# Topology Question

• Jan 18th 2009, 04:31 PM
flaming
Topology Question
On the plane $\Re^2$

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

(a) Show that $\beta$ is a basis for a topology on $\Re^2$
• Jan 18th 2009, 11:01 PM
aliceinwonderland
Quote:

Originally Posted by flaming
On the plane $\Re^2$

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

(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
$\beta_{2} = (i, j) \times (p, q) \subset \Re^2 (i < j, p

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

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

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