10

One way to define a boundary point of set S is that "every neighborhood of it contains at least one point of S and at least one point outside S".

I wonder if it's OK to replace "neighborhood of it" by "open ball centered at it"? What's the difference? A further question is why introduce the concept "neighborhood" in the first place? Its role seems very similar to open balls.

Mao
  • 109
  • 17
    Not every topological space has a metric. There might not be open balls! – vadim123 Aug 10 '16 at 04:00
  • Assuming metric space N=U{B(n,e_n) |n is in N and e_n is some constant dependent on n} so your intuition is good for metric spaces. – Jacob Wakem Aug 10 '16 at 05:05
  • @vadim123 I thought neighborhoods were defined in terms of open balls? – John Dvorak Aug 10 '16 at 07:45
  • @Jan A neighborhood of $x$ in a topological space is a set that contains an open subset that contains $x$. In metric spaces, this is the same as a set that contains an open ball centered at $x$, so you might see that being used as the "definition", but it only works for metric spaces, not general topological spaces. – Mario Carneiro Aug 10 '16 at 18:14

5 Answers5

14

In a metric space, the two definitions are equivalent, since a neighborhood of $x$ is just a set containing an open set containing $x$. In a topological space, there generally isn't a metric, and so open balls can't really be defined, and so neighborhoods are used. In topological spaces, neighborhoods are often used where open balls would be used if we were dealing with a metric space.

florence
  • 12,819
8

An open ball is a basic open set in the standard topology on metric spaces. But if the space is not metric you need the general notion of open set.

An open set is an element of some topology over some space (check the axiomatic definition of topology). A neighborhood of some point is a set that contains an open set that contains the point, i.e. if we have the topological space $(D,\mathcal T)$ and $x\in D$, then $N$ is a neighborhood of $x$ if and only if

  1. Exist some $U\in\mathcal T$ such that $x\in U$ and
  2. $U\subseteq N$ (and of course $N\subseteq D$)

In metric spaces the standard topology is induced by the metric such that if $U_{\epsilon}$ is an open ball of radius $\epsilon>0$ then $U_{\epsilon}\in\mathcal T_m$, where $\mathcal T_m$ is the topology induced by the metric in the space.

Masacroso
  • 30,417
4

Neighborhood can be seen as generalization of open balls when the concept is extended to non metric space. Concept of open spheres depends on distance defined between two points. However topological spaces allow the neighborhoods to be defined in terms of sets. A topology can then be defined as class of sets which satisfies closure under arbitrary unions and finite intersection and icludes entire set as well as empty set. For example consider $X=\{a,b,c,d\}$ the topology $T=\emptyset, \{a\},\{a,b\},\{a,b,c\}, \{a,b,c,d\}$ . Here both c and d are boundary points for $\{a,b\}$

Curious
  • 1,278
3

While the other answers cover the most obvious reason well, let me add a couple of reasons for defining and using the concept of neighborhoods that apply even in metric spaces:

  • Quite often, one needs to discuss sets that contain an open ball about a particular point, but are not open balls themselves. For example, "Let $S$ be a compact neighborhood of $x$" is easier to say repeatedly than "Let $S$ be a compact set that contains an open ball about $x$". If you need to refer to a concept enough, it is useful to have a special name for that concept. (Though I admit that many people restrict "neighborhoods" to be open, so this particular example does not apply to them - but there are other cases.)
  • Abstraction is one of the most important tools in the mathematician's toolbox. By being too specific in your definitions, you sometimes find yourself bogged down in needless details, while missing what is really important. In your example of the definition of the boundary point, the only thing that is important for that definition is no matter how close you look around the point, you see points that are inside and that are outside. The particular shape of how you are looking around the point makes no difference. So a definition that does not require a particular shape helps you to see what is important.

The latter reason may not make much sense to you without more experience, but I cannot tell you how many times I've struggled mightily to prove something, only to come back to it a few years later and discover that it is almost trivial from the right point-of-view, and all the details I fought through were unnecessary.

Paul Sinclair
  • 43,643
2

In a metric space, given a set $U$ and a point $p$, the following are equivalent:

  • $U$ is a neighborhood of $p$
  • $U$ contains an open subset that in turn contains $p$
  • For some radius $r > 0$, $U$ contains (as a subset) the open ball of radius $r$ centered at $p$.
  • For some radius $r > 0$, $U$ contains (as a subset) the closed ball of radius $r$ centered at $p$.

You may find it enlightening to try to prove the above as an exercise.

Note that the ideas of open sets and neighborhoods apply in a more general setting than metric spaces. (In particular, they apply in any topological space.) The author of your source material was probably trying to instill the good habit of thinking about open sets and neighborhoods on their own terms, in order to prepare you for thinking about more general topological spaces later on.

Community
  • 1