TITLE: A special cofinal family in $(\omega^\omega,\le)$
QUESTION [9 upvotes]: Let me start by posting a question, which is essentially a
combinatorial question. I will post below at least a short
explanation what lead me to ask this question.
Let us work with the set $\omega^\omega$ (i.e., all sequences of
positive integers) with a pointwise ordering
$$x\le y \Leftrightarrow (\forall n\in \omega) x_n\le y_n$$

Does there exist a system $\mathcal B\subseteq\omega^\omega$
  such that

$(\forall x\in\omega^\omega)(\exists y\in\mathcal B) x\le y$ (i.e., $\mathcal B$ is cofinal in $\omega^\omega$)
For any $x\in\omega^\omega$ the set $\{z\in\mathcal B; z\le x\}$ is finite.



My motivation was that I was thinking about a problem posted by Michael Greinecker in this question.
He asks there whether we can have a local base at $x$ such that no intersection of a countable subsystem of this base is a neighborhood of $x$; under some conditions on the point $x$ and the space $X$.
I wanted to start with some simple examples to get better understanding of the problem.
I was thinking about the $X$ which is obtained from topological sum of countably many copies of Alexandroff compactification of a countable discrete space, if we glue the non-isolated point from each copy into one point.
Alexandroff compactification of a discrete space is simply a convergent sequence (it is homeomorphic to $\{0\}\cup\{1/n; n=1,2,\dots\}$ with the topology inherited from the real line.)
So this space can be visualized like in the following picture. (Each ray in this picture is represents a convergent sequence and a typical basic set is marked there. Ignore the labels given there - I have simply used a picture I had prepared before for different purposes; a notation in that situation was completely different.)

Or, if you prefer another representation of this space, you can take $Y=\{n+1/k; n,k\in\mathbb N\}$ and then $X\cong Y/\mathbb N$.
This space is countably tight, since the class of countably tight spaces is closed under topological sums and quotient maps. EDIT: Alex Ravsky pointed out in a comment (now deleted), that this is unnecessary complicated - this space is countably tight simply because it is countable.  Anyway, the argument I gave would work for arbitrarily many copies, too. (Or if we are using a different countably tight space instead of the convergent sequence.) So it might still be interesting for other spaces of similar type.
This space has only one non-isolated point and it is not first countable. So this spaces fulfills the assumptions from the other question. (Hence if no family with the above properties exists, this would yield a counterexample.)
Let us denote the non-isolated point of $X$ by $\infty$.
For each sequence $x\in\omega^\omega$ we get a neighborhood of $\infty$ simply by taking all numbers greater than or equal to $x_n$ in the $n$-th copy of the convergent sequence.
Such neighborhoods comprise a local base at $\infty$.
So the cofinality condition for the system $\mathcal B$ means that if we take the neighborhoods corresponding to sequences from $\mathcal B$, they form a local base.
The second condition correspond to the assumption that no infinite countable intersection of (distinct) sets from this local base will be again an open set.

REPLY [6 votes]: If seems that there exists no family $\cal B$ with the above properties. Suppose the opposite. 
Since the family $\cal B$ is cofinal in $\omega^\omega$, the diagonal procedure shows that $\cal B$ is uncountable (moreover, maybe the minimal cardinality of $\cal B$  is related with so-called small cardinals, like $\operatorname{cof}(\omega^\omega)$). 
Now by induction we can build an non-increasing sequence $\{\cal B_n:n\in \omega\}$ of uncountable subfamilies of $\cal B$ such that for each $n$, each $x,y\in\cal B_n$ and 
each $i\le n$ holds $x(i)=y(i)$. Again by an induction we choose a sequence $\{x_n\}$ of distinct functions such that $x_n\in\cal B_n$ for each $n$. This condition implies that for each $m$ the set $\{x_n(m):n\in\omega\}$ has a finite supremum $x(m)$. Since $\cal B$ is cofinal, there exists a sequence $y\in\cal B$ such that $y\ge x$. Then $y\ge x_n$ for each $n$, which contradicts the second property of $\cal B$.