1. Dec 5, 2008

### samspotting

Say we have the set of integers, a metric space.

Does any element of this set have a neighborhood? I am confused as to which of these two casese are true:

1) N_r(p) = {q l d(p,q)<r and q is an element of R}, so in this case no element of the set of integers has a neighborhood.

2) N_r(p) = {q l d(p,q)<r and q is an element of Z}, so in this case every element of the set of integers has a neighborhood.

And a related question, if x is an interior point of a set E, then are all elements of N, the neighborhood of x which is a subset of E, also interior points of E?

2. Dec 5, 2008

### morphism

It depends. Are you considering the set of integers as a subset of the reals or as a metric space in its own right (i.e. we don't care about anything larger, and the metric in this case is the discrete metric)? N_r(p) is (1) if it's the former, and (2) if it's the latter.

As for your other question: yes, of course! Just apply the definition of "interior point" to each element of N.

3. Dec 5, 2008

### samspotting

Isnt a neighborhood always open? I don't see how a discrete set could be open.

4. Dec 5, 2008

### HallsofIvy

In a discrete set, any metric makes all sets both open and closed.

5. Dec 5, 2008

### poutsos.A

Yes a neighborhood in a metric space ,E IS an open set as the following proof shows:

Consider a neighborhood round aεE and of radius r,denoted by N(a,r).

Now to show that N(a,r) is an open set we must prove that:

for all ,y belonging to N(a,r) ,[ yεΝ(a,r) ] we can find an e >0 such that N(y,e) is a subset of N(a,r).

So let yεN(a,r), then due to the definition of a neighborhood d(a,y)<r.

Now is the crucial point of the proof to choose the right e>0.PERHAPS making the appropriate diagram will help.

Anyway we choose : e such 0<e< r-d(a,y).

Let see now if this is the right e>0.

So if this e is the right e then N(y,e) must be a subset of N(a,r).

Hence let wεN(y,e) , this implies d(w,y)<e.but e< r-d(a,y) thus d(w,y)< r-d(a,y) which implies that:

.........d(w,y) + d(a,y)< r,which by the metric properties implies d(w,a)<r which implies that,

.........wεΝ(a,r) . .................Hence N(y,e) is a subset of N(a,r)

6. Dec 9, 2008

### samspotting

I think I am confused about neighborhoods.

If the integers is the metric space, then a neighborhood of 0 with radius 3 is the set {-2, -1, 0, 1, 2}. Does the radius have to be an element of the metric space?

I am asking this, as the definition of closed is that every limit point of the set (a limit point being that every neighborhood of that point has a point in the set that is not the limit point). If we are talking about the integers, then a radius of .5 will not yield such a result. But if we are considerng the integers as a metric space in its own, then it should be closed.

A neighborhood of 0 with radius 1 will only be {0}. Doesnt this mean that 0, or any element in the set of integers not a limit point?

Thanks guys, I really appreciate this. I am self learning Rudin, and this chapter on topology is very confusing without help. I understand the proofs but I need a better feel of it.

7. Dec 9, 2008

### rodigee

It depends on your definition of neighborhood. In general, given a point x a neighborhood of x is a set such that x is in that set's interior. Some books require this set to be open. Some do not.

The radius is just a positive real number. So, no it doesn't have to be an element of the metric space. Remember, you can put metrics on things that aren't even numbers!

What do you mean by "talking about the integers" vs. "considering the integers as a metric space in its own"?

It just says that for a sequence of integers to converge to 0, that sequence must eventually stabilize to 0. For example 3,2,1,0,0,0,0,...
It looks like you're having problems with the idea of a subspace topology on some set in a metric space. Read up on that in the book or here http://en.wikipedia.org/wiki/Topological_subspace and I think you'll feel better about this Z business.

8. Dec 12, 2008

### poutsos.A

How can one prove that??

9. Dec 12, 2008

In a discrete metric space X, there is a minimum distance $\delta[/tex] such that any two points have distance greater than or equal to [itex]\delta$. Remember that in a metric space the sets $N(x; r) = \{y \in X \mid d(x, y) < r\}$ are defined to be open; then in a discrete metric space $N(x; \delta) = \{x\}$ for any $x \in X[/tex], so one-point sets are open. Thus any subset of X is open, being a union of one-point sets. Then any subset of X is closed as well, its complement being open. No; a metric on a space X is a function [itex]d \colon X \times X \to \mathbb{R}$, and the radius talks about distances between points, so it can be anything in $\mathbb{R}$ (as long as it's positive).

