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\section*{SEAR plus epsilon}

\hypertarget{sear}{}\section*{{SEAR+$\varepsilon$}}\label{sear}

\noindent\hyperlink{goal}{Goal}\dotfill \pageref*{goal} \linebreak
\noindent\hyperlink{defining_sear}{Defining SEAR+$\varepsilon$}\dotfill \pageref*{defining_sear} \linebreak
\noindent\hyperlink{conservativity_over_coshep}{Conservativity over COSHEP}\dotfill \pageref*{conservativity_over_coshep} \linebreak
\noindent\hyperlink{without_coshep}{Without COSHEP}\dotfill \pageref*{without_coshep} \linebreak
\noindent\hyperlink{without_coshep_predicatively}{Without COSHEP, predicatively}\dotfill \pageref*{without_coshep_predicatively} \linebreak


\hypertarget{goal}{}\subsection*{{Goal}}\label{goal}

The goal of this page is to prove that a non-extensional [[choice operator]] is conservative over a theory without AC. I'll take the theory without AC to be [[SEAR]], for definiteness and since that's where [[Mike Shulman|I'm]] most comfortable.

\hypertarget{defining_sear}{}\subsection*{{Defining SEAR+$\varepsilon$}}\label{defining_sear}

For the theory with a non-extensional choice operator, consider the following theory, which is a variant of the version of SEAR without fundamental equality described at [[SEAR]]. There are four basic sorts: \emph{pre-sets}, \emph{pre-relations}, \emph{elements}, and \emph{operations} (or \emph{pre-functions}). Each element belongs to a pre-set, and each pre-relation and operation has a source and target which are pre-sets. For a prerelation $\phi:A\looparrowright B$ and $x\in A$, $y\in B$, we have the assertion $\phi(x,y)$, and for an operation $f:A\rightsquigarrow B$ and $x\in A$ we have an element $f(x)\in B$. There is no predefined notion of equality of anything. Axioms 0 and 1 of SEAR are unmodified, except that the uniqueness clause of the latter is interpreted as a \emph{definition} of when two parallel pre-relations are called ``equal''. We also impose

\begin{uprop}
If $\varphi:A\looparrowright B$ is a pre-relation such that for every $x\in A$ there exists a $y\in B$ with $\varphi(x,y)$, then there exists an operation $\varepsilon_\varphi :A\rightsquigarrow B$ such that $\varphi(x,\varepsilon_\varphi(x))$ for all $x\in A$.

\end{uprop}
A \emph{set} is defined to be a pre-set $A$ equipped with an equivalence pre-relation $=_A$. A \emph{relation} between sets is a pre-relation which is \emph{extensional}, i.e. if $\varphi(x,y)$, $x'=_A x$, and $y'=_B y$, then $\varphi(x',y')$. Likewise, a \emph{function} $f:A\to B$ between sets is an operation $f:A\rightsquigarrow B$ which is extensional, i.e. if $x' =_A x$ then $f(x')=_B f(x)$. We define two functions $f,g:A\to B$ to be \emph{equal} if $f(x)=_B g(x)$ for all $x\in A$. Note that we have no notion of equality for arbitrary operations between pre-sets.

For any operation $f:A\rightsquigarrow B$ between \emph{sets}, we have a pre-relation (its \emph{graph}) defined by $\Gamma_f(x,y) \Leftrightarrow (f(x)=_B y)$, which is \emph{[[entire relation|entire]]} in the sense that for any $x\in A$, there is a $y\in B$ with $\Gamma_f(x,y)$. This pre-relation is extensional in $y$, but not in $x$ unless $f$ is actually a function (in which case $\Gamma_f$ is a \emph{[[functional relation]]} in the usual sense). Conversely, Axiom $1+\varepsilon$ says that any entire pre-relation induces an operation, and for an entire and functional relation between sets, this induced operation is a function (and is unique, in the sense of equality of functions defined above).

Now Axiom 2 can be taken to read: For any sets $A,B$ and any relation $\varphi:A\looparrowright B$, there exists a pre-set $|\varphi|$ and operations $p:{|\varphi|}\rightsquigarrow A$ and $q:{|\varphi|}\rightsquigarrow B$ such that $\varphi(x,y)$ iff there exists $z\in {|\varphi|}$ with $p(z)=_A x$ and $q(z)=_B y$. We can then define $z=_{|\varphi|} z'$ iff $p(z)=_A p(z')$ and $q(z)=_B q(z')$ to make $|\varphi|$ into a set and $p,q$ into functions that are jointly injective. Axioms 3, 4, and 5 are easy to translate.

We call this theory $SEAR+\varepsilon$. Clearly the sets, elements (with defined equality), and relations in any model of $SEAR+\varepsilon$ satisfy the axioms of SEAR. Conversely, from any model of SEAR-C we can construct a model of $SEAR+\varepsilon$ by taking the pre-sets, pre-relations, and operations to be the SEAR-C sets, relations, and functions respectively. With this interpretation axiom $1+\varepsilon$ is precisely the SEAR-C [[axiom of choice]]. The question is whether we can get $SEAR+\varepsilon$ without assuming or implying choice.

