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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 …

Perhaps this isn't what you had in mind, but the Compactness Theorem of first order logic, proved by Goedel, fits your "experimental" metaphor quite well, and variations on its theme have led to some …

This question is about the space of all topologies on a fixed set X. We may order the topologies by refinement, so that τ ≤ σ just in case every τ open set is open in σ. Equivalently, we say in this c …

This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some ti …

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 …

There is some fascinating work in the subject of cardinal characteristics of the continuum in set theory that directly relates to the concept of growth rates of functions. I believe that it is the ide …

My question is about the concept of nonstandard metric space that would arise from a use of the nonstandard reals R* in place of the usual R-valued metric. That is, let us define that a topological sp …

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 …