# Homework Help: Non discrete metric space on infinite set

1. Apr 20, 2012

### elias001

1. The problem statement, all variables and given/known data

let d be a metric on an infinite set m. Prove that there is an open set u in m such that both u amd its complements are infinite.

2. Relevant equations

If d is not a discrete metric, and M is an infinite set (uncountble), then we can always form an denumerable subset (countably infinite).

My question is i don't know how to find an open subset for a given radius delta. I know how to do it if d is a discrete metric and M is a countably infinite set.

2. Apr 20, 2012

### Citan Uzuki

Well, you seem to be very confused. Let's try to clear up a few points first:

You can find a countably infinite subset whether the metric is discrete or not, why would discreteness matter here? Also, you say uncountable as though that were a synonym for infinite. It isn't -- an infinite set may be either countable or uncountable.

The open ball of radius δ around a point x is simply the set of all points y such that d(y, x)<δ. What is confusing you about this? Also, why do you keep bring up the countable/uncountable distinction? It has nothing to do with this problem!

I get the feeling that you have just tried to memorize the concepts in your textbook without understanding them first. So the first piece of advice is to go back to the part of your text where an open set is defined and study carefully this definition. Come back when you can you can both give me the formal definition of an open set and an informal description of an open set in your own words.

3. Apr 20, 2012

### elias001

Sorry i should have been more specific.

Suppose if I am given a non discrete metric d, and an infinite set M, let M' be an countably infinite subset M' such that it only contains an even number of the elements from M, and pick an x in M', then we enclosed an open δ-ball around x, but how do we know that such an open ball would only contain the even elements from M' and would also contain the odd elements from M.

4. Apr 20, 2012

### HallsofIvy

I also am completely confused by this. I don't see why you keep talking about "countable" sets nor why you now want M' to contain "an even number" (and so finite) number of points. Neither has anything to do with a set being "open".

Finally, while you first say "an even number of elements from M" you later say "the even elements" and "the odd elements". Are you assuming your set M consists of integers? You are supposed to show this for M any infinite set, not necessarily containing the integers.

5. Apr 20, 2012

### Citan Uzuki

As improbable as it might seem, elias's latest post has given me enough information to form a plausible guess as to what he's trying to do.

elias, I think the proof strategy you're attempting is this: You want to first enumerate a countable subset $\{x_1, x_2, \ldots \} \subseteq M$, and then try to show that there is an open set which contains all of the $x_n$ where n is even and none of the $x_n$ where n is odd. You want to do this by picking the elements in such a way that the open ball of the fixed radius δ around each point of the form $x_{2n}$ does not contain any points of the form $x_{2m+1}$. And while this is easy when the metric is discrete (choose δ < 1, and then choose the $x_n$ arbitrarily), you don't have the slightest idea how to make such a selection for an arbitrary metric.

6. Apr 20, 2012

### elias001

@Citan Uzuki

Yes, that is what i am trying to ask. I don't know how to do it for the case where d is a metric that is not discrete. Basically i don't know how to choose an open set with a specific δ that in such a way that contain only the even number of elements.

Sorry, i am not very good at typing math symbols.

7. Apr 20, 2012

### Citan Uzuki

Okay, now we're getting somewhere. Unfortunately, you cannot hope to succeed using a fixed δ, as there are examples of sets M where every infinite subset A that contains all elements within δ of some element of A must contain all but finitely many elements of M. One example of such a set is $\{\frac{1}{n} : n\in \mathbb{N}\}\cup\{0\}$ So you will have to vary your strategy.

The approach I would use is to divide into two cases. First, suppose that there are two distinct accumulation points of M, say, x_1 and x_2. Prove that every open ball around an accumulation point of M must contain infinitely many elements of M (hint: prove the contrapositive). Then simply show that there exist two disjoint open balls B_1 and B_2 centered around x_1 and x_2, respectively. Then either of these open balls will satisfy the conclusion of the theorem

In the second case, we suppose that there is at most one accumulation point of M. Then every other point of M is an isolated point, so any subset whatsoever of the other points of M will be an open set (note: when proving this, you will have to choose a different δ for each point in the set), and then you can use the enumeration technique to choose an infinite subset whose complement is also infinite.

8. Apr 20, 2012

### elias001

is there any way to do it without introducing the concepts of accumlation point and limit point. The question came from a text call Real analysis, by N.L carothers. at the point of the text where the question is asked, only metric spaces, definitions of open and closed sets in terms of neighborhood definition and definitions based on convergent sequences, but limit points, closure and accumlation points and such have not been defined until the next section.

Also, does every metric space include at least or more than one accumulation point. The reason i asked is that the question's hypothesis does not include much information that can be used.

Thanks

Last edited: Apr 20, 2012
9. Apr 20, 2012

### Citan Uzuki

You can avoid introducing the names, but I don't think that there's a way to avoid introducing the concepts. I mean, the hint given for the problem is that "either the space is discrete or it's not," i.e. either it has a limit point or it doesn't. Although, since he uses this particular dichotomy and has mentioned convergence of sequences at this point, my guess is that the proof he has in mind is to show that if {x_n} is a sequence of elements converging to x with no x_n equal to x, then the elements {x_n} form a discrete subspace. Frankly, I think you should attempt my approach, since the lemmas you'll have to prove (that every open set around an accumulation point of the space contains infinitely many elements) will be stuff you have to do as problems in the next chapter anyway, and you may as well get them out of the way.

Also, let me clarify one point the author doesn't: while he casually introduces the concept of a discrete space as one equipped with the discrete metric, he then goes on to use it to refer to any set with an equivalent metric (i.e. one where every set is open). This is standard mathematical usage, but I don't think he mentions that anywhere between when he first defines a discrete space and this problem.

10. Apr 20, 2012

### elias001

Thank you for the clarification. By the way Citan Uzuki, did you use Carothers' text at some point in your education?

One other thing that is not related to my original question, when i make a thread, how do i get more math symbols on the default "quick symbol" palate to show up. I know some people might be using latex, but i don't know how to type using latex.

11. Apr 20, 2012