# What is your definition for neighborhood in topology?

As you know, Munkres-Topology and Rudin-Analysis are really widely using textbooks for undergraduates. They all define a 'neighborhood of $x$' as an open set containing $x$, so i have followed this definition for 6 months. However, surprisingly, Wikipedia defines a 'neighborhood of $x$' as a set containing an open set containing $x$.

This really makes me annoyed, since this means that whenever I find a definition referring to a neighborhood on Wikipedia, I have to check whether that definition is equivalent to my definition of a neighborhood.

Which one is widely used?

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I would never put a Wikipedia definition over the definition of any widely used textbook – Leon palafox Jan 16 '13 at 8:11
Same here, but I have to say I never used Rudin nor Munkres, and whenever I read a definition of a "neighbourhood" of some point $x\in X$, it was a set $V\subseteq X$ with $p\in V$ containing an open subset $U\subseteq X$ with $p\in U$. But then again, in my experience you nearly always read "Let $U$ be an open neighbourhood of $x$", since one can usually restrict to such and show the desired stuff there. – InvisiblePanda Jan 16 '13 at 8:15
This seems to be essentially the same question as math.stackexchange.com/q/80025 – Martin Jan 16 '13 at 8:16
There are advantages to both definitions. For example one might like to say that a space is locally compact if for every neighbourhood of a point contains a compact neighbourhood... but compact subsets are usually not open! – Zhen Lin Jan 16 '13 at 8:20
@Leon, Rand: really? I consider the average Wikipedia article much, much better than the "widely used textbook" of, say, Stewart. Do you realize that people like Terry Tao and other luminaries regularly contribute and refer to Wikipedia? – Georges Elencwajg Jan 16 '13 at 11:57

From mathworld@wolfram:

In a topological space, an open neighborhood of a point is an open set containing it. A set containing an open neighborhood is simply called a neighborhood.

In most cases proofs involve open neighborhoods, so it usually shouldn't make too much of a difference, but It does look like different text books define this differently.

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Would you please answer my comment above? – Katlus Jan 16 '13 at 9:13
@Katlus - Of course it guarantees it. by definition, if a neighborhood (wiki def) exists it contains an open set which in turn contains the point. – nbubis Jan 16 '13 at 9:17

Rudin's Real and complex analysis, third edition (1987), page 9 of the French translation (1998):

Définissons d'abord un voisinage d'un point $x$ comme un ensemble contenant un ouvert contenant le point $x$. (Let us first define a neighborhood of a point $x$ as a set containing an open set containing the point $x$.)

It appears (thanks to @Martin for this) that the English and the French versions disagree since, on page 9 of the third English edition there is a parenthetical remark defining neighborhoods:

(A neighborhood of a point x is, by definition, an open set which contains x.)

This decision of the French translator of Rudin's book to modify this definition backfires on him, later on in the book, on page 35-36 Definition 2.3(d): there, the English text defines again a neighborhood as open and mentions parenthetically that some authors use the other definition; and all of this is translated faithfully in the French edition, in contradiction with the choice made earlier on to modify Rudin's text. Traduttore, traditore...

Munkres's Topology, second edition (2000), indeed stipulates that every neigborhood is open and, immediately after the definition, signals the alternative definition (pages 96-97).

All in all, it seems that readers of Rudin's and Munkres's books might not be completely taken aback by Wikipedia's version since both these authors, while following the other convention, explicitly mention this one.

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The English and the French versions seem to disagree on that point. On page 9 of the third English edition there is a parenthetical remark defining neighborhoods: (A neighborhood of a point $x$ is, by definition, an open set which contains $x$.) – Martin Jan 16 '13 at 9:12
@Martin Indeed. Thanks for your remark, see edited answer. – Did Jan 16 '13 at 9:41
+1 for the nice addition on backfiring and especially for the last sentence of your post. – Martin Jan 16 '13 at 9:47
A very good point about Munkres' second edition-this clearly signals the OP was probably using the first edition. – Mathemagician1234 Jan 17 '13 at 4:59

Since the question asks "What is your definition?" I'll say that my book Topology and Groupoids (first edition, "Elements of Modern Topology" (1968)) uses the definition that a neighbourhood $N$ of $x$ is such that $x$ is contained in the interior of $N$. Thus in the real line, $[0,1]$ is a neighbourhood of all of its points except $0,1$.

In practical terms, the difference between the two definitions is marginal, except that the neighbourhood axioms seem simpler with the more general definition.

I still hold to the idea that for a beginner, the definition of a topology in terms of neighbourhoods is the most intuitive and easily motivated; thus for continuity it is related to $\varepsilon-\delta$ methods in analysis. Of course students have to become familiar with the open set definition as well, including that for continuity, but should not have the idea imposed that there is only one route to the useful concepts.

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Personally, I prefer the definition of it being a set that contains an open set that contains the element. When people say that a neighbourhood of $a \in X$ is just an open set that contains the element a, it takes off some cases. I'll explain it better:

We say that a is an interior point of $X \subset \mathbb{R}$ when there is a number $\epsilon>0$ such that the open interval $(a-\epsilon,a+\epsilon)$ is a subset of $X$.

The set of the interior points of $X$ is called the interior of the set $X$, represented by $int\space X$.

When $a\in int\space X$, we say that $X$ is a neighbourhood of the point a.

A set $A \subset \mathbb{R}$ is called an open set when $A = int \space A$, that is, when all the points in A are interior to A.

Being so, when a 'neighbourhood of $x$' is defined as being an open set containing $x$, it is not considering the cases in which the set in which $x$ belongs is a closed set. Let's say that we have $c<x<d$ and $x \in A = [c,d]$. Then the interior of $A$ is the open set $int \space A = (c,d)$. Also, we have that $x \in int \space A$, and so $A$ is a neighbourhood of the point $x$ - it is a set containing an open set that contains the element $x$.

In the spectrum of topological spaces, one talks about open and closed neighbourhoods - in that case, the definition of an 'open neighborhood' can then be stated as an open set containing the element.

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