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A partial answer: other examples of pretty compact spaces are uncountable powers of $\{0,1\}$ and $[0,1]$, and in general products of uncountably many non-trivial compact Hausdorff spaces. See Problem …

Topologically speaking $\mathbb{Z}$ with the topology mentioned above is just (homeomorphic to) the space of rational numbers. The space $\beta\mathbb{Q}$ has been studied a lot (not as much as $\beta …

Let $X_n$ be $\{k\in\mathbb{N}:k\ge n\}$ and let $f_n:X_{n+1}\to X_n$ be the inclusion map. The inverse limit of the system $\{X_n,f_n,\mathbb{N}\}$ is empty; the limit of the system $\{\beta X_n,\bet …

In this paper, Dimension phenomena associated with $\beta\mathbb{N}$-spaces, Ilijas Farah proved that continuous maps from $\beta\mathbb{N}^2$ (and other powers) to $\beta\mathbb{N}$ are quite simple: …

No. There is a compactification of $\mathbb{N}$ whose remainder, $K$, is ccc non-separable. So there is a continuous surjection $f$ from $\beta\mathbb{N}\setminus\mathbb{N}$ onto $K$; take a closed su …

Here is a partial answer: the Continuum Hypothesis implies that all Parovichenko spaces are soft-Parovichenko; the proof is a bit long, so I put it in a PDF-file on my website. Also, I retract my cla …

Answer to 1: In On closed subspaces of $\omega^*$ (Proc. AMS, 1993) it is shown by Dow, Frankiewicz and Zbierski that in the $\aleph_2$-Cohen model every compact zero-dimensional $F$-space of weight a …

Not without further assumptions. First create an ordered space $X$ by identifying $\langle0,\omega_1\rangle$ and $\langle1,\omega_1\rangle$ in the product $2\times(\omega_1+1)$ to a point, $\Omega$ sa …

As an answer to the bonus question: no, see K. P. Hart and J. Vermeer. Fixed-point sets of autohomeomorphisms of compact F-spaces, Proceedings of the American Mathematical Society, 123 (1995), 311–314 …

0a Correct, except for one point: is $X$ is not completely regular then there is no $\beta X$, so the third equivalence in `functionally Hausdorff' does not exist. 0b Not quite, the paper mentioned i …

Yes if the point is from $\mathbb{N}$ (it is isolated). No if the point is in $\beta\mathbb{N}\setminus\mathbb{N}$ because in that subspace every nonempty $G_\delta$-set has nonempty interior, see thi …

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 …

There is, in general, a on-to-one correspondence between closed subalgebras of $C^*(X)$ (the algebra of bounded continuous real-valued functions) and the compactifications of $X$. Closed in the sense …

For an explicit pair of disjoint open sets with intersecting closures work on the binary tree $2^{<\omega}$ of finite sequences of $0$s and $1$s. For every $x\in2^\omega$ let $A_x=\{x\mathbin{\upharpo …

$\beta\mathbb{N}$ is not a retract of a Tychonoff cube because it is not connected; it also not a retract of a Cantor cube, not even a continuous image, see problem 3.12.12 in Engelking's book. It is …