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I am reading a collection of notes from an introductory course on topology. I came across this seemingly bizarre definition and was wondering if somebody could explain the usefulness of it. It seems entirely arbitrary to me.

"If X is a topological space and S is a subset of X then a neighbourhood of S is a set V, which contains an open set U containing S i.e. $S \subset U \subset V \subset X.$"

A lot of these topological definitions just appear to be going down some sort of rabbit hole and there is no hint as to why.

saulspatz
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    So for example in $\Bbb R$ with its usual topology, $S:=[0,2]\cup (3,5)\cup {6}$ is a neighbourhood of ${1,4}$, but not of ${0}$ or ${2}$ or ${6}$. – Hagen von Eitzen Feb 14 '19 at 16:57
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    If the terminology "neighborhood" is never used again in the notes you are reading, feel free to ignore it. The more likely case is that the terminology is indeed used in several later places, in which case go look for those places and try to understand how the terminology is used. – Lee Mosher Feb 14 '19 at 16:57
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    It simply says that $V$ is a neighborhood of $S$ if it is a neighborhood of every point of $S.$ That doesn't sound so weird to me. – saulspatz Feb 14 '19 at 16:59
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    Some authors take "neighbourhood" to be "open neighbourhood" for convenience. Some may use a more complicated definition like the one in this question. Whatever the choice, the mathematical context remains the same. – GNUSupporter 8964民主女神 地下教會 Feb 14 '19 at 17:01
  • Thank you all, it seemed strange to introduce V just to introduce U containing S. It appears to simply be a strict formality that was confusing me at first glance. I have seen some other much more understandable definitions of a neighborhood, but as stated the context should remain the same.

    Any suggestions to a great topology book with some conversation/examples as opposed to definition after definition?

    – JacobCheverie Feb 14 '19 at 18:10
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    @JacobCheverie. Neighborhoods are an historical hangover. They can be weird. (-1,1) $\cup$ A, where A is any set whatsoever, is a neighborhood of 0. A could be Q or Z which has points almost infinitely far from 0. – William Elliot Feb 14 '19 at 21:23
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    This is the same definition as the one used by Bourbaki ("General Topology"). One advantage of this definition of a neighborhood is that a map $f: X\to Y$ between topological spaces is continuous at a point $x\in X$ if and only if for every neighborhood $V$ of $f(x)$ the preimage $f^{-1}(V)$ is a neighborhood of $x$. This will fail if you require neighborhoods to be open as it is customary in English-language topology literature. – Moishe Kohan Feb 15 '19 at 00:03

1 Answers1

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Given a point $x$, it is often necessary to talk about sets which contain all points sufficiently close to $x$. Thus the concept of a neighborhood. "Sufficiently close" is exactly what the concept of open set was created to describe, so this goes from the heuristic

$V$ is a neighborhood of $x$ if it contains every point sufficiently close to $x$.

to the exact

$V$ is a neighborhood of $x$ if it contains an open set about $x$.

or more symbolically

$V$ is a neighborhood of $x$ if there is some open $U$ with $x \in U \subseteq V$.

Your definition is for neighborhoods of sets instead of just points, but is equivalent to just requiring $V$ to be a neighborhood of every point in $S$.

So why not just stick with open neighborhoods, as is quite common in English mathematics, and as William Elliot apparently supports? Because there are a lot of situations where one has to talk about sets that contain every sufficiently close to $x$, but are not necessarily open themselves. While I am writing this, Moishe Cohen has provided one example. Another is the definition of "locally compact". Which do you prefer?

$S$ is locally compact if for every $x \in S$ there is a compact set $V$ and an open set $U$ with $x\in U \subseteq V$.

or

$S$ is locally compact if every $x \in S$ has a compact neighborhood.

The second definition only works when neighborhoods need not be open.

Like all terminology, "neighborhood" is just a shorthand for a useful concept, so that one does not need to specify it in detail every time it is used. Since Moishe has already brought Bourbaki up, I'll mention that Bourbaki's definition of "1" involves an estimated 24000 symbols (if I recall - it has been a very long time since I read it). One would hardly want to repeat that every time you count.

In most cases, it is sufficient to use open neighborhoods, and there are many occasions where open neighbhorhoods are required. For this reason, many authors simply choose to define neighborhoods to be open so that they don't have to say "open" each time, preferring to say "set containing a neighborhood" for those fewer occasions where having the neighborhood be open gets in the way. But this is a matter of preference, and I personally find it more convenient to use "open neighborhood" when open is necessary, and allow neighborhoods to not be open.

Paul Sinclair
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  • A space is regular when every point has a local base of closed neighbourhoods (as another example of where a wider definition makes for compact formulations). – Henno Brandsma Feb 15 '19 at 06:56
  • Thank you for the great response Paul. It would be my belief that if a neighborhood can be closed, then we should allow for the possibility. It is nice to see some examples provided, though they are beyond me right now. The reason that I was initially confused is that this is new to me and I didn't understand that it was defining a neighborhood of a set, not point as you explained, though I see now that I should have registered the subset notation. I saw an extra set as redundant. It is much clearer now. I will take a look at Bourbaki. – JacobCheverie Feb 15 '19 at 15:17