cardinal exponentiation) and let $X_0 = 2^\lambda$ (this denotes a Tychonoff product). Clearly $|X_0| > \kappa$. $X$ has a dense subset of size $\kappa$, and it has the ccc, which means that every "matching" in $X_0$ is countable. (By the way, what you're calling a matching is usually called a cellular family.) If we take $X$ to be $\kappa$ disjoint copies of $X_0$ and set $\alpha = \kappa$, then all your conditions are met.

By the way, an interesting variant of this question is obtained by asking for $\alpha < \kappa$ instead of $\alpha < |X|$. In this version, the answer is yes for some $\kappa$ and no for others. (By the argument above, we get a yes answer if there is some $\alpha$ with $\alpha < \kappa \leq 2^\alpha$. But the answer is no if $\kappa$ is a strong limit cardinal, because if a Hausdorff space has a dense subset of size $\alpha$, then its cardinality is at most $2^{2^\alpha}$.)

$\endgroup$
1
  • $\begingroup$ Thanks for your detailled answer, as well as answering the variant for $\alpha < \kappa$! $\endgroup$ Nov 22, 2019 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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