Proving $R_L \times R_L$ is completely regular. Meaning $R_L \times R_L$ is an example of a space which is completely regular, but not normal

Question

Can I please receive help/feedback on my proof for the problem below? Thank you $\def\R{{\mathbb R}}$

Prove that $\R_L \times \R_L$ is completely regular. This means $\R_L \times \R_L$ is an example of a space which is completely regular but not normal.

$\textbf{Solution:}$ To prove $\R_L \times \R_L$ is completely regular, it is enough to show that $\R_L$ is completely regular because product of two completely regular spaces is also completely regular. To show $\R_L$ is completely regular, it is enough to show $\R_L$ is normal because a normal space is completely regular.

Now, to show $\R_L$ is normal. Suppose $A$ and $B$ are two disjoint closed sets in $\R_L$. Then, note that $\R\setminus A$ and $\R\setminus B$ are open and $A\subset \R\setminus B$ and $B\subset \R \setminus A$. Given any $a\in A$, there is a basic open set $U_a := [a, a + \rho_a) \subset \R\setminus B$ for some $\rho_a > 0$. Similarly, for each $b\in B$, we can find a $\rho_b > 0$ such that $V_b:=[b,b+\rho_b) \subset \R \setminus A$. Let $U = \cup_A U_a$ and $V=\cup_B V_b$. Observe, $A\subset U$ and $B\subset V$. Lastly, we know $U$ and $V$ are disjoint. Suppose $V_a \cap V_b = [a, a+ \rho_a) \cap [b, b + \rho_b) \ne \emptyset$. Then, $\max\{a, b\} \in V_a \in V_b$. Without loss of generality, say $a=\max\{a,b\}$. Then, $a\in A$ and $a \in V_b \subset \R\setminus A$, a contradiction. Therefore, $\R_L$ is normal. Hence, $\R_L$ is completely regular. Thus, $\R_L \times \R_L$ is completely regular.

Answer 1

There are some confusing typos in the argument to show that $U$ and $V$ are disjoint, but if you meant the following, it’s fine.

• Suppose that $U_a\cap V_b=[a,a+\rho_a)\cap[b,b+\rho_b)\ne\varnothing$ for some $a\in A$ and $b\in B$. Then $\max\{a,b\}\in U_a\cap V_b$. Without loss of generality say $a=\max\{a,b\}$. Then $a\in A\cap V_b\subseteq A\cap(\Bbb R\setminus A)=\varnothing$, a contradiction.

Answer 2

Brian answered the question about your own proof, but I propose a simpler proof for the complete regularity of $\Bbb R_L$: every basic element $[a,b)$ is closed and open and so the function $f_{a,b}: \Bbb R_L\to [0,1]$, that is $1$ on $[a,b)$ and $0$ outside of it, is continuous. These suffice to prove the complete regularity.