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?
> 
> 
> 



Two questions about the "grasp" cardinal function
**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.


