# First countability

1. Aug 16, 2007

### matness

I have come up with an example when I trying to learn what first countability means
It says(from wikipedia)
In a metric space, for any point x, the sequence of open balls around x with radius 1/n form a countable neighbourhood basis $$\mathcal{B}(x) = \{ B_{1/n}(x) ; n \in \mathbb N^* \}.$$ This means every metric space is first-countable.

According to definition of nhd basis we can express every open set can be expressed as a union of elts of nhd basis But for example $$\{ B_{ sqrt{2} }(x)$$ can not be expressed in this way

Here is the question
which point did i miss?

2. Aug 16, 2007

### George Jones

Staff Emeritus
The union can be of an infinite number of sets.

3. Aug 16, 2007

### matness

But the ball is given as $$\{B_{1/n}(x)$$ i.e. center is x , meaning the biggest ball( i.e. ball with smallest n) will be the union)
and there is always a real number between 1/n and 1/n+1

definition says every open ball of x can be written as a union of elts of basis .If it has said every open ball is contained in the union it would be easier to understand .

So maybe I should ask first whether the definition i learned is correct or not

4. Aug 16, 2007

### George Jones

Staff Emeritus
Forget my last post. Without thinking, I thought the answer had something to do with writing $\sqrt{2}$ as the limit of a sequence of rational numbers.

I don't know what "union of elts of nhd basis" means. In particular, elt is not in any of my topology references.

The definitions I see say that for any x, there a countable neighbourhood basis of x such any neigbourhood of x contains at least one member of the neighbourhood basis. In your example, every member of the basis in contained in the neighbourhood that you chose.

5. Aug 16, 2007

### matt grime

elt means element, a common shorthand.

You can express the ball of radius sqrt(2) about a point as the union of elements in the given basis: just take a union over all possible sets of the form B_{1/n}(y) that are contained in B{\sqrt(2)}(x).

A clearer countable set would be the set of balls of rational radii.

6. Aug 16, 2007

### George Jones

Staff Emeritus
Yikes! I feel a bit silly now. It has benn some years since I was a student, but I don't recall hearing or seeing this term in any of the lectures for the analysis and topoplogy courses that I took.

I also had something like this in mind, but I thought matness wanted to stick to {B_{1/n}(x)}.

7. Aug 16, 2007

### matt grime

I didn't go beyond sets of the form B_{1/n}(x). matness's problem is, I think, that he assumes that when expressing a set as the union of open nbd basis elements that they all have to have the same centre. This is not the case.

8. Aug 16, 2007

### matness

yes the problem is this
and the source of problem is the example in wikpedia that is included in my first post
$$\mathcal{B}(x) = \{ B_{1/n}(x) ; n \in \mathbb N^* \}.$$

On the other hand I mostly agree with you
using basis like a cover (roughly) seems more reasonable

Last edited: Aug 16, 2007
9. Aug 16, 2007

### matness

and one more thing
in my lecture notes the definition is given as:
(X,T) a topological space and B is a subset of T
if every open nhd of a point p is union of a collection of elements of B then B is called a nhd basis for p.

Is there a problem?
I just wanted to be sure whether I have taken the notes correctly or not

Last edited: Aug 16, 2007
10. Aug 16, 2007

### CompuChip

Nope, that's the correct definition.
But note that it doesn't say that the elements in the union should all contain p (just at least one of them, of course, otherwise p is not in the union).
Also note, that usually there isn't much to construct. For example, in your original question, the best we can do is:
$$B_{\sqrt{2}}(x) = \bigcup_{B \in \mathcal{B}} B$$,
where $\mathcal{B}$ consists of all sets in the basis which are inside the left hand side (so all $B_{1/n}(y) \subseteq B_{\sqrt{2}}(x)$, which is of course true but not very practical if you'd want to draw them or write them down.

Last edited: Aug 17, 2007
11. Aug 17, 2007

### matness

$$B_{\sqrt{2}}(x) = \bigcap_{B \in \mathcal{B}} B$$
I assume you mean union( if there is anything I missed please warn me )
On the other hand I started to think the example in wiki wasn't that correct after your posts
thanks

12. Aug 17, 2007

### CompuChip

Indeed, I meant union, apparently I can't tell the difference between \cup and \cap . My apologies, it's fixed.