# Inquiry regarding neighborhood bases

I just want to see if I have the right angle here. We have a set $S$, and for each $x\in S$ there's a collection of subsets in $S$, $N(x)$ which satisfies:

1. $N \in N(x) \rightarrow x \in N$

2. $N,M \in N(x) \rightarrow \exists P \in N(x): P \subset N\cap M$

3. $x \in S \rightarrow N(x) \neq \phi$

then there's a unique topology on S such that $N(x)$ is a basis neighbourhood basis at x, for $x \in S$.

Am I right in thinking that topology is generated by the following open sets: $$\cup_{x \in S \ N \in N(x)} N$$

In which case it's unique by this very definition, right?

Thanks, I am quite rusty with topology, I took my first course in it 3 years ago.

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 I'm not sure the topology is unique: take for example $S=\mathbb R$, and for each $x\in S$, we consider $N(x):=\left\{\left[x-\frac 1n,x+\frac 1n\right],n\in\mathbb N^*\right\}$. Then $N(x)$ is a basis neighbourhood at $x$ for each $x$ for the usual topology, but also for a topology which contains for example $\left[-1,\right]$. – Davide Giraudo Sep 24 '11 at 19:43 If you have such families $N(x),x\in S$, then it might happen that there is no topology for which $N(x)$'s are neighbourhood bases. Namely take $S=\{0,1,2\}$, $N(0)=\{U\in\mathcal{P}(S):0\in U\}$, $N(1)=\{U\in\mathcal{P}(S):1\in U\}$ and $N(2)=\{S\}$; if $N(0)$ and $N(1)$ are neighbourhood bases at $0$ and $1$, then $\{0,2\}\cap\{1,2\}=\{2\}$ is open but then $N(2)$ can't be a neighbourhood basis at $2$. – LostInMath Sep 24 '11 at 19:47 @LostInMath: The nbhds don’t have to be open, so in your example the topology would be the one generated by the base $\left\{\{0\},\{1\}\right\}$. – Brian M. Scott Sep 25 '11 at 10:04 @Davide: That won’t work, because $-1$ won’t have a nbhd contained in $[-1,\to )$. – Brian M. Scott Sep 25 '11 at 10:05 @Brian: Right. I did the wrong assumption that a neighbourhood base consists of open sets by definition. Of course, this does not have to be the case when we are dealing with local bases. – LostInMath Sep 25 '11 at 11:38

Since $$\bigcup\limits_{x\in S,N\in N(x)}N$$ is simply the single set $S$, I doubt it’s what you actually had in mind.

Edit: On the (foolish) assumption that the nbhd bases were intended to be open nbhd bases, I originally wrote:

Let $$\mathscr{N} = \bigcup\limits_{x\in S}N(x),$$ the collection of all sets appearing in any of the families $N(x)$. Then $\mathscr{N}$ is a base for a topology $\mathscr{T}$ members of $-$ i.e., the open sets in the space $-$ are the unions of subcollections of $\mathscr{N}$.

Now that the wording of the problem has been confirmed, I realize that this assumption was unjustified, and what I wrote above isn’t necessarily true. Here’s a correct argument.

Let $\mathscr{T} = \{U \subseteq S:\forall x\in U\exists N\in N(x)[N \subseteq U]\}$. The first step is to prove that $\mathscr{T}$ is a topology on $S$.

1. $\varnothing \in \mathscr{T}$ vacuously, and (1) and (3) imply that $S \in \mathscr{T}$.
2. Suppose that $U,V \in \mathscr{T}$; we must show that $U\cap V \in \mathscr{T}$. To that end let $x \in U\cap V$. Then $x \in U$, so there is an $N_U \in N(x)$ such that $x \in N_U \subseteq U$, and $x \in V$, so there is an $N_V \in N(x)$ such that $x \in N_V \subseteq V$. By (2) there is $N \in N(x)$ such that $x \in N \subseteq N_U \cap N_V \subseteq U \cap V$. Since $x$ was an arbitrary element of $U \cap V$, it follows that $U \cap V \in \mathscr{T}$.
3. Now let $\mathscr{H}$ be any subfamily of $\mathscr{T}$; we must show that $\bigcup\mathscr{H} \in \mathscr{T}$. Suppose that $x \in \bigcup\mathscr{H}$; then $x \in H$ for some $H \in \mathscr{H}$. $H \in \mathscr{T}$, so there is some $N \in N(x)$ such that $x \in N \subseteq H \subseteq \bigcup \mathscr{H}$, and it follows as before that $\bigcup\mathscr{H} \in \mathscr{T}$.

