# Math Help - open closed cartesian product

1. ## open closed cartesian product

I am considering the space $\mathbb{Q}$x $\mathbb{Q}$ in $\mathbb{R}^2$.

I know that $\mathbb{Q}$ is neither open nor closed in $\mathbb{R}$, so can I say that the cartesian product $\mathbb{Q}$x $\mathbb{Q}$ is neither open nor closed in in $\mathbb{R}^2$?

2. Because $\mathbb{Q} \times \mathbb{Q} \subseteq \mathbb{R}^2$ we use the metric topology of $\mathbb{R}^2$.
Every point of $\mathbb{Q} \times \mathbb{Q}$ is a limit point of $\mathbb{R}^2$.
And every point of $\mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is a limit point of $\mathbb{Q} \times \mathbb{Q}$.

BTW: The set $\mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is pathwise connected. It is actually easy to exhibit the path between two points.

3. Originally Posted by Plato
Because $\mathbb{Q} \times \mathbb{Q} \subseteq \mathbb{R}^2$ we use the metric topology of $\mathbb{R}^2$.
Every point of $\mathbb{Q} \times \mathbb{Q}$ is a limit point of $\mathbb{R}^2$.
And every point of $\mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is a limit point of $\mathbb{Q} \times \mathbb{Q}$.

BTW: The set $\mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is pathwise connected. It is actually easy to exhibit the path between two points.
I don't understand why $\mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is path connected; can you explain your reasoning?
4. The set $\mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is the set of all ordered pairs in $\mathbb{R}^2$ that have at least one irrational coordinate. The is a point $\left( {\alpha ,\beta } \right) \in \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ which has both coordinates irrational.
Suppose that $\left( {p,q} \right) \in \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is any other point. One of its coordinates is irrational, say it’s q.
Using the three points $\left( {\alpha ,\beta } \right),\;\left( {\alpha ,q} \right)\,\& \,\left( {p,q} \right)$ construct two line segments: $l_1 (t):\left( {t\left[ {\alpha - p} \right] + p,q} \right)\,\& \,l_2 (t):\left( {\alpha ,t\left[ {\beta - q} \right] + q} \right)\;\; 0\le t \le 1$.
Both line segments are subsets of $\mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ because at least on coordinate of each point is irrational.
The first segment connects $\left( {\alpha ,q} \right)\,\& \,\left( {p,q} \right)$ and the second connects $\left( {\alpha ,q} \right)\,\& \,\left( {\alpha ,\beta } \right)$.