Definition of a Topology through neighbourhood basis?

Question

I have a question concearning the definition of a topology through neighbourhood basis. First:

Theorem 1. Let $X$ be a topological space and $x\in X$. If $\mathscr{B}(x)$ is a neighborhood basis of $x$ then:

$(i)$ $\mathscr{B}(x)\neq \varnothing$;

$(ii)$ If $U\in\mathscr{B}(x)$ then $x\in U$;

$(iii)$ If $U, V\in\mathscr{B}(x)$ there is $W\in\mathscr{B}(x)$ such that $W\subseteq U\cap V$;

$(iv)$ If $U\in\mathscr{B}(x)$ there is $V\subseteq X$ such that $x\in V\subseteq U$ and such that for every $y\in V$ there is $W\in\mathscr{B}(y)$ such that $W\subseteq V$.

The properties $(i)-(iii)$ are more or less trivial to be verified.

Question 1. How to show $(iv)$ holds?

On the other hand:

Theorem 2. Let $X$ be a nonempty set and $$\mathscr{B}:X\longrightarrow \mathscr{P}(\mathscr{P}(X)),\ x\longmapsto \mathscr{B}(x)$$ be a function such that $\mathscr{B}(x)$ satisfies satisfies $(i)-(iv)$ of the above theorem for every $x\in X$. Then there is a topology $\mathscr{T}$ on $X$ such that $\mathscr{B}(x)$ is a neighbourhood basis of $x$ for every $x\in X$.

Question 2. Is the topology $\mathscr{T}$ unique?

Obs: Theorem 2 is a very useful. For instance, we might use it to define a topology starting from a nonempty collections of seminorms on a vector space $X$ as follows: If $\mathscr{P}$ is such a family define $$\mathscr{U}:=\{U\subseteq X: \exists p_1, \ldots, p_n\in\mathscr{P}\ \textrm{and}\ \varepsilon_1, \ldots, \varepsilon_n>0; U=\bigcap_{i=1}^n B_{p_i}(0, \varepsilon_i)\}.$$ Then we might apply the above theorem to the function $x\longmapsto x+\mathscr{U}$, where $x+\mathscr{U}$ is defined the obvious, to get a locally convex topology on $X$ where $\mathscr{U}$ is the neighbourhood basis of zero. Warning: I still have to verify this but I'm almost sure that holds.

Answer 1

Call a set $A$ to be $\mathscr{B}$-open (for such an assignment) when $\forall x \in A \exists B_x \in \mathscr{B}(x) : B_x \subseteq A$. This is exactly how open sets are defined (e.g. in metric spaces): a set that is a neighbourhood for each of its points.

Then axiom (4) says that every member of a neighbourhood base set $B \in \mathscr{B}(x)$ has a $\mathscr{B}$-open subset. And this is true almost by definition of a neighbourhood of a point in a topological space.

This definition of $\mathscr{B}$-open sets is actually a topology, as can easily be verified (i) is used to see that $X$ is open, $\emptyset$ is vacuously open, intersection closed follows from (iii) and the union axiom is quite trivial. (iv) is not used for this verification, but garantuees that every set from all $\mathscr{B}(x)$ is actually a neighbourhood for that $x$ in the sense of the new topology we just defined (call it $\mathscr{T}$, the set of all $\mathscr{B}$-open sets).

The topology so defined is unique, because if $\mathscr{T}'$ is any topology such that the $\mathscr{B}(x)$ form neighbourhood bases at $x$, then if $O \in \mathscr{T}'$ then let $x \in O$. As $\mathscr{B}(x)$ is a neighbourhood base for this $\mathscr{T}'$, for some $U \in \mathscr{B}(x)$, we have that $x \in U \subseteq O$. As this holds for all $x$, $O$ is by definition $\mathscr{B}$-open and so $O \in \mathscr{T}$. Hence $\mathscr{T}' \subseteq \mathscr{T}$. On the other hand, if $O \in \mathscr{T}$, then pick $x \in O$, and we know that $O$ is $\mathscr{B}$-open, so for some $U \in \mathscr{B}(x)$, $U \subseteq O$. So as the $\mathscr{B}(x)$ is a local base for the neighbourhoods of $x$ for $\mathscr{T}'$ we now conclude that $x$ is an interior point for $O$ in the topology $\mathscr{T}'$ and as this holds for all $x \in O$, $O$ is open in $\mathscr{T}'$, or $\mathscr{T} \subseteq \mathscr{T}'$.

So having a candidate collection of neighbourhood bases, there is only one way to define what open means and then axiom (iv) garantuees that open (in this sense) neighbourhoods are indeed sitting in every candidate neighbourhood.