Understanding the theorem 13 of the paper of A.V.Arhangel'skii in 2000 
The Theorem 13 is from the paper: Sharp bases and weakly uniform bases versus point-countable bases. I cannot understand that which said:" we only need to add the family $\mathcal{B}_1$ the family of all open one-point sets of $Z$." Also I don't know why the  author command the cardinality of the space $Z$ is $\le \aleph_1$ 
What I've tried:  I think the family $\mathcal{B}_1$ is already the point-countable base of $Z$. For points of $X$, which is the set of alll isolated of $Z$, family $\mathcal{B}_1$ is point-countable; and notice that family $\mathcal{B}_1$ is also two-countable base of $Z$, so by the lemma 14, family $\mathcal{B}_1$ is point-countable at every non-isolated points of $Z$.
Could somebody help me?
 A: For my own benefit I’m going to recapitulate the early part of the argument, filling in a couple of details and making a couple of small corrections.
$\mathscr{B}$ is a $2$-countable base for $Z$, and $X$ is the set of isolated points of $Z$. $\mathscr{U}=\{V\cap X:V\in\mathscr{B}\}$. From Lemma $12$ we have a function $f:\mathscr{U}\to X$ such that $\mathscr{U}_1=\{U\setminus f(U):U\in\mathscr{U}\}$ is point-countable. The definition of $\mathscr{B}_1$ is slightly incorrect; it should be 
$$\mathscr{B}_1=\left\{V\setminus\{f(V\cap X)\}:V\in\mathscr{B}\right\}\;,$$
where $f(\varnothing)$ is some object not in $Z$. This clearly is a base at each point of $Z\setminus X$. There is a typo in the argument that $\mathscr{B}_1$ is point-countable at each point of $X$: the reference should be to Lemma $12$, and the point is that if $W\in\mathscr{B}_1$ and $W\cap X\ne\varnothing$, then $W\in\mathscr{U}_1$. That $\mathscr{B}_1$ is point-countable at each point of $Z\setminus X$ does indeed follow from Lemma $14$.
The reason that we need to replace $\mathscr{B}_1$ by $\mathscr{B}_1\cup\big\{\{x\}:x\in X\big\}$ to get a point-countable base for $Z$ is that if $x\in X$, any base for $Z$ must include the set $\{x\}$, but there is no guarantee that $\{x\}\in\mathscr{B}_1$. In fact, $\{x\}\in\mathscr{B}_1$ iff there is a $y\in X\setminus\{x\}$ such that $\{x,y\}\in\mathscr{B}$ and $f\big(\{x,y\}\big)=y$. In other words, $\mathscr{B}_1$ is point-countable, and it’s a base at every point of $Z\setminus X$, but it may not even cover $X$, let alone be a base at every point of $X$. Adding the singletons of isolated points makes it a base for the whole space and doesn’t destroy point-countability.