\hypertarget{conservativity_over_coshep}{}\subsection*{{Conservativity over COSHEP}}\label{conservativity_over_coshep}

Our goal is to prove that $SEAR+\varepsilon$ is conservative over SEAR. But since I haven't quite figured out how to do that (or, to be fair, even whether it's true), as a warm-up let's prove that the same thing is true when we add [[COSHEP]], a choice-like axiom notably weaker than full AC, to both sides.

Suppose we have a model of SEAR satisfying COSHEP, call it $V$. Recall that COSHEP means that in $V$, every set admits a surjection from a [[projective object|projective]] one. We define a model of $SEAR+\varepsilon$ as follows.

\begin{itemize}%
\item The pre-sets are the projective $V$-sets.
\item The elements are the $V$-elements.
\item The pre-relations are the $V$-relations.
\item The operations are the $V$-functions.

\end{itemize}
We will continue to qualify the notions in the old model $V$, using unqualified words such as ``set'' for the new notions defined above.

Axiom 0 of $SEAR+\varepsilon$ is obviously true. For Axiom 1, we need to translate a first-order formula in the language of $SEAR+\varepsilon$ into a formula in the language of SEAR in order to apply Axiom 1 of $V$, but the above dictionary tells us exactly how to do that. Axiom $1+\varepsilon$ is true because we have chosen the pre-sets to be the \emph{projective} sets in $V$, so any entire $V$-relation between them contains a $V$-function.

Now a \emph{set} $A$ in our putative model of $SEAR+\varepsilon$ is a $V$-set equipped with an equivalence relation $=_A$, so in particular it has a quotient $A/{=_A}$. Now any \emph{relation} $A\looparrowright B$, meaning extensional relation, induces a $V$-relation $(A/{=_A}) \looparrowright (B/{=_B})$, and conversely any such $V$-relation induces a relation $A\looparrowright B$, setting up a meta-bijection. The same is true of functions $A\to B$ and $V$-functions $(A/{=_A}) \to (B/{=_B})$. It follows that we have meta-functors $Set\to Set_V$ and $Rel\to Rel_V$ given by ``take quotients'' that are [[full and faithful functor|fully faithful]].

I claim that in fact these functors are [[essentially surjective functor|essentially surjective]] as well. For given any $V$-set $A$, by COSHEP there is a projective set $P$ and a surjection $e:P\to A$. Since $Set_V$ is a [[regular category]], it follows that $A$ is the [[quotient set]] of the [[kernel pair]] of $e$. But that means that if we call this kernel pair $=_P$, then $P$ becomes a new-style set which maps onto $A$ under the quotient functor; hence this functor is essentially surjective.

It follows that the meta-categories $Set$ and $Rel$ (in the new sense) are [[equivalence of categories|equivalent]] to the old meta-categories $Set_V$ and $Rel_V$. Since the remaining axioms of SEAR are essentially just statements about these categories, their truth in the ``new world'' follows immediately from their truth in $V$. (There's no need to worry about a meta-axiom-of-choice, since these axioms are just statements about finite numbers of objects and morphisms, so ``fully faithful and essentially surjective'' is a perfectly sufficient notion of ``equivalence.'')

Thus, we have constructed a model of $SEAR+\varepsilon$. Furthermore, its underlying SEAR-model is equivalent to $V$. (The notion of ``equivalence'' here means ``(weak) equivalence of categories of sets,'' as above. But since SEAR speaks no [[evil]], this suffices to show that exactly the same first-order statements are true in both.) It follows that the statements \emph{in the language of SEAR} which are true in this new model of $SEAR+\varepsilon$ are precisely those which are true in $V$.

Therefore, any statement in the language of SEAR which is true in all models of SEAR that underlie models of $SEAR+\varepsilon$ is, in fact, true in \emph{all} models of SEAR+COSHEP. This is the conservativity result we were looking for. Its practical upshot is that if you want to work in SEAR+COSHEP, you might as well work in $SEAR+\varepsilon$ if it's more convenient to have a choice operator, since the theorems you can prove in either case will be exactly the same.

Note that the model of $SEAR+\varepsilon$ we've just constructed also satisfies COSHEP (as it must, given its equivalence to $V$). In fact, if a model of $SEAR+\varepsilon$ admits an ``identity'' equivalence pre-relation $\equiv_A$ on every pre-set, and relative to which all pre-relations and operations are extensional, then it must satisfy COSHEP---for then every set $(A,{=_A})$ is the surjective image of the set $(A,{\equiv_A})$, which is projective by Axiom $1+\varepsilon$.

\hypertarget{without_coshep}{}\subsection*{{Without COSHEP}}\label{without_coshep}

Now [[Toby Bartels|I]] step in to say: $\mathbf{SEAR} + \varepsilon \vDash COSHEP$.

