TITLE: Two questions about the "grasp" cardinal function
QUESTION [10 upvotes]: 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?

REPLY [2 votes]: 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.