Mistake on article about Bohr compactification?

Question

$\DeclareMathOperator\b{b}\newcommand\B{{\operatorname B}}$I wish to get help understanding the content of two theorems of [Iva] that seem mutually contradictory. First some context. Let $\b(\mathbb{R})$ be the Bohr compactification of $\mathbb{R}$, and $\iota:\mathbb{R}\to\b(\mathbb{R})$ its canonical map. $\b(\mathbb{R})$ is a compact Hausdorff group, and $\iota$ is an injective group homomorphism. $\left(\b(\mathbb{R}),\iota\right)$ is not a topological compactification of $\mathbb{R}$, in the following sense: $\iota$, while having dense image, is not an embedding; it is injective and continuous, but does not have a continuous inverse.

Now the problem. As I understand, Theorem 2 of [Iva] (statement below) provides a different, now topological compactification of $\mathbb{R}$, denoted $\B\mathbb{R}$, such that $\B\mathbb{R}\setminus\mathbb{R}$ is homeomorphic to $\b(\mathbb{R}) \setminus\iota(\mathbb{R})$. On the other hand, Theorem 4 states that $\B\mathbb{R}\setminus\mathbb{R}$ is homeomorphic to a torus of dimension the continuum ($\mathfrak{c}$), that is, $\prod_{\mathfrak{c}}\mathbb{S}^1$. This would imply, by Tychonoff's theorem, that $\B\mathbb{R}\setminus\mathbb{R}$, and in consequence $\b(\mathbb{R}) \setminus\iota(\mathbb{R})$, is compact. In particular, $\b(\mathbb{R}) \setminus\iota(\mathbb{R})$ must be closed in $\b(\mathbb{R})$, so $\iota(\mathbb{R})$ must be open. Since every open subgroup is also closed, this implies that $\iota(\mathbb{R})$ is closed in $\b(\mathbb{R})$, but by density this would imply that $\iota(\mathbb{R}) = \b(\mathbb{R})$, which is known to be false, and thus, a contradiction.

If anyone can confirm there is a mistake in this article, or finds one in my reasoning, please let me know!

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Here are the statements of the Theorems as in [Iva] (save for a small change in notation).

Let $\rho_s:\mathbb{R}\times\mathbb{R}\to[0,\infty)$ denote the spherical distance in $\mathbb{R}$.

Theorem 2 [Iva]: The completion $\B\mathbb{R}$ of the uniform structure $U_B$ generated by the family of metrics of the form $$\rho_\lambda(x,y) = \rho_s(x,y) + \left|e^{i \,2\pi\lambda x}-e^{i \,2\pi\lambda y}\right| \,, \qquad \lambda\in\mathbb{R}\setminus\!\{0\}$$ is a compactification of $\mathbb{R}$. Moreover, the boundary $\B\mathbb{R}\setminus\mathbb{R}$ is homeomorphic to $\b(\mathbb{R})\setminus\mathbb{R}$.

Theorem 4 [Iva]: The boundary $\B\mathbb{R}\setminus\mathbb{R}$ is homeomorphic to a torus of continuum dimension (the cartesian product of continuum copies of the circle).

WARNING: For those interested in reading the article itself. The Bohr compactification $\b(\mathbb{R})$ is denoted there by $\mathfrak{M}(AP)$ and called Bohr compact set, while the space $\B\mathbb{R}$ is what they call the Bohr compactification.

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References:

[Iva] Ivanov, O.V.: Some remarks on the Bohr compactification of the number line. Ukr. Math. J. 38, 136–139 (1986).

Answer 1

The problem is in the proof of Theorem 2. We have two maps to begin with: the map $b:\mathbb{R}\to b\mathbb{R}$ of the Bohr compactification (called $\tau$ in the paper), and an embedding $e$ of $\mathbb{R}$ into $S^1$, say given by $x\mapsto\exp(i\pi\frac{x}{1+|x|})$ (this makes $-1$ the point at infinity). The compactification $B\mathbb{R}$ is obtained via the diagonal map $f:\mathbb{R}\to S^1\times b\mathbb{R}$ of $e$ and $b$: to get $B\mathbb{R}$ just take the closure of $f[\mathbb{R}]$ in $S^1\times b\mathbb{R}$.

The first half of the proof basically maps every $x\in b\mathbb{R}\setminus\mathbb{R}$ to the point $\langle -1,x\rangle$ in the product. However, this map is not bijective. This is essentially shown in the proof of Lemma 1. If $x<a$ then $b(x)$ is in the closure of $b\bigl[[a,\infty)\bigr]$; this implies that $\langle-1,x\rangle$ is in the closure of $f\bigl[[a,\infty)\bigr]$, that is: $B\mathbb{R}\setminus\mathbb{R}$ contains $\{-1\}\times\mathbb{R}$, and this set is disjoint from the image of the remainder $b\mathbb{R}\setminus\mathbb{R}$.

The last sentence of the proof of Theorem 2 is in error.

Addendum: in fact, Theorem 2 provides a compactification of $\mathbb{R}$ whose remainder is (homeomorphic to) $b\mathbb{R}$.