# Constructing a topology using neighbourhoods

Let $X$ be a non-empty set. I would like to show that if we define, for every $x\in X$, a non-empty family of subsets of $X$, $\mathcal V(x)$, satisfying:

• (V1) $\forall V\in \mathcal V(x), x\in V$
• (V2) $V\in \mathcal V(x), W\supset V \Rightarrow W\in\mathcal V(x)$
• (V3) $V_1, V_2\in \mathcal V(x) \Rightarrow V_1\cap V_2\in \mathcal V(x)$
• (V4) $\forall V\in \mathcal V(x) \exists W\in \mathcal V(x): \forall y\in W, V\in \mathcal V(y)$

then $$\tau=\{G\subseteq X: G\in \mathcal V(x) \ \forall x\in G\}$$ is a topology on $X$ whose neighbourhoods of a point $x$ are precisely $\mathcal V(x)$.

So far, I have showed that $\tau$ is indeed a topology (and to do so, only (V2) and (V3) are needed). Now, I have

$$V \text{ is a neighbourhood of x } \iff \exists G\in \tau: x\in G\subset V \\ \iff \exists x\in G\subseteq V: G\in \mathcal V(y)\forall y\in G \Rightarrow V\in \mathcal V(x)$$

but I am stuck in proving that every set in $\mathcal V(x)$ is a neighbourhood in $x$. Obviously (V4) must be the key, as provides that for every $V\in \mathcal V(x)$ we have $W\in \mathcal V(x)$ so that $V$ is a neighbourhood of every point of $W$. The point here is making $W$ to be a neighbourhood of every point in $W$, but I can't find the way.

• How did you show empty set and X are open? How did you show that the intersection of two open sets is open? That is not a direct result of V3. Feb 23 '18 at 20:44
• $\varnothing$ is trivially open, and $X$ is open because for every $x\in X$, there exists $V\in \mathcal V(x)$ (since $\mathcal V(x)$ is non-empty). Then, $X\supset V$, so $X\in \mathcal V(x)$. Now, if $G_1, G_2\in \tau$, pick $x\in G_1\cap G_2$. Then $G_1$ and $G_2$ belong to $\mathcal V(x)$, and by (V3), $G_1\cap G_2\in \mathcal V(x)$. Feb 23 '18 at 21:06
• This question looks a duplicate of Description of a topological space using neighborhoods.
– ChoF
Feb 23 '18 at 23:40
• The statement "non-empty families of subsets" means that, for every $x\in X$, $\mathcal V(x)$ contains at least one element. As for the intersection of two open sets, I would be glad if you could point out my mistake in the proof. Feb 24 '18 at 0:40
• I would usually expect (V4) to read more like $\forall V \in \mathcal{V}(x) \exists W \in \mathcal{V}(x) : W \subseteq V \wedge \forall y \in W, *W* \in \mathcal{V}(y)$. With the point being that then, this $W \in \tau$. Is it possible there was a typo or transcription error there? Feb 24 '18 at 0:56

Let $V\in \mathcal{V}(x)$. We want to find an open set $G$ such that $x\in G\subset V$. From $(V4)$ there is $W\in \mathcal{V}(x)$ such that for $y\in W$, we have $V\in \mathcal{V}(y)$. Let $\widetilde{W}=V\cap W$, so that we have $\widetilde{W}\subset V$, $\widetilde{W}\in \mathcal{V}(x)$ and for all $y\in \widetilde{W}$, we have $V\in \mathcal{V}(y)$.

Define $$G=\left\{y\in \widetilde{W};\text{ there is }W_{y}\in \mathcal{V}(y) \text{ with }W_y \subset \widetilde{W}\right\}.$$ I claim that $G$ is open, contains $x$ and is contained in $V$. First, let's show it's open. Let $y\in G$. Then, there is $W_y \subset \widetilde{W}$ with $W_y \in\mathcal{V}(y)$. Applying property $(V4)$ to $W_y$, we can find $Z_y \in \mathcal{V}(y)$ such that for all $z\in Z_y$, $W_y \in \mathcal{V}(z)$. Replacing $Z_y$ with $Z_y \cap W_y$, we can assume $Z_y \subset W_y$. But if $z\in Z_y$, we get $W_y \in \mathcal{V}(z)$, and $W_y \subset \widetilde{W}$, which means exactly that $z\in G$! We conclude therefore that $Z_y \subset G$, and since $Z_y \in \mathcal{V}(y)$, this implies $G\in \mathcal{V}(y)$. So $G$ is open.

It is of course a subset of $V$ (because it's contained in $\widetilde{W}$), and $x\in G$ because $\widetilde{W}\in \mathcal{V}(x)$.

• Many thanks! That was exactly what I was looking for. Just one comment: you don't have to define $\tilde W$, as (E1) guarantees that $W\subseteq V$: as $V\in \mathcal V(y)$ forall $y\in W$, then $y\in V$. Feb 24 '18 at 4:06

For $V \subseteq X$, define $i(V) := \{ x \in X \mid V \in \mathcal{V}(x) \}$. We then claim that $i(V) \in \tau$ for any $V \subseteq X$.

Thus, suppose $x \in i(V)$. Then from (V4), find a $W \in \mathcal{V}(x)$ such that for each $y \in W$, $V \in \mathcal{V}(y)$. Then this implies that $W \subseteq i(V)$, and so from (V2), we get $i(V) \in \mathcal{V}(x)$. This completes the proof of the claim.

Now, it is easy to check that if $V \in \mathcal{V}(x)$, then $x \in i(V) \subseteq V$ (using (V1) in proving the last inclusion), where we just showed $i(V)$ is open, so that implies $V$ is a neighborhood of $x$ with respect to $\tau$.

And conversely, if $V$ is a neighborhood of $x$, suppose $U \in \tau$ with $x \in U \subseteq V$. Then by definition of $\tau$, that means $U \in \mathcal{V}(x)$, and by (V2), we thus get $V \in \mathcal{V}(x)$.

(In fact, it is then easy to see that the $i$ we defined is exactly the topological interior with respect to $\tau$.)

• Many thanks for your answer! Your approach using an interior operator is very interesting. Feb 24 '18 at 4:08