3
$\begingroup$

I'm largely following the definitions of this paper, but I will replicate the relevant ones here.

I'm taking a pseudotopological space to be a set $X$ together with a relation $\rightarrow$ on the set $\mathscr{UF}X\times X$ (called "converges to" or "convergence"), where $\mathscr{UF}X$ is the set of ultrafilters on $X$, with the stipulation that for any $x\in X$ the principal ultrafilter generated by $\{x\}$ converges to $x$.

A pseudotopological space $X$ is compact if every ultrafilter converges to at least one point.

A pseudotopological space $X$ is Hausdorff if every ultrafilter converges to at most one point.

Note that in a compact Hausdorff pseudotopological space the relation of convergence is actually a function from ultrafilters to points.

Two pseudotopological spaces are homeomorphic if there is a bijection between them that preserves the convergence relation.

Given a pseudotopological space $X$ and a subset $Y$ the induced pseudotopological structure on $Y$ is just the convergence relation restricted to ultrafilters containing $Y$ and $Y$.

Given a family of pseudotopological spaces $\{X_i\}_{i\in I}$ we define the product space as a pseudotopological space whose underlying set is $\prod_{i\in I}X_i$ and where we say that an ultrafilter $\mathcal{U}$ on $\prod_{i\in I}X_i$ converges to $x \in \prod_{i\in I}X_i$ if for each $i\in I$ the induced ultrafilter on $X_i$ converges to $x_i$. One can check that an arbitrary product of compact Hausdorff pseudotopological spaces is compact Hausdorff.

A set $F\subseteq X$ is 'closed' if for every ultrafilter $\mathcal{U}$ on $X$, if $F\in \mathcal{U}$ and $\mathcal{U}$ converges to $x$, then $x\in F$. (You can check that as set is closed if and only if its complement is open in the sense of the paper linked.)


I am wondering about a generalization of a nice fact about compact Hausdorff (topological) spaces, which is that any compact Hausdorff space $X$ is homeomorphic to a closed subspace of $[0,1]^\kappa$ for some cardinal $\kappa$. So obviously the question is

Does there exist a compact Hausdorff pseudotopological space $Y$ such that any compact Hausdorff pseudotopological space $X$ is homeomorphic to a closed subspace of $Y^\kappa$ for some cardinal $\kappa$?

I have a strong hunch that this is either flat out false or sensitive to set theoretic assumptions.

An easy observation is that if there is a set sized family of compact Hausdorff pseudotopological spaces that are analogously universal then there is a single space since we can take the product of the entire family and this will be 'product universal'.

$\endgroup$
  • 3
    $\begingroup$ Can you prove that $Y = [0,1]$ doesn't work? (And if so, what does the proof look like -- what seems to be getting in the way?) $\endgroup$ – Will Brian Mar 6 at 9:41
  • 1
    $\begingroup$ Yeah but the proof isn't very satisfying. For pseudotopological spaces whose convergence relation comes from a topological space, all of these defined notions agree with the corresponding topological ones, so in particular any closed subspace of $[0,1]^\kappa$ is closed in the topological sense. Therefore the only compact Hausdorff pseudotopological spaces that arise as closed subspaces of $[0,1]^\kappa$ are compact Hausdorff topological spaces, but not all compact Hausdorff pseudotopological spaces are topological. $\endgroup$ – James Hanson Mar 6 at 12:15
5
$\begingroup$

Given such a space $Y$ we can find a space $X$ with distinct $a,b\in X$ such that every continuous $f:X\to Y$ satisfies $f(a)=f(b).$ This rules out any such embedding.

Let $\kappa$ be a regular cardinal larger than $\max(\aleph_0,|Y|)$ and take $X$ to be the pseudotopological space on $\kappa\cup \{a,b\}$ defined by $\mathcal U\to a$ for non-principal ultrafilters $\mathcal U$ that extend the generalized Fréchet filter $\{F\mid |\kappa\setminus F|<\kappa\},$ and $\mathcal U\to b$ for all other non-principal ultrafilters, i.e. if $\mathcal U$ contains a set of cardinality less than $\kappa.$ Given a continuous $f:X\to Y,$ pick $y\in Y$ with $|f^{-1}(\{y\})|=\kappa.$ There are ultrafilters $\mathcal U,\mathcal U'$ each containing $f^{-1}(\{y\})$ and such that $\mathcal U\to a$ and $\mathcal U'\to b.$ So $f(a)=y=f(b)$ as required.

Just terminology, but the property you are asking about is a small cogenerating set.

$\endgroup$
  • $\begingroup$ Am I right in thinking that $\kappa$ should be regular or have large cofinality? $\endgroup$ – James Hanson Mar 6 at 18:25
  • $\begingroup$ @JamesHanson: yes, you're right - I want to rule out maps like $\aleph_\omega\to\omega$ where each preimage has order less than $\aleph_\omega.$ I have added the word regular $\endgroup$ – Dap Mar 7 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.