# Where does this "proof" that $\Bbb{R}^\omega$ is normal in the box topology go awry?

James Munkres' Topology, 2nd Edition indicates that the space $$\Bbb{R}^\omega := \{ (x_0, x_1, x_2, ...) | x_i \in \Bbb{R}, \forall i < \omega \}$$ equipped with the box topology is completely regular, but that normality is not known.
He points to a result of Mary Ellen Rudin that the box product of countably many $$\sigma$$-compact, locally compact metric spaces is, assuming CH, not just normal but paracompact.

I thought I had a clean-looking "proof" that $$\Bbb{R}^\omega$$ in the box topology was normal, along the following lines:

Let's say we have disjoint closed sets 𝘾, 𝘿 ⊂ $$\Bbb{R}^\omega$$ in the box topology.
If 𝘼 := $$\Bbb{R}^\omega$$ \ 𝘿, then 𝘼 is open, contains 𝘾, and is disjoint from 𝘿.
Similarly, 𝘽 := $$\Bbb{R}^\omega$$ \ 𝘾 is open and disjoint from 𝘾, and it contains 𝘿.
For all 𝐱 ∈ 𝘾, let 𝐱 ∈ 𝘼(𝐱) ⊆ 𝘼 be an open box so that $$𝘼(𝐱) - 𝐱 = \prod_i (-𝑎ᵢ(𝐱), 𝑎ᵢ(𝐱))$$ with $$0 < 𝑎ᵢ( 𝐱) < 1$$ for all 𝑖 ∈ ℕ.
Similarly, for all 𝐲 ∈ 𝘿, let 𝐲 ∈ 𝘽(𝐲) ⊆ 𝘽 be an open box so that $$𝘽(𝐲) - 𝐲 = \prod_i (-𝑏ᵢ(𝐲), 𝑏ᵢ(𝐲))$$ with $$0 < 𝑏ᵢ(𝐲) < 1$$ for all 𝑖 ∈ ℕ.
By construction, 𝘼(𝐱) is disjoint from 𝘿 and 𝘽(𝐲) is disjoint from 𝘾 for all 𝐱 ∈ 𝘾, 𝐲 ∈ 𝘿.
If additionally every 𝘼(𝐱) was disjoint from every 𝘽(𝐲) for all 𝐱 ∈ 𝘾, 𝐲 ∈ 𝘿, then we could take $$𝙐 := \bigcup_{𝐱 ∈ 𝘾} 𝘼(𝐱),$$ $$𝙑 := \bigcup_{𝐲 ∈ 𝘿} 𝘽(𝐲),$$ as our disjoint open sets containing 𝘾 and 𝘿, respectively.
If for some pair 𝐱, 𝐲, 𝘼(𝐱) ⋂ 𝘽(𝐲) is nonempty, then since 𝐱 ∉ 𝘽(𝐲) and 𝐲 ∉ 𝘼(𝐱), there must be some index 𝑖 for which $$|𝑥ᵢ - 𝑦ᵢ| > \max(𝑎ᵢ(𝐱), 𝑏ᵢ(𝐲)),$$ but where $$(𝑥ᵢ - 𝑎ᵢ(𝐱), 𝑥ᵢ + 𝑎ᵢ(𝐱)) ⋂ (𝑦ᵢ - 𝑏ᵢ(𝐲), 𝑦ᵢ + 𝑏ᵢ(𝐲)) ≠ ∅.$$ But if we replaced 𝑎ᵢ(𝐱) by 𝑎ᵢ(𝐱)/3 and 𝑏ᵢ(𝐲) by 𝑏ᵢ(𝐲)/3, then it would no longer be possible for these to intersect, since $$\frac{a_i(𝐱)}{3} + \frac{b_i(𝐲)}{3} < \frac{2|x_i - y_i|}{3} < |x_i - y_i|.$$
So if we, as a precautionary measure:

• replace the open box 𝘼(𝐱) by the 1/3-scaled open box 𝘼'(𝐱) := 𝐱 + (𝘼(𝐱) - 𝐱)/3 for every 𝐱 ∈ 𝘾, and
• replace the open box 𝘽(𝐲) by the 1/3-scaled open box 𝘽'(𝐲) := 𝐲 + (𝘽(𝐲) - 𝐲)/3 for every 𝐲 ∈ 𝘿,

then every 𝘼'(𝐱) is guaranteed to be disjoint from every 𝘽'(𝐲) for all 𝐱 ∈ 𝘾, 𝐲 ∈ 𝘿, so that we can define $$𝙐' := \bigcup_{𝐱 ∈ 𝘾} 𝘼'(𝐱),$$ $$𝙑' := \bigcup_{𝐲 ∈ 𝘿} 𝘽'(𝐲),$$ as our desired disjoint open sets containing 𝘾 and 𝘿. $$\blacksquare$$

I had just about persuaded myself that this argument worked, but then I saw another post explaining that a countable box product of metric spaces need not be normal, with reference to this paper by Erik van Douwen.
So now I'm thinking that my argument "proves too much", since this line of reasoning could presumably generalize to show that any countable box product of metric spaces ought to be normal.
My question is, where specifically did I make a mistake?

• there is at least one specific index $$i$$ for which $$x_i \not \in (y_i - b_i(𝐲), y_i + b_i(𝐲))$$, and
• there is at least one specific index $$j$$ for which $$y_j \not \in (x_j - a_j(𝐱), x_j + a_j(𝐱))$$.
But if $$i \neq j$$, it might still be that $$x_j \in (y_j - b_j(𝐲), y_j + b_j(𝐲))$$ and $$y_i \in (x_i - a_i(𝐱), x_i + a_i(𝐱)).$$
So if they don't happen at the same index, I can't get that uniform factor of $$\frac{1}{3}$$ that I would need for this argument to work out correctly.