Topologies from neighborhoods

Question

Show that if each point $x$ in a set $X$ has assigned a collection $\mathscr{U}_x$ of subsets of $X$ satisfying $N-a$ through $N-d$:
N-a)If $U\in\mathscr{U}_x$, then $x\in U$;
N-b)If $U,V\in\mathscr{U}_x$,then $U\cap V\in \mathscr{U}_x$;
N-c)If $U\in\mathscr{U}_x$,then there is a $V\in \mathscr{U}_x$,such that $U\in\mathscr{U}_y$ for each $y\in V$;
N-d)If $U\in\mathscr{U}_x$ and $U\subset V$,then $V\in\mathscr{U}_x$
Then the collection $$ \tau =\left\{ G\subset X|for~each~x~in~G,x\in U\subset G~for~some~U\in \mathscr{U}_x \right\} $$ is a topology for $X$,in which the nhood system at each $x$ is just $\mathscr{U}_x$.

I've proved that $\tau$ is indeed a topology and (nhood system at each $x)\subset\mathscr{U}_x$.But I don't know how to show that $\forall U\in\mathscr{U}_x$,$U$ is a nhood of $x$, i,e.$x\in int(U)$.

Show that the $V$ from N-c) is open and the interior of $U$. — Henno Brandsma, May 01 '21 at 14:08
@HennoBrandsma I don’ think $V$ must be open or the interior of $U$.It can be a small closed ball in $U$ when $X$ is a metric space. — zik2019, May 01 '21 at 14:10
it’s at least easy to see it’s a subset of the interior. And that’s already enough. — Henno Brandsma, May 01 '21 at 14:12
@HennoBrandsma I don’t know how to show it.Would u like to give me more details.I seem to be at a dead end.Thanks! — zik2019, May 01 '21 at 14:14
I would think that in order to show $X \in \tau$, you would need an assumption such as that each $\mathscr{U}_x$ is nonempty (which in combination with N-d would imply that $X \in \mathscr{U}_x$ for each $x \in X$).. — Daniel Schepler, May 01 '21 at 15:05
@DanielSchepler yes, this is implicit. We have a nontrivial filter for each point. — Henno Brandsma, May 01 '21 at 15:11

score 0 · Accepted Answer · answered May 01 '21 at 14:56

Given $U \in \mathcal{U}_x$ define $U^\ast = \{y \in U\mid U \in \mathcal{N}_y\}$.

Then for the $V$ given by N-c) for this $U$ we have $V \subseteq U$ (as $y \in V$ implies $U \in \mathcal{U}_y$ so $y \in U$ by N-a)) and also $V \subseteq U^\ast$ by definition, so that $U^\ast \in \mathcal{N}_x$ by the enlargement axiom N-d).

Now $U^\ast$ is open in the defined topology: let $z \in U^\ast$. So $U \in \mathcal{N}_z$. Now apply N-c) to this $U$ and $z$ to get a $V$ in $\mathcal{N}_z$ which obeys $\forall y \in V: U \in \mathcal{N}_y$, so in particular $y \in U^\ast$ for all $y \in V$ so $V \subseteq U^\ast$ and so $U^\ast \in \mathcal{N}_z$ by N-d) again. As $z$ was arbitrary in $U^\ast$, $U^\ast$ is open in this topology by definition. It follows that $x \in U^\ast \subseteq \mathrm{int}(U)$, as required.

This argument is a part of my larger answer here if you're interested in full details both ways. — Henno Brandsma, May 01 '21 at 14:59

Topologies from neighborhoods

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