0
$\begingroup$

Problem

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.

$\endgroup$
2
$\begingroup$

(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 $\tau$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.