# Real Line closed interval topology Sorgenfrey plane

Let $\mathbb{R}_l$ be the real line with the left closed interval topology (= the topology on $\mathbb{R}$ with base consisting of all intervals $[a, b)$ with $a < b$). We will consider $\mathbb{R}_l \times \mathbb{R}_l$ with the product topology (= Sorgenfrey plane). This problem shows that this space is not normal. The argument is by contradiction. So, assume that the Sorgenfrey plane is normal.

Let $L \subset \mathbb{R}_l \times \mathbb{R}_l$ be given by $L = {(t, -t) | t \in \mathbb{R}}$

Let $D = \mathbb{Q} \times \mathbb{Q} \subset \mathbb{R}_l \times \mathbb{R}_l$ be the set of all points with both coordinates rational.

a. Show: if $A \subset L$, then A is a closed subset of the Sorgenfrey plane.

Now assume that for each $A \subset L$ we can choose (and fix) open subsets $U_A, V_A$ in $\mathbb{R}_l \times \mathbb{R}_l$ such that $A \subset U_A, L - A \subset V_A$, and $U_A \cap V_A = \varnothing$. Define a function $$\vartheta : P(L) \to P(D)$$

by setting $\vartheta(A) = U_A \cap D$.

b. Show that $\vartheta(A)$ determines $A$. Hint: think of sequences in NE quadrant.

c. Why does this show that $\vartheta$ is an injection?

d. Why does this contradict a result of Cantor?

(a) Just show that each point of $\Bbb R_\ell^2\setminus L$ has an open nbhd disjoint from $L$; this is completely trivial.
(b) What you’re being asked to show is that if $\theta(A)=\theta(B)$, then $A=B$ or, equivalently, that if $A,B\subseteq L$ and $A\ne B$, then $\theta(A)\ne\theta(B)$. Suppose that $a\in A\setminus B$. Then $a\in U_A\cap V_B$, and therefore $\theta(A)\cap V_B\ne\varnothing$; why? And this in turn implies (why?) that $\theta(A)\setminus\theta(B)\ne\varnothing$ and hence that $\theta(A)\ne\theta(B)$.
(d) How many subsets does $D$ have? How many subsets does $L$ have?