The reason is that every preset \emph{does} admit an identity prerelation as in the last paragraph above; let $x \equiv_A y$ if $x \sim_R y$ for every reflexive prerelation $R: A \looparrowright A$ (or even for every equivalence prerelation). This will work also in $\mathbf{ETCS} - AC + \varepsilon$ by [[quantification|quantifying]] over prefunctions $f: A \to \mathcal{P}1$ and using the kernel of $f$ (relative to the standard equality on truth values) as the equivalence relation. (Of course, $\mathbf{SEAR}$ is capable of using this method too.) It will still work in versions with intuitionistic logic, but not (as far as I can see) in $\mathbf{CETCS} - COSHEP + \varepsilon$ (following \href{http://www.math.uu.se/~palmgren/cetcs.pdf}{Palmgren}), where one cannot take power sets or quantify over subsets.

[[Mike Shulman]]: Ah, you're right. I actually thought of that briefly, but then I didn't immediately see how to prove the following.

\begin{theorem}
\label{equiv-exten}\hypertarget{equiv-exten}{}
Define $x \equiv_A y$ if $R(x,y)$ for every equivalence pre-relation $R: A \looparrowright A$. Then every operation and every pre-relation is extensional with respect to these identity relations.

\end{theorem}
\begin{proof}
Suppose $f:A\rightsquigarrow B$ is an operation. Define $K_f:A\looparrowright A$ by $K_f(x,x')$ iff $f(x)\equiv_B f(x')$. Then $R$ is an equivalence pre-relation, so if $x\equiv_A x'$, then $K_f(x,x')$, and hence $f(x)\equiv_B f(x')$.

Similarly, suppose $\varphi:A\looparrowright B$ is a pre-relation. Define $R_\varphi:A\looparrowright A$ by $R_\varphi(x,x')$ iff $\varphi(x,y)\Leftrightarrow\varphi(x',y)$ for all $y\in B$. Then $R_\varphi$ is an equivalence pre-relation, so if $x\equiv_A x'$, then $R_\varphi(x,x')$ and so $\varphi(x,y)\Leftrightarrow\varphi(x',y)$ for all $y\in B$.

\end{proof}
Of course, as you point out, this is very impredicative. Does that mean that your original \href{http://golem.ph.utexas.edu/category/2009/09/towards_a_computeraided_system.html#c026817}{suggestion} would only acceptable be for someone who either (1) accepts COSHEP or (2) doesn't accept (quantification over) powersets?

\emph{Toby}: That may be so; I hadn't thought through all of the implications, so I was hoping otherwise. However, I'm not sure that $\varepsilon$ has to work as in Axiom $1 + \varepsilon$; how about this?

\begin{uprop}
Add a symbol (actually several symbols) to the language for an operation that assigns to each set $A$:

\begin{itemize}%
\item a set $A'$,
\item an injective function $i_A: A \to A'$,
\item and an element $\epsilon_A$ of $A'$;

\end{itemize}
then add the axiom that $\epsilon_A$ belongs to the image of $i_A$ if (hence iff) $A$ is inhabited.

\end{uprop}
With excluded middle, this is a theorem with $A + 0^A$ for $A'$ as in the discussion at [[choice operator]]. But even this may be too strong; I really want to say that $\epsilon_A$ takes values in $A_\bot$ (the set of subsingleton subsets of $A$) with $\epsilon_A$ an inhabited subsingleton if (hence iff) $A$ is inhabited, but the existence of $A_\bot$ is itself impredicative (in the presence of function sets) since $1_\bot$ is the set of truth values.

\hypertarget{without_coshep_predicatively}{}\subsection*{{Without COSHEP, predicatively}}\label{without_coshep_predicatively}

Okay, second try. Define \textbf{constructive SEAR} to use [[intuitionistic logic]] and consist of Axioms 0, 1B, 2, and 4 of SEAR (no power sets) together with

\begin{itemize}%
\item The existence of [[quotient sets]] (i.e. $Set$ is an [[exact category]])
\item The existence of [[disjoint unions]] (i.e. now $Set$ is a [[pretopos]])
\item The existence of [[dependent products]] (i.e. now $Set$ is a $\Pi$-[[Π-pretopos|pretopos]]).

\end{itemize}
Now modify it as above, removing basic equality and adding a basic notion of ``operation,'' to obtain \textbf{CSEAR+$\varepsilon$}. Note that the axiom of quotient sets becomes redundant, given the definition of ``set'' as a preset with an equivalence prerelation.

The same argument as above should show that CSEAR+$\varepsilon$ is conservative over CSEAR+COSHEP. (Just as Bounded SEAR is equivalent to ETCS, CSEAR+COSHEP is equivalent to Palmgren's CETCS.) However, the converse argument above fails since we cannot define $\equiv_A$ by quantifying over relations (note that avoiding this requires not only throwing out powersets, but restricting axiom 1 to [[bounded quantification]]).

Now: is CSEAR+$\varepsilon$ conservative over CSEAR?

[[!redirects SEAR plus epsilon]] [[!redirects SEAR plus ?]] [[!redirects SEAR plus ?]] [[!redirects SEAR+epsilon]] [[!redirects SEAR+?]] [[!redirects SEAR+?]] [[!redirects SEAR epsilon]] [[!redirects SEAR ?]] [[!redirects SEAR ?]]



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