Is this proof for a topological basis ok?

[Damascus Road]

On the plane $$R^{2}$$ let,

B= {(a,b) x (c,d) $$\subset$$ $$R^{2}$$ | a < b, c < d }

a.) Show that B is a basis for a topology on $$R^{2}$$.


This means I have to show that every x in $$R^{2}$$ is contained in a basis element, and that every point in the intersection of two basis elements is contained in a basis element in that intersection.

So:

Each x $$\in$$ $$R^{2}$$ $$\subset$$ B, since B is the set of all open rectangles on $$R^{2}$$.

If x $$\in$$ B1$$\bigcap$$B2, then x $$\in$$ B3 $$\subseteq$$ B1$$\bigcap$$B2, since the intersection of open rectangles is also an open rectangle.

This B is a basis.

How's it look?

 

[mighty2000]

Your proof looks good! To make it a bit more rigorous, you can explicitly state that B is a collection of open subsets of R^{2} and satisfies the two conditions for being a basis: 1) every point in R^{2} is contained in at least one basis element, and 2) for any two basis elements B1 and B2, and any point x in their intersection, there exists a basis element B3 such that x is contained in B3 and B3 is a subset of B1 and B2. This shows that B is a basis for a topology on R^{2}. Great job!