A topology generated by neighborhood

Question

Let $X=\{a,b,c,d\}$ be a space.

We are given the neighborhood of the elements of $X$as:

$N(a)=\{X,\{a\},\{a,b\},\{a,c,d\}\}$

$N(b)= \{X,\{b\}, \{a,b\}, \{b,c,d\}\}$

$N(c)= \{X,\{a,c,d\},\{b,c,d\}, \{c,d\}\}$

$N(d)= \{X,\{a,c,d\},\{b,c,d\}, \{c,d\} \}$

How can we find the topology on $X$?

Is it true that we take union of these neighborhoods in order to generate the topology on $X$?

Answer 1

A set $O$ is open iff for all $x \in O$, $O \in \mathcal{N}(x)$ and we can easily test this.

$\emptyset$ is open because there are no $x$ to test.

$\{a\}$ is open as $\{a\} \in \mathcal{N}(a)$.

$\{b\}$ is open as $\{b\} \in \mathcal{N}(b)$.

$\{a,b\}$ is open as their union, e.g.

$\{c,d\}$ is open as $\{c,d\} \in \mathcal{N}(c)$ and $\{c,d\} \in \mathcal{N}(d)$.

Now we know the minimal open subset $O_x$ that contains $x$ for every $x \in X$. This easily determines the open sets of $X$, as the unions of the set $O_x$, (which are $\{a\}, \{b\}, \{c,d\}$). So we also get $\emptyset, \{a,b\}, \{a,c,d\}, \{b,c,d\}$ and $X$ as extra open sets and we have the full topology.

Answer 2

Brute-force method in a finite case: Denote the generated topology you aim to describe by $\tau_{N's}$. Keep taking (finite) unions and intersections of the sets in the union of all neighbourhoods given, until no new set pops up. Put all these sets together and call this collection $T_{N's}$. Add the empty set and $X$ if necessary. Now you have a candidate topology, say $T_{N's}$. By your construction it must be that $T_{N's}\subset \tau_{N's} $. It remains to show $T_{N's}\supset\tau_{N's}$, which follows if you can show that $T_{N's}$ is a topology containing the generating sets (using the definition $\tau_{N's}:=\ $smallest topology containing all sets in the given neighborhoods).