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

Question

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.)

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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'.

Answer 1

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.