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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 statement is analogous to the general result that, for Tychonoff spaces, compactness is equivalent to pseudo-compactness plus realcompactness, see the beginning of section 3.11 in Engelking's Gene …

For every $K$ the subspace $X_K$ of all functions bounded by $K$ is homeomorphic to the Cantor set, as $X_K=\{1,2,\ldots,K\}^\mathbb{N}$ and the metric induces the product topology. It follows that $\ …

The first thing to to with a definition like that is test it against some familiar examples. The sequence $\langle 2^{-n} : n\in\mathbb{N}\rangle$ converges to $0$ in $\mathbb{R}$ and it would seem de …

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.

Using @Anonymous' second comment one can show that your property characterizes locally compact scattered spaces. For if $X$ is locally compact Hausdorff and not scattered then it contains a closed den …

No, your $\beta_2S$ is $\beta S$, the Čech-Stone compactification of $S$. As an example let $S=\omega$ and $P=\omega+1$ (with order topology, so it is just the converging sequence). Let $g$ be the cha …

A partial answer. for your specific example, the algebra $B$ generated by the rational intervals, the answer is negative. This algebra is countable and dense in $2^\mathbb{R}$. It is countable and ze …

Here's a counterexample. Let $B$ be a Bernstein set in the plane, so $B$ and its complement intersect every uncountable closed subset of $\mathbb{R}^2$. Let $X$ be a metric compactification of $B$, wi …

Assume $f:\mathbb{R}^2\to\mathbb{R}$ is continuous, closed, and open. It is indeed surjective, because the range has to be closed and open. Since the map is open the set $f^\gets(0)$ is nowhere dense: …

The proofs rely, in the background, on Urysohn's Lemma, which follows from the Principle of Dependent Choices but is not provable without some Choice. It is false in the ordered Mostowski model, see G …

A partial answer to the question about the Gleason cover of the unit interval: it can be found everywhere in every compact extremally disconnected space. The point is: in a compact extremally disconne …

To make the answers concrete: the Boolean algebra of clopen subsets of the Cantor set is the unique countable atomless Boolean algebra. Its completion is the regular-open algebra of the Cantor set, wh …

You can consult Problem 8.5.13 in Engelking's General Topology. It deals with Dieudonné complete spaces (Tychonoff spaces that have a complete uniformity). Part (d) shows that every paracompact space …

To answer the explicit question: the extent of every $\Sigma$-product of $\mathbb{N}$ is countable. H. H. Corson showed in Normality in subsets of product spaces, Amer. J. Math 81(1959), 785–796 that …