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.

Let $X$ be a topological space, and $x \in X$ be a point. There are two prevalent conventions on how to define a neighborhood of $x$:

1) A neighborhood of $x$ is any open subset $W \subset X$ such that $x \in W$. (This convention is used in Munkres's book for example.)

2) A neighborhood of $x$ is a subset $W \subset X$ such that there exists an open set $A$ such that $x \in A \subset W$. (For example this is the definition in Bourbaki's General Topology)

Thus, every neighborhood in the sense of 1) is a neighborhood in the sense of 2), but not vice-versa.

One often needs to show that a neighborhood of a point $x$ in the sense of 2) is actually open, and often this is a non-trivial verification from the given context. An example that I can come up with now (and this example was a motivation for asking this question) is the following:

Let $G$ be a topological abelian group, and $H$ a subgroup of $G$ which is also a neighborhood of $0 \in G$ in the sense of 2). Then one can show that $H$ is in fact an open set.

[[The trick is to observe that for a given $g \in G$, the map $\phi_{g} :G \rightarrow G$ given by for $x \in G, \phi_{g}(x) = g + x$ is a homeomorphism, and so any neighborhood of a point $g \in G$ is of the form $g + U$, where $U$ is a neighborhood of $0$. Thus, for an $h \in H$, $h + H$ is a neighborhood of $h$, and moreover, $h + H \subset H$ because $H$ is a subgroup.]]

The above proof shows that sometimes proving that a neighborhood in the sense of 2) is open is not completely trivial, while a neighborhood in the sense of 1) is always open. At times like these, I wonder why the second definition of a neighborhood is used at all. But, I have learned topology primarily from Munkres, and so I might be ignorant of the advantages of definition 2).

So, what do you think are some of the advantages of using 2) as a definition for a neighborhood of a point $x \in X$, where $X$ is a topological space?

(This might be a duplicate question. But, I searched a little bit and could not find a question that was exactly similar to this one. So, excuse me if I have asked something that was already asked.)

share|improve this question
10  
I assume that you're aware that you can define topological spaces with the notion in 2), see here. The nice thing about it is that it is well adapted to arguments involving filters (Bourbaki's way of dealing with convergence) and uniform spaces (which in turn were motivated by topological groups and topological vector spaces). I prefer to add open to neighborhood when I intend Munkres's definition 1). –  t.b. Nov 7 '11 at 23:29
7  
I like the second definition just because it's nice to be able to say "take a compact neighborhood $U$ of $x$". Adding the word "open" every once and a while seems like less work than expanding this. –  Dylan Moreland Nov 7 '11 at 23:55
    
I seem to remember that if you go with (2) you can formulate some notion of "local continuity" that fails with (1). Since "locally continuous" doesn't seem to appear in Munkres, perhaps my memory serves me well. –  Joe Johnson 126 Nov 7 '11 at 23:56
    
Thanks. As I said, I have learnt topology primarily from Munkres, and so I am not aware of the advantages of using 2) as a definition. I asked this question so that I could begin to get an appreciation for definition 2). –  Rankeya Nov 7 '11 at 23:59
    
Also, I understand that picking out a particular example where 2) seems to lead to a bit more additional work than 1) is not really saying much. –  Rankeya Nov 8 '11 at 0:02

4 Answers 4

up vote 10 down vote accepted

I don't follow your argument; if anything, it strikes me as an argument for definition 2). You're saying that it may be non-trivial to show that a neighbourhood in the sense of 2) is a neighbourhood in the sense of 1). But why is the definition that contains more non-trivial properties the better one? It's easier to add the non-trivial properties when they're there than to subtract them out when they're not. If we use definition 2), we can easily express both concepts by speaking of "neighbourhoods" and "open neighbourhoods". If we use definition 1), we have to speak of "sets containing neighbourhoods" and "neighbourhoods" -- that's slightly more complicated.

Also, one often doesn't care whether the neighbourhood is open. For example, if a function is constant on a neighbourhood of $x\in\mathbb R$, its derivative at $x$ is zero. If you formulate this in terms of open neighbourhoods and for some reason you have that a function is constant on some closed interval, you have to insert the rather artificial step of specifying an open interval within that closed interval on which the function is constant. Or you could say that if a function is constant on a set containing a neighbourhood of $x\in\mathbb R$, its derivative at $x$ is zero, but that involves the complication with "sets containing neighbourhoods" again.

share|improve this answer
    
Thank you for this point of view. –  Rankeya Nov 7 '11 at 23:54

So, after all the great comments and answers (especially t.b.'s comment) I went and actually borrowed Bourbaki's General Topology from the library. After reading pages 18, 19, 20, I really have been able to get a better understanding of why definition 2) of a neighborhood is beneficial. As t.b., Andre and Joriki mention, it allows us to uniquely define a topology on a set X (proved as Proposition 2 in Bourbaki's book).

