open closed cartesian product

**TXGirl** · May 12th 2008, 03:19 AM

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 $\mathbb{R}^2$ ?

**Plato** · May 12th 2008, 07:36 AM

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 $\mathbb{Q} \times \mathbb{Q}$ .

Can you answer your questions now?
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.

**TXGirl** · May 12th 2008, 10:24 AM

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 $\mathbb{Q} \times \mathbb{Q}$ .

Can you answer your questions now?
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?

**TXGirl** · May 12th 2008, 10:33 AM

I found a proof:

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

Thanks a bunch Plato...

**Plato** · May 12th 2008, 11:09 AM

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 $\left( {\alpha ,q} \right)\,\& \,\left( {\alpha ,\beta } \right)$ .
Therefore the set is clearly pathwise connected.

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