12

Given a topological space $(X,\tau)$ and a point $x\in X$ we can define a fundamental system of neighborhoods of $x$ (or perhaps a neighborhood base at $x$), say $\mathscr{N}(x)\subseteq2^X$, by every neighborhood $U$ of $x$ contains an element of $\mathscr{N}(x)$ and the elements of $\mathscr{N}(x)$ are themselves neighborhoods of $x$ (I'm taking a neighborhood of $x$ to be a set containing an open set containing $x$). For example, $\{(-1/n,1/n)\}_{n\in\mathbb{N}}$ and $\{[-1/n,1/n]\}_{n\in\mathbb{N}}$ are both fundamental systems of neighborhoods of $0\in\mathbb{R}$ with the standard topology.

So right, we can go from a topological space to a fundamental system of neighborhoods at a point. I've seen this addressed in certain situations before, but how can you go from having a fundamental system of neighborhoods at every point to a topology? It seems to me you would take finite intersections and arbitrary unions of the various fundamental systems of neighborhoods. But then $\{[x-1/n,x+1/n]\}_{n\in\mathbb{N},x\in\mathbb{Q}}$ would not generate the standard topology on $\mathbb{R}$.

In short, how do you go from no topology and just a bunch of sets (which you would wish to call a bunch of fundamental system of neighborhoods) to a topology? Do you just take finite intersections and arbitrary unions of your sets?

danzibr
  • 712
  • 1
    Why wouldn't your example generate the standard topology on $\mathbb R$? Which open sets would it be missing? – AndreasT Mar 05 '13 at 14:07
  • 1
    It would be missing nothing, but it would include, for example, closed intervals. – danzibr Mar 05 '13 at 14:28
  • 1
    @danzibr Here is a short explanation: A fundamental system of neighborhoods ${F_p}$ produces a neighborhood system ${ N_p }$, where $N_p = { A \subseteq \mathbb{R} \mid F \subseteq A \text{ for some } F \in F_p }$. We declare a subset $U \subseteq \mathbb{R}$ open if $\forall p \in U$ we have $ U \in N_p$, i.e. $U$ is a neighborhood to each point inside it. The interval $[a,b]$ is not open because, for example, it is not a neighborhood of $a$ (it doesn't contain intervals of the form $[a-1/n, a+1/n]$). So the topology you get is the standard topology. – Ra1le May 24 '21 at 15:10

2 Answers2

8

Given for each $x \in X$ a nonempty set $U_x \subseteq \mathcal{P}(X)$ which satisfies
(1) $x \in \bigcap U_x$
(2) $V \in U_x, V \subseteq W \Longrightarrow W \in U_x$
(3) $V,W \in U_x \Longrightarrow V \cap W \in U_x$
(4) $\forall V \in U_x \,\exists W \in U_x: \forall y \in W: V \in U_y$
then there is exactly one topology on X s.t. for all $x \in X$: $U_x$ is the set of neighborhoods of $x$. A subset $O \subseteq X$ is open, iff it is a neighborhood of each of its elements - i.e. $\forall x \in O: O \in U_x$.

Both fundamental systems of neighborhoods, you mentioned above, are filter bases for the same $U_x$ which is obtained by adding all supersets in order to satisfy condition (2).

Dune
  • 7,397
  • In (3) you certainly mean $V\in U_y$? And you don't have to require $W\subseteq V$ since it follows from $\forall y\in W:V\in U_y$. – Stefan Hamcke Mar 05 '13 at 14:23
  • I just noticed... Don't you have to require that the intersection of two neighborhoods is also a neighborhood? You need it so the topology will be closed under taking finite intersections. – Stefan Hamcke Mar 05 '13 at 14:26
  • Ah, yes of course. The $U_x$ are ment to be filter, so they should be closed under finite intersections. – Dune Mar 05 '13 at 14:38
  • 1
    Sorry, I don't mean to be bugging you, but your answer would be perfect if you added that $U_x$ must be non-empty, otherwise there would be a point $x$ such that every $O$ containing it would not be open. – Stefan Hamcke Mar 05 '13 at 17:01
  • That's perfect now ;-) – Stefan Hamcke Mar 05 '13 at 17:14
  • 1
    Done ;) Interestingly my source (Bourbaki, General Topology) also forgot to mention, that these sets have to be nonempty (which is obviously necessary). – Dune Mar 05 '13 at 17:16
  • @Dune: Alright, think I got it. Do you want to mention in $(4)$ that $W\subseteq V$? And instead do you want $W\in U_y$? That condition is supposed to say that every neighborhood of $x$ contains an open set containing $x$, right? Oh, and sorry for the delayed response. – danzibr Mar 13 '13 at 01:57
6

Your neighbourhood base $\mathcal V_x$ has to be non-empty and satisfy three properties:

(1) Each $V\in\mathcal V_x$ has to contain $x$.
(2) If $V_1,V_2\in\mathcal V_x$ then there is $V_3\in\mathcal V_x:V_3\subseteq V_1\cap V_2$.
(3) For each $V\in\mathcal V_x$ there is a $W\in\mathcal V_x$ such that for each $y\in W$ exists a $Y\in\mathcal V_y:Y\subseteq V$.

Then you can define $\mathcal U_x:=\{U\subseteq X\mid\exists V\in\mathcal V_x,V\subseteq U\}$ which satisfies (1) - (4) in Dune's answer. You can then proceed by defining a topology such that $\mathcal U_x$ is the neighborhood filter at $x$.

The $\left\{\left[x-\frac1n,\ x+\frac1n\right]\right\}_{n\in\mathbb{N}}$ neighborhood base would lead to the same neighborhood filter as $\left\{\left(x-\frac1n,\ x+\frac1n\right)\right\}_{n\in\mathbb{N}}$, since $\left[x-\frac1n,\ x+\frac1n\right]$ contains $\left(x-\frac1n,\ x+\frac1n\right)$, and $\left(x-\frac1n,\ x+\frac1n\right)$ contains $\left[x-\frac1{2n},\ x+\frac1{2n}\right]$.

Stefan Hamcke
  • 27,733
  • The "You can then proceed to define a topology ..." in the unclear step (unclear to me at least). Do you take finite intersections and arbitrary unions of all the elements of the fundamental systems of neighborhoods? – danzibr Mar 05 '13 at 16:14
  • 1
    @danzibr: See the last paragraph of my answer. The answers of Stefan H. and mine are essentially the same. The only difference is: while Stefan's answer deals with fundamental systems of neighborhoods, my answer deals with entire neighborhoods. – Dune Mar 05 '13 at 16:27
  • @Dune: You still have to change (4) to $V\in U_y$ at the end of the line. – Stefan Hamcke Mar 05 '13 at 16:32
  • @StefanH.: Yes you are right - I've corrected it now. Sorry for the confusion :) – Dune Mar 05 '13 at 16:50