# Countable collection

1. Feb 10, 2009

### tomboi03

Show that the countable collection
{(a,b) x (c,d) | a<b and c<d and a,b,c,d are rational}
is a basis for R2.

I was wondering... if i have to use the definition of a basis in order to solve this?
soo... meaning.. a basis:
Axioms:
1. for each x $$\in$$X, there is at least one basis element B containing x.
2. If x belongs to the intersection of two basis elements B1 and B2, then there is a basis element B3 containing x such that B3$$\subset$$B1$$\cap$$B2.

right? or am i wrong?

Thank You,
Jonnah Song

2. Feb 10, 2009

### CompuChip

Yes, I think that's exactly what you're supposed to do.
In my experience, topology often makes small steps which either involve checking that the definition is satisfied, or applying the correct theorem / lemma.

So let p = (x, y) be a point in R^2 and start by finding an element in the basis which contains p;

3. Feb 10, 2009

### WWGD

Or, if you want a more general result, prove that if the collection B is
a basis for X , and B' is a basis for Y , then BxB' is a basis for XxY.
It may be a good idea to try both exercises. (Note that I am not saying--
it is actually false -- that the product topology on a product XxY is the same
as the product of the topologies, i.e., the order topology on XxY is not necessarily
the same as the product of the order topologies of X and Y. A specific example
is that of the order topology on R; order intervals in R^2 are not the product
of order intervals in R . An example when it is true, is for the discreet
topology.)