# Thread: What is the Interoir of the set QxQ?

1. ## What is the Interoir of the set QxQ?

Given the metric space (M,d) where M = RxR and d(x,y) = the Euclidean metric, what is the interior of the subset of M, QxQ?

I realize it is the empty set because there are points that lie in the open ball of any element of QxQ that are not actually in the set QxQ.

What I am struggling with is giving a formal, reasonable justification of why the interior is the empty set.

Any help or suggestions to get me thinking in a good direction would be greatly appreciated. Thank you so much, in advance!

2. Originally Posted by rgriss1
Given the metric space (M,d) where M = RxR and d(x,y) = the Euclidean metric, what is the interior of the subset of M, QxQ?

I realize it is the empty set because there are points that lie in the open ball of any element of QxQ that are not actually in the set QxQ.
Each point of $Q\times Q$ is a boundary point of $Q\times Q$.
No interior point of a set can be a boundary point of the set.
Therefore $Q\times Q$ has no interior.

3. Thank you very much. That is an excellent start.

4. Originally Posted by rgriss1
Given the metric space (M,d) where M = RxR and d(x,y) = the Euclidean metric, what is the interior of the subset of M, QxQ?

I realize it is the empty set because there are points that lie in the open ball of any element of QxQ that are not actually in the set QxQ.

What I am struggling with is giving a formal, reasonable justification of why the interior is the empty set.

Any help or suggestions to get me thinking in a good direction would be greatly appreciated. Thank you so much, in advance!
To prove what Plato has said, try to prove that if $D,D'$ are dense subsets of the topological spaces $X,X\prime$ then $D\times D'$ is dense in $X\times X'$ under the product topology. Now since $\mathbb{R}-\mathbb{Q}$ is dense in $\mathbb{R}-\mathbb{Q}$ we know that $\left(\mathbb{R}-\mathbb{Q}\right)\times\left(\mathbb{R}-\mathbb{Q}\right)$ is dense in $\mathbb{R}\times\mathbb{R}$. Similarly, $\mathbb{Q}\times\mathbb{Q}$ is dense in $\mathbb{R}\times\mathbb{R}$ and so given any $(x,y)\in\mathbb{R}$ for any neighborhood $N$ of $(x,y)$ there exists points of both $\mathbb{Q}\times\mathbb{Q},\left(\mathbb{R}-\mathbb{Q}\right)\times\left(\mathbb{R}-\mathbb{Q}\right)$ and since the latter is a subset of $\mathbb{R}\times\mathbb{R}-\mathbb{Q}\times\mathbb{Q}$ it follows that $(x,y)\in\partial\left(\mathbb{Q}\times\mathbb{Q}\r ight)=\partial\left(\left(\mathbb{R}-\mathbb{Q}\right)\times\left(\mathbb{R}-\mathbb{Q}\right)\right)$. Since $(x,y)$ was arbitrary it follows that $\partial\left(\mathbb{Q}\times\mathbb{Q}\right)=\p artial\left(\left(\mathbb{R}-\mathbb{Q}\right)\times\left(\mathbb{R}-\mathbb{Q}\right)\right)=\mathbb{R}\times\mathbb{R }$. And, as Plato pointed out we have that no interior point can be a boundary point, and so $\left(\mathbb{Q}\times\mathbb{Q}\right)^{\circ}=\l eft(\left(\mathbb{R}-\mathbb{Q}\right)\times\left(\mathbb{R}-\mathbb{Q}\right)\right)^{\circ}=\varnothing$

Note: The product topology coincides with the usual topology on $\mathbb{R}^2$