Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer
    
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 $0$.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.