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The answer is no, if we interpret separable as "has a countable dense subset" (Fedor Petrov's answer appears to have interpreted it as possessing a countable base). Consider the Moore plane, or Niemyt …

I think you are asking about cardinal functions on metric spaces. First, the property you call $\aleph_0$-compact is more usually known as Lindelöf. As you say, it is well-known that for a metric spac …

The fact you are probably looking for is that, for any Baire space $X$ (e.g. a completely metrizable space or a compact Hausdorff space) the inclusion map $\mathfrak{R}(X) \rightarrow \mathfrak{B}o(X) …

In effect, you are asking if $\bigvee\limits_{i \in \omega}B_i = \bigcup\limits_{i \in \omega}B_i$, where the left hand side is the closure of the union, which is the join/supremum of the family $\{B_ …

The answer is no, essentially because $\mathbb{R}$ embeds as a locally compact open subspace of $\beta\mathbb{R}$, and $\mathbb{R}$ is not an F-space. In detail, for the purposes of this answer I wi …

There is nothing to say in the general case, but in the case of a continuous dcpo $D$ there is a well-known construction of certain bases for the Scott topology on $D$. We say $d$ is way below $e$, …

This is not really research level, and is probably better suited to math.stackexchange, but here's an answer anyway. I will take as given that you know the notation $\sigma(E^*,E)$ for the weak-* topo …

Yes. In the next paragraph I will show that if $X$ is a Fréchet space (without requiring separability) then $X'_c$ with the compact-open topology is a $k$-space. As you note, this implies sequentialit …

The answer is yes, following from Klee's extension of Keller's theorem on homeomorphisms of infinite dimensional compact convex sets: Klee, V. L., Some topological properties of convex sets, Trans. Am …

Fremlin's measure theory textbook is a good reference for these things. I am splitting things up into the Boolean algebra part and the real-valued functions part. Complete Boolean algebras: The way …

Here is an answer to your question about monadicity, as it's too long for a comment. I will not fill in every detail, so if you follow along there will be several definitions that need to be expanded …

Since questions recirculate to the front page forever if left unanswered, I will amalgamate the comments into an answer, which I've made community wiki. Firstly, by the characterization of the Cantor …

It is not true that $B$ is necessarily a measure algebra. The counterexample is due to Michel Talagrand, who constructed a Maharam algebra that is not a measure algebra. Maharam, D., An algebraic cha …

Here is some information on the history of the result, which was actually proven before Shirota's 1952 theorem. It was proved in 1948 by Marczewski and Sikorski as Theorem VI in: Marczewski, E.; Sikor …

The hyperstonean case can be dealt with using a result from Fremlin's Measure Theory. For every hyperstonean space $X$, we can find a semi-finite measure $\mu$ defined on the sets with the Baire prope …