Two questions about the "grasp" cardinal function

Question

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:

1. $g(\mathcal{T})\leq w(\mathcal{T})$. (Obvious).
2. $|\mathcal{T}|\leq w(\mathcal{T})^{g(\mathcal{T})}$. (Obvious).
3. $g(\mathcal{T})\leq g(D(w(\mathcal{T})))$ where $D(\lambda)$ is discrete space of cardinality $\lambda$.
4. If $Y$ is a subspace of $X$ and $w(Y)=w(X)$ then $g(Y)\leq g(X)$.

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

   1. $\operatorname{cf}(\kappa)\leq g(\kappa)\leq \kappa$.
   2. 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?

Answer 1

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.

Answer 2

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

$$ V[G]\vDash X=\bigcup\nolimits_{{\alpha}<{\omega}_1} X_{\alpha}. $$