# Neighbourhood of a point

1. Sep 20, 2011

### gliteringstar

neighbourhood of a point
A set T (a subset of R) is called neighbourhood of a point s if for any positive number e>0,the open interval (p-e,p+e) is a subset of T

It is also given that any open interval is neighbourhood of all its points.

let us take an example

(2,3) is an open interval and according to the definitions given above it is neighbourhood of all its points

so it will be a neighbourhood of any of its points,say, 2.1 if we take any positive number,say .1

then 2.1 belongs to (2,2.2) which inturn is a subset of (2,3)

what if we take our positive number to be .2

in this case 2.1 belongs to (1.8,2.4)which isn't a subset of (2,3)

is there a restriction that we need to choose a very small positive number?

2. Sep 20, 2011

### gliteringstar

"what if we take our positive number to be .2

in this case 2.1 belongs to (1.8,2.4)which isn't a subset of (2,3)

is there a restriction that we need to choose a very small positive number?"

there is a mistake, the open interval obtained should be (2.1-.2,2.1+.2)= (1.9,2.3) instead of (1.8,2.4).
But the doubt remains as it is- (1.9,2.3) still isn't a subset of (2,3)

3. Sep 20, 2011

### Jamma

Firstly, this really is the wrong section for this question, you want to put it in the topology and geometry section really.

Secondly, your definition seems to be wrong:

"neighbourhood of a point
A set T (a subset of R) is called neighbourhood of a point s if for any positive number e>0,the open interval (p-e,p+e) is a subset of T"

I think you meant (s-e,s+e) in those brackets, you haven't said what p is yet. The other problem is that I think you want there to EXIST a positive number e>0, not demand it works for all e>0. Otherwise, you require (s-e,s+e) to be in T for e as large as you like, so your T will have to be the whole number line.

4. Sep 20, 2011

### Jamma

So I've criticised your question, and it may be more instructive to actually show you the correct definition and see why things like (x,y) are now neighbourhoods of all their points:

Def: A set T (a subset of R) is called neighbourhood of a point p if there exists a positive number e>0 such that the open interval (p-e,p+e) is a subset of T"

Let's look at your example of (2,3), all points which are bigger than 2 but less than 3. If x is in (2,3), then x-2 and 3-x are both positive numbers. Taking the smallest of these numbers (call it e), we see that (x-e,x+e) is contained in (2,3) [this equates to locating x in the interval, and then putting an interval around it that stretches all the way down to 2, or up to 3, depending on which is closer). Of course, you could take a smaller interval than this e.g. (x-e/2,x+e/2), but the one we have chosen is fine. So (2,3) is a neighbourhood of any point x living in it.

All of this can be generalised to spaces other than R. If we have some notion of distance on a space X (called a metric), we say that:

"A set T (a subset of X) is called neighbourhood of a point p if there exists a positive number e>0 such that the ball B(x,e) is a subset of T"

Here, the ball B(x,e) means the subset of points in X which are within distance e of x.

In your case, there is a notion of distance on R (which is just the absolute value in the difference between the values) and what you said is just a general case of the above (a ball around x of radius e here will be (x-e,x+e)).

All of this is leading on to the subject of topology!

5. Sep 20, 2011

### gliteringstar

Thanks a lot!!

The wonderful explanation given by you has cleared all my doubts regarding positive number epsilon :)

So the bottomline is no matter whatever be our x whether it is in the middle of our set or near the end-points,there always will exist a positive number e,such that this new interval around x shall always lie within the open set !!
With (2,3) as the example ,if our x is very close to 2 but not 2 (since it is an open-set),there will always exist a positive number howsoever small it is that gives the interval about x so that it falls within the set entirely!This makes the open set neighbourhood of all its points!

If we take an example of a set such as [2,3),then this set is neighbourhood of all its points except 2 because here no positive number e(epsilon)can exist (howsoever small) that gives an interval about 2 which makes it lie entirely within the set!

And i am really sorry for having it posted here,in this section.Next time ,i'll do it in the right one! thanks once again!..:)

6. Sep 21, 2011

### Jamma

Yep, you've definitely got it!

A nice way to think of it is if you were a little number, living inside the interval (2,3), you will always have friends around you that are also in (2,3), no matter where you are in it. It's like every point in there is surrounded by other points- these sets are often called "open" to express this- they are the sets for which they are neighbourhoods of all points in them.

In R, all of the open sets are just intervals, or unions of them (and we say that the empty set is open too). You might like to check that if you take the union of as many open sets as you like, then it will still be open, and that if you take the set of points which are in both of any two open sets that is still open. This applies also to the more general definition I gave you. So then you have a system of open sets satisfying:

1) all of the space X and the empty set are open
2) if you intersect any two open sets, that set is still open
3) if you take the union of any number of open sets, that is still open

These 3 conditions are what is needed to say that your open sets form a "topology" for you space. This is one of the motivating reasons for the above definition- and once you have that you can do an awful lot!