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.