# Definition of neighborhood and open set in topology

I am a Physics undergrad, and just started studying Topology. How do you define neighborhood and open set in Topology.Wikipedia gives a circular definition.

An open set is defined as follows.

In topology, a set is called an open set if it is a neighborhood of every point

While a neighborhood is defined as follows:

If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$, which includes an open set $U$ containing $p$

which itself contains the term open set.

How do you define it exactly?

• What book are you using to learn Topology? – davidlowryduda Jun 13 '12 at 11:05
• The whole point about topology is that you define open sets in any way you like, subject to a handful of constraints. There's no single fixed definition of an open set. Of course, commonly used sets (like the complex numbers for example) have standard topologies, with much tighter definitions of what constitutes an open set. – user22805 Jun 13 '12 at 11:06
• @mixedmath I was going through Geometry, Topology and Physics by Nakhara just to get an idea. But I need a more mathematical text which does not go into depth. I have been told Armstrong is good. What do you think? How about Munkres.? – user23238 Jun 13 '12 at 11:08
• @ramanujan_dirac: I'd always back Munkres up in a fight about introductory topology books. But it does have a certain level of rigor. – davidlowryduda Jun 13 '12 at 11:12
• I support Munkres book. – rschwieb Jun 13 '12 at 11:13

One defines a topology on a set by specifying the open sets.

Let $X$ be a set. If $\tau$ is a family of sets with the following properties, it is called a topology.

• $X$ and $\varnothing$ are in $\tau$
• Any (possibly infinite, even uncountably infinite) union of sets in $\tau$ is in $\tau$.
• The intersection of any finite number of elements of $\tau$ is in $\tau$.

We call the sets in $\tau$ the open sets. You can see that the collection of open sets in, for example, $\mathbb{R}^2$ has exactly this set of properties.

A neighborhood of a set $S$ is a set $P$ that contains an open set $U$ so $S\subset U\subset P$.

• So I can go on to say that If I can construct this family $\tau$ as prescribed above, then every set in this family is open? That although some of the sets in $\tau$ may also be closed, they are at least all open? – Relative0 Jan 15 '17 at 0:24
• @Relative0 Yes. It is possible for a set to be both open and closed. – Potato Jan 15 '17 at 22:57
• Thanks Potato but I wasn't quite asking that. What I wanted to know is if I can just take some arbitrary set of subsets of $X$, make sure this set of sets is closed under Unions of Intersections of subsets along with the $\emptyset$ and then just say - Wallah! They are all open just because I say so! – Relative0 Jan 21 '17 at 3:18
• @Relative0 Yes, if you find that collections of sets and check they satisfying these axioms, then they define a topology, and the sets are all open in that topology. – Potato Jan 21 '17 at 3:29

One of the problems of introducing students to topology is that the open set axioms are often taken as THE definition of a topology, when they are quite unintuitive, though extremely useful in the long run. I argue that the neighbourhood definition, while somewhat cumbersome, has the advantage of being closely related to ideas from analysis, and has a historical basis; it is of course as follows:

A neighbourhood topology on a set $X$ assigns to each element $x \in X$ a non empty set $\mathcal N(x)$ of subsets of $X$, called neighbourhoods of $x$, with the properties:

1. If $N$ is a neighbourhood of $x$ then $x \in N$.

2. If M is a neighbourhood of $x$ and $M \subseteq N \subseteq X$, then $N$ is a neighbourhood of $x$.

3. The intersection of two neighbourhoods of $x$ is a neigbourhood of $x$.

4. If $N$ is a neighbourhood of $x$, then $N$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$.

Then one says a function $f: X \to Y$ is continuous wrt neighbourhoods on $X$ and $Y$ if for each $x \in X$ and neighbourhood $N$ of $f(x)$ there is a neighbourhood $M$ of $x$ such that $f(M) \subseteq N$. The open set definition of continuity is then justified as being equivalent to this definition in terms of neighbourhoods.

One also says a set $U$ in $X$ is open if $U$ is a neighbourhood of all of its points.

THEN one can develop the open set axioms and show that one can recover the neighbourhoods.

Students should be aware that there are many approaches to the notion of topology, whose advantages should be compared. There should be no "take it or leave it" approach, but students should be encouraged to form a judgement, in terms of the character of the theory and its methods. And see which definition is appropriate in which cases.

June 14: The above approach is taken in my book Topology and Groupoids, in order to motivate the definition of open set.

November 17, 2016. Peter Freyd writes in the Introduction to his book Abelian Categories

"If topology were publicly defined as the study of families of sets closed under finite intersection and infinite unions a serious disservice would be perpetrated on embryonic students of topology. The mathematical correctness of such a definition reveals nothing about topology except that its basic axioms can be made quite simple. And with category theory we are confronted with the same pedagogical problem. ......

A better (albeit not perfect) description of topology is that it is the study of continuous maps; and category theory is likewise better described as the theory of functors. Both de­scriptions are logically inadmissible as initial definitions, but they more accurately reflect both the present and the historical motivations of the subjects."

