Two questions about the "grasp" cardinal function For a topology $\mathcal{T}$ on a set $S$, where $\mathcal{T}$ does not have a finite base, I define the grasp $g(\mathcal{T})$ to be the least infinite cardinal $\kappa$ such that $\mathcal{T}$ has a base $\mathcal{B}$ with $|\mathcal{B}|= w(\mathcal{T})$ , ($w$ is the usual weight function), satisfying $$\mathcal{T}=\{\bigcup V : V\in [\mathcal{B}]^{\leq \kappa}\}.$$ 
That is, every open set is the union of at most $\kappa$ members of $\mathcal{B}$, and $|\mathcal{B}|$ is minimum among all bases.
The following can be shown by elementary means: 


*

*$g(\mathcal{T})\leq w(\mathcal{T})$. (Obvious).

*$|\mathcal{T}|\leq w(\mathcal{T})^{g(\mathcal{T})}$. (Obvious).

*$g(\mathcal{T})\leq g(D(w(\mathcal{T})))$  where  $D(\lambda)$ is discrete space of cardinality $\lambda$.

*If $Y$ is a subspace of $X$ and $w(Y)=w(X)$ then $g(Y)\leq g(X)$.

*When $\kappa$ is an infinite cardinal with the discrete or the order topology then 


*

*$\operatorname{cf}(\kappa)\leq g(\kappa)\leq \kappa$.

*If $\kappa$ is a singular strong limit then $g(\kappa)=\operatorname{cf}(\kappa)$.


(This last one helps to distinguish $g$ from other topological cardinal functions.)
I have two questions.

Question 1: Referring to (2) above, $g(T)$ is not necessarily the least cardinal $\lambda$ such that $|\mathcal{T}|\leq w(\mathcal{T})^\lambda$ because if we assume $2^{\omega}=2^{\omega_1}$ and let $\mathcal{T}$ be the discrete topology on $\omega_1$ then $|\mathcal{T}|=\omega_1^{\omega}$ but by (5)(i) $g(\mathcal{T})=\omega_1$. So is there an example like this in $\mathsf{ZFC}$?  
Question 2: By (5)(i) we have $\operatorname{cf}(2^{\omega})\leq g(D(2^{\omega}))\leq 2^{\omega}$. What values for $g(D(2^{\omega}))$ other than $2^{\omega}$ are consistent?

