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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 $\mat …

The metric induces the product topology, so the group $G$ is compact. The direct sum of $\mathbb{Z}$ many copies of $G'$ is a countable dense subgroup, but not the whole group, so it is not closed.

Or, you can embed $X$ into a Tychonoff cube $[0,1]^\kappa$, where $\kappa$ is the weight of $X$, say. The Tychonoff cube embeds in the corresponding power of the unit circle, say by embedding $[0,1]$ …

Q1. Consider the space of rationals $\mathbb{Q}$. It is the direct limit of the family of finite unions of convergent sequences (including their limits), ordered by inclusion, as a set is closed iff i …

You can use the universality property with the following Boolean group as codomain: $B$ is the measure algebra over the unit interval (the quotient of the $\sigma$-algebra of Lebesgue measurable sets …

According to your first reference your space is dense with respect to the density topology; this implies that, in $Y$, the closure of $Y\cap(0,\infty)$ is $Y\cap[0,\infty)$; the latter set is not open …