Third axiom proof I feel stuck proving the third axiom for the topology. I proved the first two. The intersection does not seem obvious to me, and I have spent some time trying to prove it but have ran out of luck, and would really appreciate some help.
Prove Theorem 5.11:
Let X be a non-empty set and let there be assigned to each point p $\in$ X a class $\mathcal A_p$ of subsets of X satisfying the following axioms:
[$\mathbf A_1$] $\mathcal A_p$ is not empty and p belongs to each member of $\mathcal A_p$
[$\mathbf A_2$] The intersection of any two members of $\mathcal A_p$ belongs to $\mathcal A_p$
[$\mathbf A_3$] Every superset of a member of $\mathcal A_p$ belongs to $\mathcal A_p$
[$\mathbf A_4$] Each member $\mathcal N $$\in$ $\mathcal A_p$ is a superset of a member $\mathsf G$ $\in$ $\mathcal A_p$ such that $\mathsf G$ $\in$ $\mathcal A_g$ for every g $\in$ $\mathsf G$.
Then there exists one and only one topology $\tau$ on X such that $\mathcal A_p$ is the $\tau$-neighborhood system of the point p $\in$ X.
My candidate topology consists of the class $\mathcal A_p$ of subsets of X satisfying the four given axioms and the empty set.
If there is another topology I should be using, please let me know.
 A: Let $B=\{\mathsf{G}\subseteq X:\forall g\in\mathsf{G},\, \mathsf{G}\in\mathcal{A_{g}}\}$. We want to show that $B$ satisfies the axioms of a topological basis, then it will generate the desired topolgy $\tau$. Indeed, for every $p\in X$, the $\tau$-neighbourhood system of $p$ is precisely $\mathcal{A_{p}}$ (this can be seen by applying $[\mathbf{A_{3}}]$ and $[\mathbf{A_{4}}]$).
By $[\mathbf{A_{1}}]$ and $[\mathbf{A_{4}}]$, $B$ covers $X$. Now suppose $U,V\in B$ ; we want to show that $U\cap V\in B$. For every $p\in U\cap V$,  we know that $U,V\in\mathcal{A_{p}}$ (by definition of $\tau$), so $U\cap V\in\mathcal{A_{p}}$ by $[\mathbf{A_{2}}]$. Thus, $U\cap V\in B$. Therefore $B$ is a topological basis generating a topology $\tau$ with the desired properties.
For uniqueness, suppose that $\tau'$ is another topology on $X$ which has the property that, for every $p\in X$, the $\tau'$-neighbourhood system of $p$ is $\mathcal{A_{p}}$. Thus every $\mathsf{G}\in B$ is open in $\tau'$, so $\tau\subseteq\tau'$. 
On the other hand, suppose $\mathcal{N}$ is open in $\tau'$. Note that, for every $p\in\mathcal{N}$, the $\tau'$-neighbourhood system of $p$ contains $\mathcal{N}$, so $\mathcal{N}\in\mathcal{A_{p}}$. Thus, $\mathcal{N}\in B$, so $\tau'\subseteq\tau$. Therefore, $\tau=\tau'$ and we have uniquness.
A: The $\tau$ you have defined is perhaps the obvious choice for the topology, but it is not actually the one you want. To see why, we need to think about the definition of neighborhood of a point in a topological space.
Definition: If $(X,\tau)$ is a topological space and $p\in X$, then $N\subseteq X$ is a neighborhood of $p$ if there exists $U\in\tau$ such that $p\in U$ and $U\subseteq N$.
Notice that a neighborhood of a point need not belong to the topology! For example, $[0,1]\times[0,1]\subseteq\Bbb R^2$ is a neighborhood of $(1/2,1/2)$ when $\Bbb R^2$ is given the usual topology, because $(1/3,2/3)\times(1/3,2/3)\subseteq [0,1]\times[0,1]$ is open.
With this in mind, you can re-examine the axioms. Axiom 1 is clear - every point belongs to every neighborhood of it, and every point in the space has at least one neighborhood (the whole space). Axiom 2 says that the intersection of two neighborhoods is a neighborhood - this is because if $N_1$, $N_2$ are neighborhoods of $p$, then there exist $U_1\subseteq N_1$ and $U_2\subseteq N_2$ containing $p$ which are open, and $p\in U_1\cap U_2\subseteq N_1\cap N_2$. Because $U_1\cap U_2$ is open, $N_1\cap N_2$ is a neighborhood of $p$. Axiom 3 makes sense, because if $N$ is a neighborhood of $p$ and $U\subseteq N$ is an open containing $p$, then any superset of $N$ also contains $U$.
The last axiom is the one which really defines the open sets: it specifies that a neighborhood of a point $p$ is precisely a set $N$ containing $p$ such that there exists $U\subseteq N$ which contains $p$ and is a neighborhood of every point in it. The only sets which are neighborhoods of every point they contain are precisely the opens. Clearly opens satisfy this property, but conversely, if $U$ is a neighborhood of all its points, then $U = \bigcup_{x\in U} U_x$, where $U_x\subseteq U$ is an open containing $x$ inside $U$. As this is a union of arbitrary opens, it is itself open.
See if you can use this intuition to define the appropriate $\tau$ and prove the results.
A: We define $U$ as an open set if for each $p\in U$ there is $V \in \mathcal A_p$ such that $V \subset U$. Lets consider two open sets $U_1$ and $U_2$. If $x\in U_1 \cap U_2$ then there are $V_1,V_2\in \mathcal A_x$ such that $V_j \subset U_j$. Just take $V = V_1\cap V_2$ and you will have $V \in \mathcal A_x$ and $V \subset U_1 \cap U_2$. Therefore $U_1 \cap U_2$ is open set.
