On the plane $\displaystyle \Re^2$

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

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

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- Jan 18th 2009, 04:31 PMflamingTopology Question
On the plane $\displaystyle \Re^2$

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

(a) Show that $\displaystyle \beta$ is a basis for a topology on $\displaystyle \Re^2$ - Jan 18th 2009, 11:01 PMaliceinwonderland
My attempt to this problem is as follows:

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

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

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

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

There are several cases of intersections, we find the basis element satisfying the above.

Let

$\displaystyle \beta_{1} = (a, b) \times (c, d) \subset \Re^2 , (a < b, c<d ) $

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

For instance, $\displaystyle a \leq i \leq b \leq j$ and $\displaystyle c \leq p \leq d \leq q$, and if x belongs to the above $\displaystyle \beta_{1} \cap \beta_{2}$, then x belongs to the below $\displaystyle \beta_{3}$

$\displaystyle \beta_{3} = (i, b) \times (p, d) \subset \Re^2 , (i < b, p<d ) $

In each cases, $\displaystyle \beta_{3}$ can be described as

$\displaystyle \beta_{3} = (l, m) \times (n, o) \subset \Re^2 , (l < m, n<o ) $.

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