Search Results

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 …

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 …

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 …

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 $X$. …

See Problem 3.12.24(c) in Engelking's General Topology, or Glicksberg, Stone-Čech compactifications of products. If $a$ is in the product take $b$ in the product that differs everywhere from $a$. …

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 …

Also, I retract my claim in the comments that all compactifications with $\omega_1+1$ as a remainder are soft. … It is true, in ZFC, that $\omega_1+1$ is soft-Parovichenko but "all compactifications with remainder $\omega_1+1$ are soft" is equivalent to $\mathfrak{t}>\omega_1$. …

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 …

The spaces in this answer are pseudocompact.

More generally: if $X$ is normal and $A$ is closed in $X$ then, by the Tietze-Urysohn theorem, the closure in $\beta X$ of $A$ is $\beta A$. In the example above $X=\mathbb{R}$ and $A=\mathbb{R} \setm …

Another possibility is to use proximities - or equivalently (totally bounded) uniformities: in the metric case one defines $A$ and $B$ to be 'close' (usually denoted $A\mathrel\delta B$) if $d(A,B)=0$ …

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. …