0
$\begingroup$

The definition of neighbourhood, according to Wikipedia:

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

$p \in U \subseteq V$.

Note that the neighbourhood V need not be an open set itself.

But why should the neighbourhood of some point x be considered containing an open set, rather than closed?

$\endgroup$
  • $\begingroup$ Are you asking about the source of the terminology or about proving that a neighbourhood of some point x is open ? $\endgroup$ – Belgi Sep 24 '15 at 9:26
  • $\begingroup$ @Belgi I'm asking, why does the definition of neighbourhood say that the neighbourhood is an open set? $\endgroup$ – mavavilj Sep 24 '15 at 9:26
  • $\begingroup$ because $(a,b)\subset[a,b]$ but the converse isn't true $\endgroup$ – JMP Sep 24 '15 at 9:29
  • $\begingroup$ @JonMarkPerry And why does one want to consider the smaller set of the two? $\endgroup$ – mavavilj Sep 24 '15 at 9:35
  • $\begingroup$ not all sets have closed neighbourhoods around open ones $\endgroup$ – JMP Sep 24 '15 at 9:38
7
$\begingroup$

To set things straight:

A neighborhood of $x$ is any set containing an open neighborhood of $x$. An open neighborhood of $x$ is an open set containing $x$.

$\endgroup$
  • 2
    $\begingroup$ While I agree that this is the (much) better convention, there are also people - like Rudin - who define neighbourhoods to be open. Probably the OP read such a source. $\endgroup$ – Daniel Fischer Sep 24 '15 at 10:31
  • $\begingroup$ I don't agree that it is the "generally accepted definition" - the whole point of the term 'neighborhood' is that when you say "a nbd of $x$" you mean points that are "close to x" in a sense made precise by the topology. As a result, is actually no point in introducing the term 'neighborhood', if whenever you want to indicate an open set containing some point, you are required to add the adjective "open". $\endgroup$ – uniquesolution Sep 24 '15 at 11:05
  • $\begingroup$ So my comment is also written for nothing... we agree that nbd's are open. $\endgroup$ – uniquesolution Sep 24 '15 at 11:06
  • 4
    $\begingroup$ @uniquesolution No. Some authors define neighbourhoods to be open. Others (the vast majority of authors I've read) define a neighbourhood of $x$ to be any set $N$ with $x\in \overset{\Large\circ}{N}$. The latter definition has several advantages. The neighbourhoods of a point form a filter. You can speak of closed neighbourhoods and compact neighbourhoods of points. Occasionally, you need to say "open neighbourhood" when with the other convention "neighbourhood" would have sufficed. But on the whole, I find it much more convenient to not demand neighbourhoods be open. $\endgroup$ – Daniel Fischer Sep 24 '15 at 11:12
  • 1
    $\begingroup$ I was brought up in an environment where when you say "neighborhood" it is understood that the nbd is open, and when you want to indicate something else, close, compact, connected, you add the appropriate adjective. So we don't agree that nbd's are open, that's fine. $\endgroup$ – uniquesolution Sep 24 '15 at 11:15
2
$\begingroup$

The word "neighbourhood", was introduced into mathematics in order to express the notion that some points are close, or near, to others. Just as you do not identify all the inhabitants of your country as your neighbours, so should not all points of a topological space be identified as neighbours. However, Topology is somewhat more lenient than Geography, so it allows you to declare an entire space as a neighborhood of each point.

The notion of "nearness", or "closeness", between points in a general topological space is more abstract than it is when there is a metric that induces the topology. A metric introduces the notion of distance between points, and so it is easy to say when is a point $x$ close to a point $y$. Clearly if $d(x_1,y_1)<d(x_2,y_2)$, we understand that as much as $x_2$ is near $y_2$, $x_1$ is even nearer to $y_1$. Now forget the metric, and suppose you only have a topology on some set $X$. Pick two points in $X$, $x$ and $y$. How can you use the topology on $X$ in order to introduce a notion of "nearness" between $x$ and $y$? Well, all you have is two points and a family of subsets of $X$ satisfying a few axioms of topology. What can you do? You could ask if any of the sets in the family contain both of the points. Think about the family of sets as a family of geographical regions; if you find a region containing both your points, you would like to think of them as near to each other. Of course, the entire region always contains both of them, but that's not very helpful. Suppose you have two different regions containing both points. Then you can form even a potentially smaller region containing both of them – by intersecting both regions. So the more regions containing the points $x$, $y$, the closer they are to each other. What if both points belong to all the regions at once? Then depending upon the choice of regions, you can sometimes deduce that the points are equal. For example, if your topology has some property of separation.

The "regions" are of course the open sets of the topology. The nearness of points is expressed by the word 'neighbourhood'. That's why it is reasonable to think about neighbourhoods as open sets. That said, it should be emphasized that although the notion of being neighbours is naturally linked to the notion of being near or close to each other, the choice of the term 'open sets' is quite arbitrary, and we could just as well develop the theory of topology with 'closed sets'.

$\endgroup$
0
$\begingroup$

Because you want it to "surround" the point. If you accepted any closed set as a neighbourhood you could end up with only the point being the neighborhood.

Even if you require it to contain an disc of positive radius (which is meaningless in general topology, you could of course require it to contain an open set, but that seem to overcomplicate it) containing the point you would have an assymetry, the neigborhood of $a$ wouldn't need to be a neighborhood of all points in it since you would have points on the boundary of the neighborhood (again the problem with surrounding the other points).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.