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The set of all topologies on a given set $X$ admits a lattice structure under the refinement relation $\tau\leq\sigma$, whereby every $\tau$ open set is open with respect to $\sigma$, meaning that $\s …

To see that $S$ is countably compact, suppose that we have a countable open cover $\mathcal{U}$. Notice that $S$ is the union of the nested chain of subspaces $$(\omega_1+1)\times (\alpha+1)\cup (\alp …

The product of ultafilters $F_\lambda$ for $\lambda<\kappa$ is defined on $\kappa\times X$, not $X^\kappa$, and it is defined relative to a fixed ultrafilter $\mu$ on the index set $\kappa$. Namely, f …

One of the standard topologies to consider would be the lower-cone topology, whose basic open sets are the lower cones $i{\downarrow}=\{j\mid j\leq i\}$. In this topology, the open sets are exactly th …

Here is a counterexample to Q2, with your stated extra condition. Let $X$ consist of the half-open unit interval $(0,1]$ on the $x$-axis in the plane, together with the full unit interval $[0,1]$ at h …

It seems to me that in the general setting of a metric space, what one learns from the sampling data will be precisely the bounds provided by the instances of the triangle inequality that must be obey …

Regarding the dense embedding, perhaps this is helpful. Statement 1 can be taken as a definition of density, which makes the connection with topology by means of the lower-cone topology. Theorem. Supp …

The way you set this up, it might not be dense, since you only have that $p'$ forces that $f$ is a function from $\omega$ to the ordinals. Perhaps other incompatible conditions force that $f$ is not a …

This is more of a comment than an answer, since it is not a perfect fit. But I just thought I would mention the following paper, which is concerned not with continuous reducibility, but computable red …

Yes. The space of rational numbers $X=\mathbb{Q}$ is an instance. We can view $X$ as a countable union of countably many disjoint copies of $\mathbb{Q}$. Any nonempty subset $A$ of those copies (t …

I like this question a lot. The answer is no. Let $X$ consist of two disjoint copies of the real line. This is $T_2$ and homogeneous, in the sense that for any two points, there is a homeomorphism t …

Yes, just take a copy of the ordinal $\omega^\omega$ in the reals. This has Cantor Bendixson rank exactly $\omega$. One way to see this is first to understand how to make a closed set last for exact …

Not necessarily. Let $X$ be an uncountable set with the discrete topology, and let $\mathcal{E}$ be the collection of singletons, which is a base for the topology, since every set is a union of single …

This is a partial answer. I would like to note merely that this is not possible in a countable space. Theorem. In any countable Hausdorff space $X$ and any open cover $U$, there is a refinement of $U …

The $\frak{c}$-long line is $T_2$, connected and size continuum $\frak{c}$, but has $2^{\frak{c}}$ many open sets, since there is a size continuum discrete subset. The more familiar $\omega_1$-long …