What I have been able to distill from all the comments and reading the first few pages of Bourbaki is that an important advantage of defining a neighborhood of a point $x \in X$ as a subset of $X$ that contains an open set containing $x$ is this simple consequence:

Any set that contains a neighborhood of $x$ is also a neighborhood of $x$. (This was mentioned by Joriki in his answer)

In fact note that if we define neighborhood as in 2) and denote by $B(x)$ the set of all neighborhoods of a point $x \in X$, then $B(x)$ satisfies the following properties:

(i) Any subset of $X$ that contains an element of $B(x)$ is an element of $B(x)$.

(ii) The intersection of two elements of $B(x)$ is an element of $B(x)$.

(iii) Every element of $B(x)$ contains $x$.

(iv) If $V \in B(x)$, then there exists $W \in B(x)$ such that for all $y \in W, V \in B(y)$.

Proposition 2 from Bourbaki now says the following: Let $X$ be a set, and for all $x \in X$ let $B(x)$ denote a collection of sets satisfying 1) - 4). Then there exists a unique topology on $X$ with respect to which the set $B(x)$ is precisely the collection of neighborhoods of $x$.

If one tries to prove this proposition, one will see that (i) is crucial in the proof of the fact that one can construct a unique topology just using neighborhoods (what t.b., Andre and Joriki have mentioned above), and it fails if we use definition 1) for a neighborhood. Note that conditions (ii), (iii), (iv) are satisfied by neighborhoods if defined in the sense of 1) (i.e., what people call open neighborhoods).

So, definition 2) puts open sets and neighborhoods on an equal footing. In the Introduction to General Topology, Bourbaki says that one can either take neighborhoods or open sets as the more primitive concept.

I also see why in a first course on topology, my professor did not decide to spend too much time explaining this issue.

(I do not mean to answer my own question this way. I am merely repeating what others have already said, as writing something out often helps me understand the concept better. So for those of you who have posted answers, please don't be offended.)

share|improve this answer

Let $X$ and $Y$ be two topological spaces. Denote by $\mathcal{V}_x$ the neighborhoods of $x \in X$. Do the same for $y \in Y$. Then, a function $f: X \to Y$ is continuous at a point $x \in X$ iff $$ f^{-1}\left(\mathcal{V}_{f(x)}\right) \subset \mathcal{V}_x. $$

The prototype for the concept of topological spaces is that of metric spaces. There, the natural concept is not that of "open set". In my opinion, the natural concept to make the link to the topology is that of a neighborhood base. The balls centered at $x$ are a neighborhood base at $x$. The same is true if you restrict yourself to balls of radius $\varepsilon_n > 0$, where $\varepsilon_n \rightarrow 0$.

With the concept of neighborhood, an open set is a set that is a neighborhood of all of its points. This is how one defines "open sets" in a metric space!! And this is only possible if the definition of "neighborhood" does not require that a set be open.

As you said, the fact that one has to worry about a set being or not open might be irrelevant, and yet, hard to prove. For example, for topological vector spaces or even topological groups, the topology is determined by the family of neighborhoods of a certain point. Linear transformations or homomorphisms are continuous iff they are continuous at a certain point. I see many professors struggling to prove that certain sets are open, when all they needed to do was to prove that they have non-empty interior!!

share|improve this answer

A few comments actually:

I assume that when you say $A \subset W$ in (2) you allow the case $A = W$ (if not, then you can take a space with one point and discrete topology and then by definition (2) the whole space is not a neighborhood of itself).

The point is axiomatically you define a topology on a space by defining (rather declaring) what are the open sets. Now you can declare ALL open sets or you can declare a small sub collection $B$ of open sets (called basis) and then ask for topology defined by that basis.

The distinction between Munkres and Bourbaki is that, Munkres defines the topology by defining the collection of open sets, Bourbaki implicitly defines a basis and then says a set W is open is every point has a basis element contained in W.

So both definitions are actually equivalent (by definition arbitrary unions of opens are open so neighborhood in the sense of Bourbaki is obliviously open).

So your group theoretic example is a tautology. What is interesting is to show that if $H \subset G$ has the claimed property then actually it is closed too.

share|improve this answer

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.