Timeline for Is this internalization of a bijection between a set and its powerset possible?
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21 events
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| Oct 16, 2021 at 20:25 | comment | added | Farmer S | @ZuhairAl-Johar, it would be enough to assume that $M$ models ZF + "$X$ and $Y$ have no amorphous subsets". If both $X$ and $Y$ had amorphous subsets then one would have to match them up in some way I suppose, the plausibility of which it seems might depend on the circumstances. | |
| Oct 16, 2021 at 18:27 | comment | added | Zuhair Al-Johar | @FarmerS, is choice necessary for your proof? I mean would the same work over any model $M$ of $\sf ZF$? | |
| Oct 10, 2021 at 17:00 | comment | added | Zuhair Al-Johar | @FarmerS, possibly we need to do that in some large cardinal hypothesis, and possibly one that is incompatible with choice? | |
| Oct 8, 2021 at 12:31 | comment | added | Farmer S | @Zuhairal-Johar I don't see that either. | |
| Oct 7, 2021 at 16:39 | comment | added | Zuhair Al-Johar | @FarmerS, can we get a partial resut of $j$ obeying the condition you proved and also obeying existence of a set $_2\mathsf{Back}_j(S) $ for every set $S$? | |
| Oct 7, 2021 at 10:19 | comment | added | Zuhair Al-Johar | I think this needs to be put as an answer to a partial result related to my question or to Noah's version. | |
| Oct 7, 2021 at 9:20 | comment | added | Farmer S | Remark: Note that in $M[G]$, $j$ induces a bijection $j^+:\mathcal{P}(X)\cap M[G]\to \mathcal{P}(Y)\cap M[G]$ (i.e. $j^+(S)=j[S]$), and we have $j^+[(\mathcal{P}(X))\cap M]=\mathcal{P}(Y)\cap M$, so $j$ induces a bijection $k:\mathcal{P}(X)\cap M\to\mathcal{P}(Y)\cap M$ in this way. So we can ask whether $k$ also has the same property, i.e. whether $k[S]\in M$ and $k^{-1}[S]\in M$ for each $S\in M$. If so, it would give a strengthening of the desired property in the original question. But it's not obvious to me how to extend the forcing argument to do this (or to do the original question). | |
| Oct 7, 2021 at 9:15 | comment | added | Farmer S | Finally note that for every $S\in\mathcal{P}(X)\cap M$ and every condition $p$, $S$ is a finite variant of some $p$-suitable $S'$, so by density we will get some such $S'$ with $j[S']\in V$, but then it follows that $j[S]\in V$. It is completely analogous going in reverse, i.e. for $j^{-1}$. | |
| Oct 7, 2021 at 9:14 | comment | added | Farmer S | It is now easy to find a condition $p_1\leq p$ with $p_1=(f,P_1,Q_1,F_1)$ for some $Q_1,F_1$. Note that $p_1$ forces that $\dot{j}[S]=\bigcup (F_1``D)$, and in particular forces $\dot{j}[S]\in\check{V}$... | |
| Oct 7, 2021 at 9:13 | comment | added | Farmer S | And e.g. given $S\in\mathcal{P}(X)\cap M$, and a condition $p=(f,P,Q,F)$, say that $S$ is $p$-suitable iff for each $B\in P$, either $B\subseteq S$ or $B\cap S=\emptyset$ or both $B\cap S$ and $B\backslash S$ are infinite. So if $S$ is $p$-suitable then we get a refinement $P_1$ of $P$ given by $P_1=\{B\cap S\bigm|B\in P\}\cup\{B\backslash S\bigm|B\in P\}$, and all elements of $P_1$ are infinite, and note that $S$ is the union of some (finite) subset $D\subseteq P_1$... | |
| Oct 7, 2021 at 9:11 | comment | added | Farmer S | If $G$ is $M$-generic then $j=\bigcup_{p\in G}f^p$ is a bijection $j:X\to Y$ as desired, i.e. $j[S]\in M$ for each $S\in\mathcal{P}(X)\cap M$, and $j^{-1}[S]\in M$ for each $S\in\mathcal{P}(Y)\cap M$. This is a straightforward density calculation. The inifiniteness of the sets $B\in P$ and $C\in Q$, and finiteness of the functions $f$, ensure that $j$ will end up being a bijection... | |
| Oct 7, 2021 at 9:10 | comment | added | Farmer S | The conditions are pairs $(f,P,Q,F)$ where $f$ is a finite partial injective function $A\to Y$ where $A\subseteq X$, $P$ is a partition of $X$ into finitely many subsets, each of which is infinite, $Q$ is a partition of $Y$ into finitely many subsets, each of which is infinite, $F:P\to Q$ is a bijection, and $f$ respects $F$, i.e. $f(b)\in F(B)$ for each $b\in A$ and $B\in P$ with $b\in B$. And $(f',P',Q',F')\leq(f,P,Q,F)$ iff $f'$ extends $f$, $P'$ refines $P$, $Q'$ refines $Q$, and $F'$ refines $F$, in that if $B\in P$ and $C\in Q$ and $C\subseteq B$ then $F'(C)\subseteq F(B)$... | |
| Oct 7, 2021 at 9:09 | comment | added | Farmer S | Actually I noticed now that the version I asked about is consistent; in fact if $M\models\mathrm{ZFC}$ then one can force it over $M$: Fix infinite sets $X,Y\in M$. Consider the following forcing $\mathbb{P}$, which attempts to build such bijection with this property... | |
| Oct 7, 2021 at 5:12 | comment | added | Zuhair Al-Johar | I don't know, but I guess it"s unknown too. However, if the above hold then it would follow from it. | |
| Oct 6, 2021 at 21:26 | comment | added | Farmer S | Is the answer known for the version where you just demand that for each set $S$, we have both $j[S]$ and $j^{-1}[S]$ are also sets? | |
| Oct 6, 2021 at 16:52 | comment | added | Zuhair Al-Johar | @NoahSchweber, Your version, albeit more general, is very interesting, well to me at least. | |
| Oct 6, 2021 at 9:12 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Oct 6, 2021 at 5:37 | comment | added | Noah Schweber | Interesting question! For what it's worth, even the following version of this question (which I'll phrase model-theoretically for clarity) is unclear to me: suppose $M$ is a countable transitive model of $\mathsf{ZF}$ and $X,Y\in M$ are infinite. Must there be an external bijection $j:X\rightarrow Y$ such that for each $S\in M$ both $$_2\mathsf{Forth}_j(S):=\{\{j(m),j(n)\}:\{m,n\}\in S\}$$ and $$_2\mathsf{Back}_j(S):=\{\{m,n\}:\{j(m),j(n)\}\in S\}$$ are in $M$ as well? | |
| Oct 6, 2021 at 4:56 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Oct 6, 2021 at 4:47 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Oct 6, 2021 at 4:41 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |