Thread: open closed cartesian product

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

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

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

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

Originally Posted by Plato

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

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

I don't understand why $\displaystyle \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is path connected; can you explain your reasoning?

I found a proof:

PlanetMath: $\mathbb{R}^2 \setminus C$ is path connected if $C$ is countable

Thanks a bunch Plato...

The set $\displaystyle \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is the set of all ordered pairs in $\displaystyle \mathbb{R}^2$ that have at least one irrational coordinate. The is a point $\displaystyle \left( {\alpha ,\beta } \right) \in \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ which has both coordinates irrational.
Suppose that $\displaystyle \left( {p,q} \right) \in \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ is any other point. One of its coordinates is irrational, say it’s q.
Using the three points $\displaystyle \left( {\alpha ,\beta } \right),\;\left( {\alpha ,q} \right)\,\& \,\left( {p,q} \right)$ construct two line segments: $\displaystyle l_1 (t):\left( {t\left[ {\alpha - p} \right] + p,q} \right)\,\& \,l_2 (t):\left( {\alpha ,t\left[ {\beta - q} \right] + q} \right)\;\; 0\le t \le 1$.
Both line segments are subsets of $\displaystyle \mathbb{R}^2 \backslash \left( {\mathbb{Q} \times \mathbb{Q}} \right)$ because at least on coordinate of each point is irrational.
The first segment connects $\displaystyle \left( {\alpha ,q} \right)\,\& \,\left( {p,q} \right)$ and the second connects $\displaystyle \left( {\alpha ,q} \right)\,\& \,\left( {\alpha ,\beta } \right)$.
Therefore the set is clearly pathwise connected.