Search Results

Here is an argument that if $A$ is a countable subset of $\beta \omega \setminus \omega$, then every compactification of $X = \beta \omega \setminus A$ is zero-dimensional. Suppose $K$ is a compactifi …

Let $\omega$ denote the natural numbers and let $N$ be a countably infinite discrete subset of $\beta \omega \setminus \omega$. If $X = \beta \omega \setminus N$, then $X$ is not locally compact, and …

Of course, a necessary condition is that $X$ have no non-trivial convergent sequences. If $X$ is realcompact, then that condition is also sufficient. For that, it is enough to show that for realcomp …

I have been asked to add my comment as an answer, so here it is. A dense subset of $\omega^*$ cannot be countably tight. The reason is that if $x$ is any element of $\omega^*$, there is an open subs …

To get a countably compact space $X$ such that there is a continuous surjection $f \colon X \to [0,1]$ but if $C$ is a compact subset of $X$, $f(C) \ne [0,1]$, let $X$ be a Bernstein-type set in $\ome …