$\mathbb{R}_{\rm Sorg.}$ is paracompact but $\mathbb{R}_{\rm Sorg.} \times \mathbb{R}_{\rm Sorg.}$ is not

Question

I’m trying to solve this problem: $\def\bbR{\mathbb{R}} \def\RSorg{\bbR_{\rm Sorg.}} \def\calR{\mathcal{R}} \def\calB{\mathcal{B}} \def\calC{\mathcal{C}} \DeclareMathOperator{\range}{range}$

Prove that the Sorgenfrey line $\RSorg$ is paracompact but $\RSorg \times \RSorg$ is not.

Where $\RSorg$ is the topological space over $\bbR$ with base set $\calB := \{ [a,b) \; : \; a < b \}$.

This is how I proved that $\RSorg$ is paracompact.

Some notations

Whenever $\calR$ is a set of sets I let

$\range(\calR) := \cup_{U \in \calR}{U}$

Just to make the proof easier to understand, $\calB$ stands for “basis”, $\calC$ stands for “covering” and $\calR$ stands for refinement.

Begin of the proof

Let $\calC$ be an open covering of $\RSorg$.

Let $x_0 \in \bbR$, I define $\calR(x_0,\omega)$ for every ordinal $\omega$ by transfinite recursion in the following way:

Let $\calR(x_0,1) := \{ [x_0,x_0 + \delta_0) \}$ where $\delta_0 > 0$ is such that ($\delta_0 < 1$ and) $\exists V \in \calC \; : \; [x_0,x_0 + \delta_0) \subset V$.

Now let $\calR(x_0,\lambda)$ be defined for every ordinal $\lambda < \omega$, I let $$\calR^{-}(x_0,\omega) := \bigcup_{\lambda < \omega}{\calR(x_0,\lambda) }$$ Clearly there exists a $\beta \in (x_0,\infty]$ such that $\range(\calR^{-}(x_0,\omega) ) = [x_0,\beta)$

If $\beta = \infty$ I let $\calR(x_0,\omega) := \calR^{-}(x_0,\omega)$, otherwise there exists a $h > 0$ and a $V \in \calC$ such that $[\beta,\beta + h) \subset V$, I then let

$$\calR(x_0,\omega) := \calR^{-}(x_0,\omega) \cup \{ [\beta, \beta + h) \}$$

Now suppose by contradiction that for every ordinal $\omega$ $\range(\calR(x_0,\omega)) \neq [x_0,\infty)$, then because $\omega \mapsto \sup(\range(\calR(x_0,\omega)))$ is increasing there would be an injective “function” (I’m not sure if “function” is the appropriate term since the domain isn’t a set but a class) from the class of ordinals to the reals, which would imply that $\bbR$ has cardinality strictly greater than the one of the continuos, which is a contradiction.

Therefore, there exists $\omega$ such that $\range(\calR(x_0,\omega)) = [x_0,\infty)$. I then let

$$\calR(x_0) := \calR(x_0,\omega)$$

Now let $x_1 < x_0 - \delta_0$, I define $\calR(x_1,\omega)$ in a very similar way.

I let $\calR(x_1,\omega) := \{ [x_1,x_1 + \delta_1) \}$ where $\delta_1$ is such that $0 < \delta_1 < \delta_0$ and $\exists V \in \calC$ such that $[x_1,x_1+\delta_1) \subset V$.

(Observe that $[x_1,x_1+\delta_1) \cap U = \varnothing \;\, \forall U \in \calR(x_0)$, this will imply that $\calR(x_1)$ has at least two elements, which implies that the “first” set of $\calR(x_1)$ and the “last” set of $\calR(x_1)$ are distinct, which is in fact crucial.)

Now I define $\calR(x_1,\omega)$ in the same way as the case $\calR(x_0,\omega)$, the only difference is that whenever I let $\calR(x_1,\omega) := \calR^{-}(x_1,\omega) \cup \{ [\beta,\beta + h) \}$, I impose that $h < \delta_0$.

Now let $\omega$ be the smallest ordinal such that $\beta := \sup(\range(\calR(x_1,\omega))) > x_0$, then by definition clearly $\beta < x_0 + \delta_0$, I define $\calR(x_1) := \calR(x_1,\omega)$.

The only two distinct non-disjoint sets in $\calR(x_0) \cup \calR(x_1)$ are $[x_0,x_0+\delta_0)$ and $[\alpha,\beta)$ for a certain $\alpha \in (x_1,\beta)$ ($[\alpha,\beta)$ is the “last” set of $\calR(x_1)$ and $[x_0,x_0+\delta_0)$ is the “first” set of $\calR(x_0)$.)

Now let $x_2 < x_1 - \delta_1$, I define $\calR(x_2)$ in the same way as I have defined $\calR(x_1)$ but now $\calR(x_0)$ is replaced by $\calR(x_1)$, $x_0$ is replaced by $x_1$ and $\delta_0$ is replaced by $\delta_1$, then the only two distinct disjoint sets in $\calR(x_1) \cup \calR(x_2)$ are $[x_1,x_1+\delta_1)$ and $[\alpha,\beta)$ for a certain $\alpha \in (x_2,\beta)$ ($[\alpha,\beta)$ is the “last” set of $\calR(x_2)$ and $[x_1,x_1+\delta_1)$ is the first set of $\calR(x_1)$.)

Moreover, for any $U \in \calR(x_2)$ and $V \in \calR(x_0)$, $U \cap V = \varnothing$.

I keep going in this way, defining $\calR(x_i)$ for all $i = 0,1,\dots$ I also let $x_i = -i$ so that $x_i \to -\infty$ and I define

$$\calR := \bigcup_{i = 0}^{\infty}{\calR(x_i)}$$

$\calR$ has the following properties:

1. $\calR \subset \calB$

2. $\calR$ is a cover of $\bbR$, meaning that $\forall x \in \bbR \; \exists U \in \calR \; : \; x \in U$

3. $\forall U \in \calR \; : \; \exists V \in \calC \; : \; U \subset V$

4. $\forall x \in \bbR \exists U \in \calR \; : \;$ there are at most two sets $V \in \calR$ such that $U \cap V \neq \varnothing$ (one of which is $U$ itself).

Property 3) tells me that $\calR$ is a refinement of $\calC$, property 4) tells me that $\calR$ is locally finite, therefore $\RSorg$ is paracompact.

End of the proof

Is the proof correct? How can I prove that $\RSorg \times \RSorg$ is not paracompact? Can I prove it similar to how I have proved that $\RSorg$ is paracompact?