Topology Question

Quote:

Originally Posted by flaming

On the plane $\Re^2$

$\beta = \{(a, b) \times (c, d) \subset \Re^2 | a < b, c<d \}$

(a) Show that $\beta$ is a basis for a topology on $\Re^2$

My attempt to this problem is as follows:

To show $\beta$ is a basis, we need to show that

(1) For each $x \in \Re^2$ , there is at least one basis element $\beta$ containing x.

Let x be $(i,j) \in \Re^{2}, i,j \in \Re$ . Then there is a basis element containing x such that $(i - \frac{1}{n}, i + \frac{1}{n}) \times (j - \frac{1}{n}, j+\frac{1}{n})$ , n is a positive integer.

(2) If x belongs to the intersection of two basis elements $\beta_{1}$ and $\beta_{2}$ and then there is a basis element $\beta_{3}$ containg x such that $\beta_{3} \subset \beta_{1} \cap \beta_{2}$ .

There are several cases of intersections, we find the basis element satisfying the above.
Let
$\beta_{1} = (a, b) \times (c, d) \subset \Re^2 , (a < b, c<d )$
$\beta_{2} = (i, j) \times (p, q) \subset \Re^2 (i < j, p<q)$

For instance, $a \leq i \leq b \leq j$ and $c \leq p \leq d \leq q$ , and if x belongs to the above $\beta_{1} \cap \beta_{2}$ , then x belongs to the below $\beta_{3}$
$\beta_{3} = (i, b) \times (p, d) \subset \Re^2 , (i < b, p<d )$

In each cases, $\beta_{3}$ can be described as
$\beta_{3} = (l, m) \times (n, o) \subset \Re^2 , (l < m, n<o )$ .

Thus, $\beta$ is a basis for $\Re^{2}$ .