 A: Proposition: It is consistent that $cf (2^{\omega})=g(D(2^{\omega}))<2^{\omega}$.
Proof:  Assume GCH in the ground model. Let    $P=Fn(\omega_{\omega_1},2)$.
Since $2^\omega=\omega_{\omega_1}$ in $V^P$ 
it is enough to show that $g(D(2^{\omega})=\omega_1$ in the generic extension.
Let $G$ be a generic filter.
In $V[G]$
for ${\alpha}<\omega_1$ let
$$
\mathcal B_{\alpha}=\mathcal P(\omega_{\alpha})\cap V[G\cap Fn(\omega_{\alpha},2)].
$$
Since 
$$
|\mathcal B_{\alpha}|= (2^{{\omega}_{\alpha}})^{V[G\cap Fn(\omega_{\alpha},2)]}=
{\omega}_{{\alpha}+1},
$$
the family
$$
\mathcal B=\bigcup\{\mathcal B_{\alpha}:{\alpha}<\omega_1\}
$$
has cardinality ${\omega}_{{\omega}_1}$. 
Since $\mathcal B$ clearly contains all the singletons of $D({\omega}_{{\omega}+1})$,
it is enough to show that every $X\subset {\omega}_{{\omega}+1}$
is the union of $\omega_1$ many elements of $\mathcal B$.
Assume that 
$$
1\Vdash \dot X\subset {\omega}_{{\omega}_1}.
$$
For each ${\xi}\in {\omega}_{\omega_1}$ pick a maximal antichain $D_{\xi}$ in 
$P$ such that the element of $D_{\xi}$ decides if ${\xi}\in \dot X$ or not.
Define a function $f:{\omega}_{{\omega}_1}\to {\omega}_{{\omega}_1}$ in $V$
as follows: 
$$
f({\xi})=\min\{{\alpha}<\omega_1: D_{\xi}\subset Fn({\omega}_{\alpha},2)\}.
$$
Let 
$$
Y_{\alpha}=\{{\xi}<{\omega}_{{\alpha}}:f({\xi})\le {\alpha}\},
$$
and let  $\dot X_{\alpha}$ be an $Fn({\omega}_{\alpha},2)$-name of a subset of $Y_\alpha\subset {\omega}_{\alpha}$
such that for all ${\xi}\in Y_{\alpha}$ and for all $d\in D_{\xi}$
we have 
$$
d\Vdash {\xi}\in \dot X \text{ iff }d\Vdash {\xi}\in \dot X_{\alpha}.
$$
and 
$$
d\Vdash {\xi}\notin \dot X \text{ iff }d\Vdash {\xi}\notin \dot X_{\alpha}.
$$
Then in   $V[G]$ we have $X\cap Y_{\alpha}=X_{\alpha}$ and $X_{\alpha}\in \mathcal B_{\alpha}$.
Since $\bigcup\nolimits_{{\alpha}<{\omega}_1} Y_{\alpha}={\omega}_{\omega_1}$ we have
$$
V[G]\vDash X=\bigcup\nolimits_{{\alpha}<{\omega}_1} X_{\alpha}.
$$
A: Proposition (with Z. Szentmiklóssy): It is consistent that 
$\omega_1=cf (2^{\omega})<g(D(2^{\omega}))=2^{\omega}$.
Proof:
Assume GCH in the ground model. 
For ${\alpha}<{\omega}_1$ let 
$$
P({\alpha})=Fn({\omega}_{{\alpha}+1}\times {\omega}_{{\omega}_1+1},2;{\omega}_{{\alpha}+1}),
$$
and 
$$
P=\prod_{{\alpha}<{\omega}_1}P({\alpha}),$$
where the product is  the product with full support.
So   $P$ is just an Easton-forcing, hence it preserves all the cardinals,
and $2^{{\omega}_{{\alpha}+1}}={\omega}_{{\omega}_1+1}$ for all ${\alpha}<{\omega}_1$
in $V^P$.
Claim: For each  ${\beta}<{\omega}_1$
there is an almost disjoint family 
$\mathcal F_{\beta}\subset [{\omega}_{{\beta}+1}]^{{\omega}_{{\beta}+1}}$ 
with $|\mathcal F_{\beta}|={\omega}_{{\omega}_1+1}$ in $V^P$.
Proof of the Claim:
Write $P^{\beta}=\prod_{{\beta}\le {\alpha}<{\omega}_1}P({\alpha})$ and
$P_{\beta}=\prod_{{\alpha}<{\beta}}P({\alpha})$.
Since $P^{\beta}$ is ${\omega}_{{\beta}+1}$-complete, and  forcing 
with $P({\beta})$ introduces ${\omega}_{{\omega}_1+1}$ new subsets of ${\omega}_{{\beta}+1}$ , 
we have
$$
V^{P^{\beta}}\models 2^{{\omega}_{\beta}}={\omega}_{{\beta}+1}\land 
2^{{\omega}_{{\beta}+1}}\ge {\omega}_{{\omega}_1+1},
$$
hence in the model $V^{P^{\beta}}$ we have 
an almost disjoint family 
$\mathcal F_{\beta}\subset [{\omega}_{{\beta}+1}]^{{\omega}_{{\beta}+1}}$ 
with $|\mathcal F_{\beta}|={\omega}_{{\omega}_1+1}$.
Since $P=P^{{\beta}}*P_{\beta}$, we have such an $\mathcal F_{\beta}$ even in  $V^P$.
So we proved the Claim. Q.E.D
Let $Q=P*Fn ({\omega}_{{\omega}_1},2)$.
Since $P$ is $\sigma$-complete, we have 
$$
V^P\models 2^{\omega}={\omega}_1\land  ({\omega}_{{\omega}_1})^{\omega}={\omega}_{{\omega}_1},
$$
and so 
$$
|2^{\omega}|^{V^Q}\le ((|Q|^{\omega})^{\omega})^{V^P}={\omega}_{{\omega}_1}.
$$
So $2^{\omega}={\omega}_{{\omega}_1}$ in $V^Q$.
Assume on the contrary that 
$g(D({\omega}_{{\omega}_1}))={\omega}_{\alpha}<{\omega}_{{\omega}_1}$ in $V^Q$.
Let $\mathcal  B$  be a base witnessing $g(D({\omega}_{{\omega}_1}))={\omega}_{\alpha}$.
Since every $F\in \mathcal F_{\alpha}\subset [{\omega}_{{\alpha}+1}]^{{\omega}_{{\alpha}+1}}$
is the union of at most ${\omega}_{\alpha}$-many elements of $\mathcal B$, so 
 $F$ contains at least one element  $B_F\in \mathcal B$ of cardinality ${\omega}_{{\alpha}+1}$.
But the elements of $\mathcal F_{\alpha}$ are almost disjoint  (i.e.
the cardinalities of the intersections $\le {\omega}_{\alpha}$), so 
$B_F\ne B_{F'}$ for $F\ne F'$. 
Thus $|\mathcal B|\ge |\mathcal F_{\alpha}|>{\omega}_{{\omega}_1}$. Contradiction.
