Let $X$ be a set. A neighbourhood filter system $\mathcal F$ on $X$ is a rule that assigns to each element $x \in X$ a family $\mathcal F_x \subset \mathcal P(X)$ such that

(1) if $x \in X$, then $\mathcal F_x \neq \emptyset$

(2) if $x \in X$ and $\mathcal A \in \mathcal F_x$, then $x \in A$

(3) if $x \in X$, $A \in \mathcal F_x$ and $B \in \mathcal P(X)$ are such that $A \subset B$, then $B \in \mathcal F_x$

(4) if $x \in X$ and $A,B \in \mathcal F_x$, then $A \cap B \in \mathcal F_x$

(5) if $x \in X$ and $A \in \mathcal F_x$, then there is $B \in \mathcal F_x$ such that $B \subset A$ and $B \in \mathcal F_y$ for all $y \in B$.

If $\mathcal F$ is a neighbourhood filter system and if we define

$\tau=\{A \in \mathcal P(X) :\space \text{for all} \space x \in A, A \in \mathcal F_x\} \cup \{\emptyset\}$, show that $\tau$ is a topology.

The attempt at a solution

I have to prove three properties

(i) $X \in \tau$ (ii) arbitrary union of open sets is open (iii) finite intersection of open sets is open

I am pretty lost and confused with the exercise, for example, I have to prove that $X \in \tau$, but $X \in \tau$ if and only if for all $x \in X$, $X \in \mathcal F_x$. None of the five axioms describe $\mathcal F_x$ so how can I know that $X \in \mathcal F_x$?

The same goes for (ii) and (iii), suppose $U_i \in \tau$ for $i \in I$, I want to prove that $\bigcup_{i \in I} U_i \in \tau$, which means, for every $x \in \bigcup_{i \in I} U_i$, $\bigcup_{i \in I} U_i \in \mathcal F_x$, again, I have no idea how to use the axioms to check this.

I would really appreciate help with the problem.


(i) Note that for every $x\in X$ we have $\mathcal{F}_x\neq \emptyset$ by (1), so let $A\in\mathcal{F}_x$ for some $A \subseteq X$. So we use (3) and directly get $X\in\mathcal{F}_x$. Since $x\in X$ was arbitrary we get $X \in \mathcal{F}_x$ for all $x\in X$ which means $X\in \tau$.

(ii) Suppose $\{U_i:i\in I\}$ is a family of sets of $\tau$. To show shat $V:=\bigcup_{i\in I}U_i \in \tau$ we pick any $x \in V$ and show that $V \in \mathcal{F}_x$. So if $x\in V =\bigcup_{i\in I}U_i$ there is an $i\in I$ such that $x\in U_i$. Since $U_i$ is open we get $U_i \in \mathcal{F}_x$ by the definition of $\tau$. Again using (3) and the fact that $U_i \subseteq V$ to get $V \in \mathcal{F}_x$.

(iii) Let $A, B\in \tau$ and we want to show that $A\cap B\in \tau$. Again, take any $x\in A\cap B$ and show that $A\cap B \in \mathcal{F}_x$. Since we know that $A, B\in \tau$ we get $A, B\in \mathcal{F}_x$. So we use (4) and get $A \cap B\in \mathcal{F}_x$. So we have shown that for all $x\in A\cap B$ we have $A\cap B \in \mathcal{F}_x$ which implies $A\cap B \in \tau$ by definition of $\tau$.


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