10. Dec 13, 2008

### poutsos.A

Hallsofivy said and i quote:

" in a discrete set,any metric makes all sets both open and closed"

He did not say discrete metric space but discrete set.

There is a great difference between discrete metric space and discrete set.

How can one define discrete set 1st of all??

Last edited: Dec 13, 2008
11. Dec 13, 2008

I'm going by the definition Wikipedia gives (since I have not heard of a discrete set elsewhere): A set which is made up of only isolated points is a discrete set.

Then that's even easier: A discrete set has only isolated points, and x is an isolated point if it has a neighborhood that doesn't contain other points, that is, {x} is an open set. Then in a discrete set, any one-point set is open, so any subset of X is open, being a union of one-point sets, and any subset of X is closed as well, its complement being open.

Embarrassing mistake: My earlier post characterizing a discrete space is also incorrect (Wikipedia calls that uniformly discrete), but nonetheless it consists of only isolated points. (My textbook (by Munkres) defines x to be an isolated point if {x} is open.)

Last edited: Dec 13, 2008
12. Dec 14, 2008

### HallsofIvy

But that still uses "discrete space" (a set with the discrete topology) and not "discrete set" which was the term I used. I did that accidently. I was really thinking "finite set". And, of course, even then I was wrong since the "indiscreet topology" (the only open sets being the empty set and the entire set itself) does not make every set open.

13. Dec 14, 2008

### poutsos.A

Hallsofivy after all your original statement that ,in a discrete set any metric makes all sets both open and closed seams to be correct, but i would like to see adriank to give a more formal proof of that and not rather intuitive one ,as he did

14. Dec 14, 2008

It really is a formal proof though, but where it starts depends crucially on how you define the space you're talking about. Give me your precise definition of a discrete set, or whatever, and I'll show you more formally that every subset is open.

15. Dec 15, 2008

### poutsos.A

using your definition of a discrete set to prove that {x} is open ,where x belongs to the set,we must prove :

For all , y, yε{x} there exists an r >0 such that a N(y,r) ( a neighborhood of center y and radius,r) is a subset of {x},or equivalently:

For all y, yε{x} there exists an r>0 such that ,if zεN(y,r) then zε{x}.

That is what i mean when i say rather formal proof

Taking as a definition for an isolated point the following :
x is isolated in a set A iff there exists an r>0 such that N(x,r) $$\cap$$ [ A-{x}]=Φ

Where Φ is the empty set

16. Dec 15, 2008

Alright then: I'll say a discrete set is one such that every point is an isolated point.

Let $X$ be a discrete set; I will show that every subset of $X$ is open in $X$. For any $x \in X$, since $x$ is an isolated point, there is a number $r > 0$ such that $N(x; r) \cap (X - \{x\}) = \emptyset$. This shows that $N(x; r) \subseteq \{x\}$, so $\{x\}$ is open in $X$.

Then for any subset $A \subseteq X$, $A = \bigcup_{x \in A} \{x\}$ is a union of open sets, so $A$ is open in $X$.

It follows that $A$ is also closed, since $X - A$ is an open subset of $X$.

17. Dec 15, 2008

### poutsos.A

How :N(x,r ) $$\cap$$ (X-{x} ) = Φ show that: N(x,r) $$\subseteq$$ {x}????

18. Dec 15, 2008

It's pretty simple. If $y \in N(x; r)$, then $y \notin X - \{x\}$ because $N(x; r) \cap (X - \{x\})$ is empty. Therefore $y \in \{x\}$.

19. Dec 15, 2008

### poutsos.A

Do you thing ,what you just wrote and it is correct ,should be hided away from the whole proof??as pretty simple??Then in the name of simplicity we should have no proofs in mathematics .or lets say a 10 lines proof should be contracted into to a one line proof

20. Dec 15, 2008

Whether I'd include the detail that I say is pretty simple in that proof really depends on the kind of reader I'm aiming the proof for. Do I have to prove to you that $A = \bigcup_{x \in A} \{x\}$ as well?