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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 …

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 …

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 …

The proof that I can think of applies a Baire-category argument to the normal topology of the real line. As in your argument for realcompactness you need to look at zero-sets that are subsets of the $ …

The following is due to Alan Dow: In any model obtained by adding $\aleph_2$ many Cohen reals to a model of $\mathsf{CH}$ the statement is false. We force with $\mathbb{P}=\operatorname{Fn}(\omega_2, …

The answer to this question contains two locally compact zero-dimensional spaces whose Cech-Stone compactification is not zero-dimensional.That may put a limit on what can be said about components of …

In topological language: for any closed subset, $F$, of $\beta\mathbb{N}\setminus\mathbb{N}$ the family $I_F=\{A:A^*\cap F=\emptyset\}$ is an ideal. Your property translates into: the closed set $F$ i …

If $\kappa=\aleph_0$ then yes: every ultrafilter is $\aleph_0$-complete. If $\kappa>\aleph_0$ then no, if $\lambda X$ is nonempty. Split $X$ into countably many sets $\{X_n:n\in\mathbb{N}\}$, of the s …

The answer to Q1 is even more `no': Kunen showed that there are x and y such that x cannot be mapped to y and y cannot be mapped to x, see this review. This has been strengthened by Rudin and Shelah. …

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 …