Suppose that corresponding to each point $p$ of a set $X$, there is a family $B_p$ of subset of $X$ satisfying (NB1), (NB2) and (NB3) above. Then $$T=\{G\subset X: \text{for each }x \in G\text{, there is }B \in B_x\text{ such that }x \in B \subset G\}$$ is a topology on $X$ in which $B_p$ is a neighborhoods base for $p$.
Proof
Let $$n_p =\{N\subset X:N\supset B\text{ for some }B \in B_p\}$$ It is easy to see that $n_p$ satisfies (N1), (N2) and (N3) of proposition 6. (N4) follows from (NB3), so that by Proposition 6, the families $n_p$ yield a topology $T$ on $X$ in which $n_p$ is the neighborhood system for each $p$. By the construction of $n_p$ from $B_p$ it is clear that the latter is an neighborhood of $p$.
In a space $X$, a set $D\subset X$ is called dense in $X$ if $\mathrm{cl} D=X$. If for a set $D$ in $X$, $\mathrm{cl} D \neq X$, then $U=X\setminus \mathrm{cl} D$ is a nonempty open set such that $U \cap D = \emptyset$. In oter words, if every nonempty open set $U$ meets $D$, then $D$ is dense in $X$. The converse is clear. Thus we get...
Do you have any answer for this proposition? to make that answer be specific and easy to understand, help me... thank you.