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Noah Schweber
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Extending models of topological set theory

$\mathsf{GPK_\infty^+}$ is an alternative set theory in which we have comprehension for formulas which are positive in a certain sense; see the SEP article for more detail (or this MO post, which contains a nice description of one particular model of $\mathsf{GPK_\infty^+}$). Every model $X$ of $\mathsf{GPK_\infty^+}$ carries a natural topology $\sigma_X$, generated by a basis whose elements correspond to the elements of $X$ itself: for each $x\in X$, we make $\{y\in X: y\not\in x\}$ a basic open set.

Now in $\mathsf{ZF}$-land, we have a dichotomy between sets and proper classes provided by size: roughly, a class is proper iff it surjects onto the ordinals. However, in a model $X$ of $\mathsf{GPK_\infty^+}$ we have a natural trichotomy amongst "$X$-classes:" actually an $X$-set, not an $X$-set but $\sigma_X$-closed, and not even $\sigma_X$-closed. I'm curious about whether $X$-classes in this second category are genuinely "more ($X$-)set-like;" this question is my attempt to make this vague idea precise.


If $M,N$ are models of $\mathsf{GPK_\infty^+}$ with $M$ a submodel of $N$, say that $N$ is a nice extension of $M$ if

  • for each $m\in M$ we have $\{x\in M: M\models x\in m\}=\{y\in N: N\models y\in m\}$ (that is, $N$ is an end extension in the usual sense); and

  • $M$ is closed in $\sigma_N$.

Now say that a model $X$ of $\mathsf{GPK^+_\infty}$ is rich iff all $\sigma_X$-closed sets are $\sigma_X$-basic closed sets. I believe the usual construction of a model of $\mathsf{GPK^+_\infty}$ from a weakly compact cardinal produces a rich model, but not all models are rich (e.g. no countable model can be rich, and $\mathsf{GPK_\infty^+}$ is after all "just" a first-order theory). However, I don't immediately see an obstacle to every model being "potentially rich:"

Is there an $M\models\mathsf{GPK_\infty^+}$ with no rich nice extension?

We could also look for a "single culprit" set:

Is there an $M\models\mathsf{GPK_\infty^+}$ and a $\sigma_M$-closed $A\subset M$ such that no nice extension $N\supseteq M$ has an $n\in N$ with $A=\{x\in M: N\models x\in n\}$?

I don't have any ideas in either direction with respect to either question. On general grounds of orneriness I suspect that their answers are "yes" and "no" respectively, but this isn't really based on anything.

An analogous question can be posed for pretty much any "topological" set theory. My specific focus on $\mathsf{GPK_\infty^+}$ is simply due to the fact that it is the topological set theory which seems most natural to me at the moment; I would be interested in answers for (non-contrived) other choices of topological set theories as well.


EDIT: It occurs to me that there is a natural candidate "good result," contra my ornery expectations. Given subclasses $A,B$ of a model $\mathcal{X}\models\mathsf{GPK_\infty^+}$, say that $A$ is internally Wadge reducible to $B$ ($A\le_\mathcal{X}B$) iff there is some set function $f\in\mathcal{X}$ such that $A=f^{-1}[B]$. Internal Wadge reducibility (unlike definable Wadge reducibility) plays well with $\mathsf{GPK_\infty^+}$, in the sense that the latter proves (in the appropriate sense) that anything internally Wadge reducible to a set is again a set. This motivates the hope that the following could be equivalent for subclasses $A,B\subseteq\mathcal{X}$, for at least some wide class of $\mathcal{X}$s:

  • $A\le_\mathcal{X} B$.

  • In every nice extension $\mathcal{Y}$ of $\mathcal{X}$, if $B$ is a set then $A$ is a set.

While it's not my main question, I'd be interested in any comments re: the plausibility of this situation as well.

Noah Schweber
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