• What guarantees that $\mathcal{N}(x) \neq \varnothing$, for any $x\in X$? (See math.stackexchange.com/questions/404659) – kjo May 28 '13 at 13:27
• @kjo: Good point! I am pleased to say my book includes the crucial words "a non-empty set". – Ronnie Brown Jun 4 '13 at 16:06
• I have added the "non empty" condition. – Ronnie Brown Jul 24 '15 at 17:56
• Nice answer. And thank you so much for making your book available online! – evaristegd Sep 1 '18 at 15:28
• I'm glad the online availability is appreciated, and I know "clickability" via hyperref can be very useful. Making accessible both versions of the book was a kind of moral duty in view of the way writing the book and so trying to clarify my understanding as I drafted versions led me into so many research paths. – Ronnie Brown Sep 8 '18 at 20:34

To complement the other answers, which tell you what the normal definition of open set in a topology, I'll give another possibility for the definition of neighbourhood in a metric space (note that this won't make sense for general topological spaces, but I think it's what's motivating the definition of open set you gave).

For a point $p$ in a metric space $(X,d)$, say that a subset $U\subset X$ is a neighbourhood of $p$ if there exists $\varepsilon>0$ such that $B(p,\varepsilon)=\{x\in X:d(x,p)<\varepsilon\}$ is a subset of $U$. Now the definition of open set you've given agrees with the usual one for metric spaces.

Usually, you define a set to be open in a space $X$ if and only if it is in the topology $T$ of $X$.

For example, you can take $X = \mathbb R$ and endow it with what is called the trivial topology, $T = \{ \varnothing, X = \mathbb R\}$. Then the only open sets in this space are $\varnothing$ and the whole space $\mathbb R$.

A more common example is $X = \mathbb R$ with the standard topology. This is what you might be familiar with: you start with the set of all intervals $(a,b)$ and then define the topology to be the set of all unions of these intervals.

Then you can define a neighbourhood of a point $x$ to be a set $N$ such that there exists a set $O \in T$ such that $x \in O \subset N$.

• If you would like to read some introductory material, have a look at Bert Mendelson's "Introduction to Topology". It's a very nice book and very easy to read. – Rudy the Reindeer Jun 13 '12 at 11:12

After the topology is defined, all open sets are known.

A "neighborhood of $x$" is a set containing an open set containing $x$. Usually I'm only thinking about open neighborhoods, but I guess some people find convenient uses for the more flexible definition of a not necessarily open neighborhood.

In this sense, $[0,1]$ is a neighborhood of $0.5$, but it is not a neighborhood of $0$.

The definition of a topology (in the most abstract point of view) is given by the collection of the open sets. It has to verify that

• a union of open sets is open
• a finite intersection of open sets is open
• $X$ and $\varnothing$ are open

exemple : the set of all subsets of $\mathbb{R}$ such that their complement is finite is a (weird) topology on $\mathbb{R}$.

Then you define a neighborhood $V$ of a point $x$ as a set containing an open set $O$ which contains $x$ (obviously, open sets are neighborhood of each of their elements).

This blog gives a good introduction to topology, neighborhoods, open sets etc.

There is also an alternative approach midway between open sets and the neighbourhood systems in Ronnie Brown's answer, one that is quite natural from an order theoretic point of view. Specifically, we could consider "neighbourhood relations" $$\prec$$ on subsets $$\mathcal{P}(X)$$, where $$M\prec N$$ signifies that $$N$$ is a neighbourhood of $$M$$, i.e. of all points $$x\in M$$. The axioms would be much the same as for neighbourhood systems, with an extra infinite union axiom like with open sets, specifically
\begin{align} \tag{0}\emptyset\prec\emptyset\qquad&\text{and}\qquad X\prec X.\\ \tag{1}M\prec N\qquad&\Rightarrow\qquad M\subseteq N.\\ \tag{2}K\subseteq L\prec M\subseteq N\qquad&\Rightarrow\qquad K\prec N.\\ \tag{3}L\prec M\text{ and }L\prec N\qquad&\Rightarrow\qquad L\prec M\cap N.\\ \tag{4}L\prec N\qquad&\Rightarrow\qquad\exists M\ (L\prec M\prec N).\\ \tag{5}\forall\lambda\in\Lambda\ (M_\lambda\prec N)\qquad&\Rightarrow\qquad\bigcup_{\lambda\in\Lambda}M_\lambda\prec N. \end{align}
Note axiom (4) for neighbourhood relations looks a little nicer than the corresponding axiom for neighbourhood systems. In domain theoretic terms, (4) is just "interpolation", while (1) and (2) are just saying that $$\prec$$ is "auxiliary" to $$\subseteq$$. Open sets are precisely the elements of $$\mathcal{P}(X)$$ on which $$\prec$$ is reflexive or, again in domain theoretic terms, the "compact" elements of $$\mathcal{P}(X)$$.
You could also consider uniform neighbourhoods, where $$M\prec N$$ signifies that $$N$$ contains an $$\epsilon$$-neighbourhood of $$M$$, for some $$\epsilon>0$$. These would satisfy the same axioms except that (5) would only apply to finite collections, i.e. $$\tag{5'}L\prec N\text{ and }M\prec N\qquad\Rightarrow\qquad L\cup M\prec N.$$ These would also satisfy an additional axiom for complements, namely $$\tag{6}M\prec N\qquad\Rightarrow\qquad X\setminus N\prec X\setminus M.$$ These are the axioms defining "proximity relations" - see https://ncatlab.org/nlab/show/proximity+space or https://en.wikipedia.org/wiki/Proximity_space. So neighbourhood relations are really just a slight variant of proximity relations.
One advantage of this approach would be the unification of continuity and proximal continuity(=uniform continuity under suitable conditions). Specifically, $$f$$ is continuous or proximally continuous iff its inverse preserves $$\prec$$, i.e. $$M\prec N\qquad\Rightarrow\qquad f^{-1}[M]\prec f^{-1}[N].$$