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$D(\mathbb{Q})$ is not isomorphic to $C(X)$ (as an $\mathbb{R}$-algebra) for any topological space $X$. For both $D(X)$ and $C(X)$, you can recover the (extended) uniform norm from the $\mathbb{R}$-al …

1-7 together are pretty strong. You aren't going to have a procedure for doing this without most of the powerset topologies being indiscrete. First note that if $X$ is a set and $\tau$ is at topology …

No this is not possible. There might be an easier proof than this but I don't know one off the top of my head. First note that a profinite set is scattered if and only if no non-empty closed subspace …

No. Consider the subset $X$ of $\mathbb{R}^2$ consisting of the union of line segments beginning at $(0,0)$ and ending at $(1,2^{-n})$ for $n\geq 0$ or $(1,0)$. Let $x=(0,0)$ and $y=(1,0)$ and conside …

The basic form of Brouwer's fixed point theorem does still hold. Fix an uncountable $\kappa$ and a continuous function $f:[0,1]^\kappa \to [0,1]^\kappa$. For any $X \subseteq \kappa$, let $\pi_X : [0, …

There was a request by Jean-Baptiste Vienney on the Category Theory Zulip for an answer to this question giving an exact characterization. As was already mention in Taras's answer, the possibility of …

A nice recent paper of Bezhanishvili and Kornell (which actually references this MathOverflow question specifically) has shown that the existence of products fails very badly in $\mathrm{Top}_{\mathrm …

I believe $[x]_\varepsilon$ is always $\sigma$-compact. First consider a slightly modified definition where we replace 'diameter $<\varepsilon$' with 'diameter $\leq \varepsilon$' and call the corresp …

Let $A=\{0,2,4,\dots\}$ be the even numbers and let $B=\{1,3,5,\dots\}$ be the odd numbers. Topologize $A\cup \beta B$ so that $A$ is a sequence limiting to a unique point in $\beta B \setminus B $. T …

I think that maybe I should have thought about this a little harder before posting, because it seems like there is a fairly easy positive answer: Let $X=[0,1]$ with a standard enumeration of the rati …

The statement is false in any unit sphere of an infinite dimensional Banach space. A compactness argument together with the form of Dvoretsky's theorem stated here (theorem 1.2), gives: Propositio …

Recall that a metric space is perfect if it has no isolated points. Also recall that a Polish space is a separable completely metrizable space. Finally recall that Baire space is the space of sequence …

The answer to both of your main questions is no assuming the existence of a weakly compact cardinal larger than $M$. In A general construction of hyperuniverses Forti and Honsell showed how to build a …

Given any topological space $X$ and subset $F\subseteq X$, define the Cantor-Bendixson sequence of $F$ in $X$ as: $F^{(0)} = F$ $F^{(\alpha +1)} = F^{(\alpha)} \setminus \{x \in F^{(\alpha)} : x \te …

This is only a partial answer. ZF + DC + 'every Banach-Mazur game is determined' is inconsistent. Let $X$ be the set of all functions of the form $f: \alpha \rightarrow \{0,1\}$, with $\alpha$ an ord …