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Let $X$ be a set. A set ${\cal C}\subseteq {\cal P}(X)$ is said to be a cover of $X$ if $\bigcup {\cal C} = X$ and $X\notin {\cal C}$.

If ${\frak U}$ and $\frak{W}$ are collections of covers of a set, we define the property ${\frak U}$ choose ${\frak W}$ as follows:

${\frak U} \choose {\frak W}$: For each ${\cal U}\in {\frak U}$ there is ${\cal W}\subseteq {\cal U}$ such that ${\cal W}\in{\frak W}$.

We consider the following kinds of open covers of a topological space $X$: An open cover ${\cal U}$ is said to be:

  1. a large cover if every $x\in X$ is contained in infinitely many members of ${\cal U}$;
  2. an $\omega$-cover if every finite subset of $X$ is contained in some member of ${\cal U}$, but $X\notin{\cal U}$;
  3. a $\tau$-cover if it is large and for all $x,y\in X$ either $\{U\in {\cal U}: x\in U, y\notin U\}$ is finite or $\{U\in {\cal U}: y\in U, x\notin U\}$ is finite;
  4. a $\gamma$-cover if ${\cal U}$ is infinite and every $x\in X$ belongs to all but finitely many members of ${\cal U}$.

Let $\Omega, \text{T}, \Gamma$ denote the collection of $\omega$-, $\tau$- and $\gamma$-covers, respectively.

A topological space $X$ is called a $D$-space if if whenever one is given a neighborhood $N(x)$ of $x$ for each $x\in X$, then there is a closed discrete subset $D\subseteq X$ such that $\{N(x): x\in D\}$ covers $X$.

I'm interested in implications (or counterexamples - spaces having one property but not the other) between the 3 properties

  • $D$;
  • $\Omega \choose \text{T}$;
  • $\Omega \choose \Gamma$.

(One implication is trivial: $\Omega \choose \Gamma$ implies $\Omega \choose \text{T}$ because for every space $X$ we have $\Gamma\subseteq \text{T}$.)

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  • $\begingroup$ $\{X\}$ is always an $\omega$-cover, so no space is either $\Omega \choose T$ or $\Omega \choose \Gamma$. $\endgroup$ – Ramiro de la Vega Jul 29 '15 at 18:30
  • $\begingroup$ Oops - I have to exclude this "pathological" cover -- edited the post accordingly. Thanks for noticing $\endgroup$ – Dominic van der Zypen Jul 30 '15 at 6:35
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Aurichi proved that Every Menger space is D. The last two properties imply Menger, so they imply D. On the other hand, Menger's property does not imply $\Omega \choose \text{T}$, for example since the latter is equivalent to $S_{fin}(\Omega,\text{T})$ (Details are available here) and thus implies $S_{fin}(\Omega,\Omega)$, which is the same as Menger's property in all finite powers. In summary, we have the following implications $${\Omega \choose \Gamma}\Rightarrow{\Omega \choose \text{T}}\Rightarrow D$$ and the last implication is not reversible.

The question whether the first implication is actually an equivalence is one of the most notorious and important open problems in the field of selection principles.

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