Prove that the product $\mathbb{R}\times \mathbb{R}$, each given the upper-limit topology, is not normal.

Question

Consider the upper limit topology on $\mathbb{R}$. It is required to prove that the product topology on $\mathbb{R^2}$ is not normal. As a hint I'm asked to consider the two sets $E=\{(x,-x):x\in\mathbb{Q}\}$ and $F=\{(x,-x):x\in\mathbb{R\setminus Q}\}$. I guess that these two sets should be closed in $\mathbb{R^2}$ but I cannot find a way to prove it. Or is my guess wrong? Can someone please help me prove it? Thanks.

Answer 1

Let $D=\{(x,-x):x\in\mathbb R\}$ (called the anti-diagonal). Clearly $D$ is closed in $\mathbb R^2$, when $\mathbb R$ is given the standard topology. Hence $D$ is closed in $\mathbb R^2$, when $\mathbb R$ is given the upper-limit topology, as the latter topology is stronger.

It easily follows that every subset of $D$ is relatively open (as a subset of $\mathbb R^2$, when $\mathbb R$ is given the upper-limit topology). To see that every singleton $\{(x,-x)\}$ is relatively open in $D$, note that $\{(x,-x)\}=D\cap ([x,x+1)\times[-x,-x+1))$ and $[x,x+1)\times[-x,-x+1)$ is open. Since every singleton is relatively open, it follows that every subset of $D$ is relatively open (i.e., open as a subset of $D$, when $D$ is given the induced topology). Thus $D$, as a subspace of $\mathbb R^2$ (when $\mathbb R$ is given the upper-limit topology) has the discrete topology. It follows that every subspace of $D$ is a also relatively closed. But since $D$ itself is closed, we have that every subset of $D$ is closed in $\mathbb R^2$ (when $\mathbb R$ is given the upper-limit topology).

In particular both $E$ and $F$ are closed. This question is a duplicate (twice, as I indicated in a comment) having received two answers (from the same user, H.B.), both answers using Jones' Lemma (which is one of the standard proofs answering your question). I prefer to write a slightly different proof, using the Baire category theorem.

Suppose there were disjoint open sets $U$ and $V$ with $E\subseteq U$ and $F\subseteq V$, separating $E$ and $F$ in $\mathbb R^2$ (when $\mathbb R$ is given the upper-limit topology). For each $x\in\mathbb R\setminus\mathbb Q$ fix $\varepsilon_x>0$ such that $([x,x+\varepsilon_x)\times[-x,-x+\varepsilon_x))\subseteq V$. For each $n\ge1$ let $F_n=\{x\in\mathbb R\setminus\mathbb Q:\varepsilon_x>\frac1n\}$. Then $R\setminus\mathbb Q=\bigcup_{n\ge1}F_n$. Since $R\setminus\mathbb Q$ is of the second category (as a subspace of $\mathbb R$ with the usual topology, where "category" is understood as in the Baire category theorem), there must be an $n\ge1$ such that the (usual) closure of $F_n$ contains some interval $(a,b)$ (with $a<b$). You may verify then that the set $\Bigl(\bigcup_{a<t<b}((t,t+\frac1n)\times(-t,-t+\frac1n))\Bigr)\subseteq V$. Take any rational $q\in(a,b)$. There is some $\varepsilon>0$ such that $([q,q+\varepsilon)\times[-q,-q+\varepsilon))\subseteq U$. But $[q,q+\varepsilon)\times[-q,-q+\varepsilon)$ must intersect $\bigcup_{a<t<b}((t,t+\frac1n)\times(-t,-t+\frac1n))$, hence $U$ intersects $V$, which completes the proof that $\mathbb R^2$ (when $\mathbb R$ is given the upper-limit topology) is not normal. (Note also that the latter space is known as the Sorgenfrey plane, and that $\mathbb R$ with the upper-limit topology is known as the Sorgenfrey line.)