Nonexistence of a 'product universal' compact Hausdorff pseudotopological space?

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'.

• 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?) – Will Brian Mar 6 '19 at 9:41
• 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. – James Hanson Mar 6 '19 at 12:15

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.
• Am I right in thinking that $\kappa$ should be regular or have large cofinality? – James Hanson Mar 6 '19 at 18:25
• @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 – Dap Mar 7 '19 at 8:49