It remains to show that $\mathscr{T}$ is unique. Suppose that $\mathscr{U}$ is a topology such that for each $x \in S$, $N(x)$ is a nbhd base at $x$ with respect to $\mathscr{U}$. Then for any $U \in \mathscr{U}$ and any $x \in U$ there must be some $N \in N(x)$ such that $x \in N \subseteq U$, which of course means that $U \in \mathscr{T}$. This shows that $\mathscr{U} \subseteq \mathscr{T}$.

Suppose now that $U \in \mathscr{T}$, so that for each $x \in U$ there is an $N_x \in N(x)$ such that $x \in N_x \subseteq U$. $N_x$ is a $\mathscr{U}$-nbhd of $x$, so there must be some $V_x \in \mathscr{U}$ such that $x \in V_x \subseteq N_x$. But then for each $x \in U$ we have $x \in V_x \subseteq U$, so $U = \bigcup\limits_{x \in U} V_x$. $U$ is therefore a union of members of the topology $\mathscr{U}$, so $U \in \mathscr{U}$, and hence $\mathscr{T} \subseteq \mathscr{U}$.

Putting the pieces together, we have $\mathscr{U} = \mathscr{T}$, and it follows immediately that $\mathscr{T}$ is unique.

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 This question from Bredon's textbook: Topology and geometry 1993 edition, question 5, page 8. Iv'e copied word for word the question. – MathematicalPhysicist Sep 25 '11 at 9:46 @MathematicalPhysicist: I wrote that on the assumption that the $N(x)$ were supposed to be open nbhd bases. Can you confirm that Bredon uses the term neighborhood more generally, so that a nbhd of $x$ is simply a set with $x$ in its interior? – Brian M. Scott Sep 25 '11 at 10:08 The definition of neighbourhood basis at x, is that its a collections of subsets of S containing x such that each neighbourhood of x in S contains some element of this basis and each element of the basis is a neighbourhood of x. A set is called a neighbourhood if there's an open set $U \subset N$ with $x \in U$. – MathematicalPhysicist Sep 27 '11 at 5:30

(Edited in response to a comment by jdc)

In your list of axioms the equivalent of the "topological triangle inequality" is missing. It says the following: For any neighborhood $U$ of the point $x$ there exists a neighborhood $V$ of $x$ such that $U$ is also a neighborhood of all points $y\in V$. If one expresses this in terms of your neighborhood bases $N(x)$ one arrives at the axiom

$4$. For any $N\in N(x)$ there is an $M\in N(x)$ such that for all $y\in M$ there is an $N'\in N(y)$ with $N'\subset N$.

The answer by Brian M. Scott above produces a topology ${\mathscr T}$ related to the family $\bigl(N(x)\bigr)_{x\in S}$ alright, but the given $N\in N(x)$ need not be neighborhoods of $x$ with respect to this topology. He then proves that if they are indeed neighborhoods then everything is in order.

Consider the following example: In $S:={\mathbb R}^2$ let $N(z)$ consist of all crosses $+$ with center $z$. The family ${\mathscr N}:=\bigl(N(z)\bigr)_{z\in S}$ violates axiom $4$ above. Now open sets in the topology ${\mathscr T}$ defined from ${\mathscr N}$ extend in two dimensions at all of their points. Therefore no $N\in N(0)$ can contain an open neighborhood of $0$.

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I was trying to figure this out as well; I noticed the extra axiom in Bourbaki. But Brian M. Scott's answer above seems to me to work, which confuses me even more. Do you know what's going on here? – jdc Nov 9 '12 at 9:10
@jdc: See my edit. – Christian Blatter Nov 10 '12 at 9:18
That makes much sense. Thank you. – jdc Nov 10 '12 at 14:04

You need the set of neighbourhoods that contain $x$ to be a filter, $\mathcal N_x$, on $X$ such that

$$x \in \cap \mathcal N_x = \bigcap_{A \in \mathcal N_x}A, \tag1$$

i.e. $x$ is in every element of $\mathcal N_x$.

And a local basis then needs to be a filter base $\mathcal B_x$ with the same intersection property, $x \in \cap \mathcal B_x$.

Indeed it's possible (but probably not a good idea) to describe all of topology in terms of filters: Suppose we have an ordered pair where $\langle X,(\mathcal N_x)_{x \in X}\rangle$, where each $\mathcal N_x$ satisfies $(1)$, then we can declare a $U \subseteq X$ to be open when it's the case that $x \in U \implies U \in \mathcal N_x$, so from the set of neighbourhood filters we can recover the topology.

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