If $\Bbb R$ has the half-open interval topology. Then is $\Bbb R \times \Bbb R$ not normal?

Question

Problem: Show that if $\Bbb R$ is given the half-open interval topology, then $\Bbb R\times\Bbb R$ is not normal.

I can't start solving this problem. Please help me.

Answer 1

We follow the standard argument.

First, we observe that the product topology on $\mathbb{R} \times \mathbb{R}$ is $\beta = \{ E_{1} \times E_{2} : E_{1} , E_{2} \subseteq \mathbb{R} \text{ are open sets }\}$ is a basis for the product topology.

To show that $\mathbb{R} \times \mathbb{R}$ is not a normal space under this topology, it suffices to find disjoint closed sets, $A_{1} , A_{2} \subseteq \mathbb{R} \times \mathbb{R}$ such that every pair of open neighborhoods of $A_{1}, A_{2}$ respectively intersect non-trivially.

Consider the anti-diagonal $-\Delta = \{ (x,-x) : x \in \mathbb{R} \}$, and $S = \{ (x,-x) : x \in \mathbb{Q} \}$.

$S$ and $-\Delta - S$ are closed, but cannot be separated by disjoint open neighborhoods.

The rough idea to see that $S$ is closed is that the basis for the product topology is rectangles which contain its bottom and left boundaries but not its top and right, considering the complement of $-\Delta$, we get that this is open because any point $(x,-x + t) \in (-\Delta)^{c}$ is such that $(x,-x+t) \in [x - \frac{t}{4} , x + \frac{t}{4}) \times [-x + \frac{3}{4}t , -x + \frac{5}{4} t) \subseteq (-\Delta)^{c}, \forall t \in \mathbb{R}$. If $(x, -x) \in (S^{c}) , x \in \mathbb{R} - \mathbb{Q}$, then $(x,-x) \in [x,x+1) \times [x,x+1) \subseteq S^{c}$

Similarly, we can argue that $-\Delta - S$ is closed by considering the complement.

$S$ and $-\Delta - S$ are not separable for the following reason. Suppose that $N_{1} \supseteq S $ and $N_{2} \supseteq \Delta - S$ are disjoint open neighborhoods. The rough argument is as follows:

$K_{n} = \{ x \in \mathbb{R} - \mathbb{Q} \cap [0,1] : [x,x + \frac{1}{n}) \times [-x,-x + \frac{1}{n}) \subseteq N_{2}\}$.

$[0,1] - \mathbb{Q} = \cup_{n \in \mathbb{N}} K_{n}$.

Baire Category Theorem implies $\overline{K_{n}} \supseteq (a,b)$ for some $n \in \mathbb{N}, a < b\in \mathbb{R}$, and $(a,b) \subseteq [0,1]$.

We need to show that $(x,-x + \epsilon) \in N_{2} , \forall x \in (a,b), \epsilon \in (0,\frac{1}{n})$. To show this, we need to find $y \in K_{n} : (x,-x + \epsilon) \in [y,y + \frac{1}{n}) \times [-y , -y + \frac{1}{n})$. Then $x \in [y , y + \frac{1}{n})$ and $-x + \epsilon \in [-y , -y + \frac{1}{n}) \iff x \in [ y - \frac{1}{n} + \epsilon, y + \epsilon)$ so $x \in [y , y + \epsilon)$ for $\epsilon < \frac{1}{n}$. Since $K_{n}$ is dense in $(x-\epsilon ,x] \cap (a,b)$ (which is non-empty) we can choose the desired $y$.

Then we need to show that $(r,-r) \in S$, $r \in (a,b)$ is a limit point of $N_{2}$. Let $\{x_{k}\} \subseteq (a,b)$ converge to $r \in \mathbb{Q}$. Then $ \{ (x_{k} , -x_{k} + \frac{1}{2k}) \} \subseteq N_{2}$ and $(x_{k} , -x_{k} + \frac{1}{2k}) \rightarrow (r,-r)$.

We may conclude that $N_{1} \cap N_{2} \ne \emptyset$ because any open neighborhood of $(r,-r)$ contains points of the form $(x_{k} , -x_{k} + \frac{1}{2k}) \in N_{2}$.