Proving a particular product topology has an uncountable discrete subspace. If I generate a topology for $\mathbb{R} \times \mathbb{R}$ by taking products of intervals $[a,b)$, how can I demonstrate that this topology has an uncountable discrete subspace?
Intuitively I think I understand how this new space will have an uncountable subspace, as it will contain a copy of $\mathbb{R}$. However, I'm having a difficult time writing out my ideas, and addressing the discreteness of the subspace.
I also know that if the subspace is discrete then any element of the subspace exists in an open subset which contains no other point of the subspace.
 A: Consider the so-called antidiagonal $D = \{(x,-x): x \in \mathbb{R}\}$. Prove that $([a,a+1) \times [-a,-a+1)) \cap D = \{(a,-a)\}$.
It is clear that $(a, -a)$ is a member of $([a,a+1) \times [-a,-a+1)) \cap D$, which is a basic open set of $\mathbb{R} \times \mathbb{R}$ (where $\mathbb{R}$ has the "lower limit" or Sorgenfrey topology) and of $D$ as well, so the right to left inclusion is clear.
Suppose that $(x,y) \in ([a,a+1) \times [-a,-a+1)) \cap D$. So $y = -x$, as $(x,y) \in D$ and also $x \ge a$ and $-x = y \ge -a$. The latter gives $x \le a$, so in total we have $x= a$, $y = -a$ as required.
So now for every $(a,-a)$ in $D$ we have shown that $\{(a,-a)\}$ is open in $D$ (as the intersection of an open subset of the whole product intersected with $D$), so $D$ is discrete (as a subspace), and also closed.
A: To expand on Henno Brandsma's answer, let $D=\{(x,-x):x\in\mathbb{R}\}\subset\mathbb{R}\times\mathbb{R}$.  Then $D$ is uncountable, and the claim is that it is discrete as a subspace of $\mathbb{R}\times\mathbb{R}$ with your topology.  To show this, you want to show that for each $(x,-x)\in D$, the singleton set $\{(x,-x)\}$ is open in the subspace topology on $D$.  That is, you want to find some open set $U\subset\mathbb{R}\times\mathbb{R}$ such that $U\cap D=\{(x,-x)\}$.  An example of such a subset is $U=[x,x+1)\times[-x,-x+1)$.  To show that this works, first note that clearly $(x,-x)\in U\cap D$.  Conversely, if $(y,z)\in U\cap D$, then $z=-y$ since $(y,z)\in D$.  Since $(y,z)=(y,-y)\in U$, we have the inequalities $x\leq y<x+1$ and $-x\leq -y<-x+1$.  But $x\leq y$ implies $-y\leq -x$, so $x=y$ since we also have $-x\leq -y$.  Thus our point $(y,z)$ must be $(x,-x)$.
