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I made a comment earlier, but let me try converting it to an answer. It's similar in flavor to Ivan's. By a back and forth argument, all countable dense subsets of $\mathbb{R}$ are homeomorphic to $\m …

There aren't any. Hausdorff spaces are sober spaces. If $X, Y$ are sober, then every frame map $\mathcal{O}(Y) \to \mathcal{O}(X)$, i.e., every poset map between their topologies that preserves finite …

I'll go ahead and say that Polish spaces are an interesting and almost sui generis class. There is a rich literature of applications to and from descriptive set theory, with layers of "pathology" hier …

Certainly if $U$ is a finite-dimensional inner product space, then $\text{Isom}(V, U)$ is paracompact (being a closed subspace of $\text{Lin}(V, U) \cong U^n$). If $U$ has countably infinite dimensi …

Yes. See Kannan and Rajagopalan - Constructions and applications of rigid spaces, I (MSN), particularly their construction 2.2.4, which gives a strongly rigid connected Hausdorff space $Y$. In particu …

The statement in the answer by Dominic that the interval topology of the product poset is the product topology of the interval topologies is incorrect. The argument that the product topology contains …

I think the premise of the question is mistaken because $\mathbf{F}(f)$ need not be continuous for a poset map $f: P \to Q$, i.e., $\mathbf{F}$ is not functorial. I was about to ask about this in a co …

No, it does not. If it did, then $\mathbf{Comp}$ would be a reflective subcategory of the total category $\mathbf{Top}$, and hence would be total itself. Now, total categories admit all small limits, …

I won't swear it's my absolute favorite, but today I learned of a nice proof due to Clementino and Tholen who take as their starting point the closed-projection characterization of compactness, viz. t …

Edit: Will Sawin has pointed out some difficulties with this answer. I'm going to leave this up for at least a while, in case anyone has any ideas about repairing it. Or perhaps this could be a cautio …

As for question 2.: it's hardly recent, but since I didn't see it mentioned in this thread, let me mention that the Tychonoff theorem for locales does not require the axiom of choice and is a piece of …

To me, the second definition of local compactness is much to be preferred for the simple reason that such locally compact spaces $X$ are exponentiable in $Top$, meaning that $X \times -: Top \to Top$ …

One way of answering is just by applying Zorn's lemma. By the Cantor-Schroeder-Bernstein theorem, we really only have to work one cardinality at a time; that is, for each cardinal $\kappa$, find a m …

Following the example of my comment, with $X = \mathbb{R}^2$ and $$f((x_1, y_1), (x_2, y_2)) = ((x_1 - x_2)^{1/2} + (y_1 - y_2)^{1/2})^2,$$ one sees that $f((0, 0), (0, 1/2)) = 1/2$ and $f((0, 1/2 …

Note: This answer had been sent to Ali in private email (the question had been closed while I was composing it). Now that the question has been reopened, the answer may be given here. (This has been e …