1
$\begingroup$

Try to understand the definition of the property $D$ in one paper of J.E.Vaughan's

Firstly, let us see the Definitions(see page 238 of the paper):

A countably infinite discrete set $A \subset X$ has property $D$ in $X$ provided there exists a discrete family of open sets $\{U_a : a \in A\}$ such that $U_a \cap A = \{a\} $ for all $a\in A$.

and

A space $X $ is said to have property $D$ provided that every countably infinite discrete set in $X$ has property $D$ in $X$.

And in the page 239 of this paper, the author said that a regular space $X$ has property $D$ if and only if every pair of disjoint closed sets, one of which is a discrete sequence, can be separated by disjoint open sets.

However, as we know, in the regular space, any countable discrete subspace is strongly discrete. So, my question is this:

Is any regular space has property $D$?

(Therefore we can omit the condition that "every pair of disjoint closed sets, one of which is a discrete sequence, can be separated by disjoint open sets. ")

Thanks for any help and any hint.

$\endgroup$
2
$\begingroup$

Note that when Jerry says discrete set, he means closed, discrete set. (He makes this clear just before Definition 1.) Similarly, when he says that the family $\{U_a:a\in A\}$ is discrete, he means that it’s pairwise disjoint and locally finite. The deleted Tikhonov plank is an example of a $T_3$-space that is not a $D$-space: the set $\{\omega_1\}\times\omega$ is a closed, discrete set, but if $U_n$ is an open nbhd of $\langle\omega_1,n\rangle$ for $n\in\omega$, every open nbhd of $\langle\alpha,\omega\rangle$ meets infinitely many of the $U_n$’s for every sufficiently large $\alpha<\omega_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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