Timeline for Can this kind of Mereology be synonymous with Set Theory?
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16 events
| when toggle format | what | by | license | comment | |
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| Jan 28 at 7:19 | vote | accept | Zuhair Al-Johar | ||
| Jan 28 at 7:11 | comment | added | paste bee | Yep, that's synonymous with MK. The specifics of the argument are a little different (now $\{\{\varnothing\}\}$, which is $\{b\}$, is unchanged, and $\{\varnothing,\{\varnothing\}\}$, which would be $\{a,b\}$, has to become something else, since $b = \{a,b\}$) but it's the exact same idea, and indeed the fact that they're definably different means we can just say $a$ corresponds to $\varnothing$ and not get any problems. | |
| Jan 28 at 7:01 | comment | added | Zuhair Al-Johar | You see there is a sense in which Mereology enables strong visualization into internal set structures, and sometimes can enable mush consize arguments. Anyhow, I'm seeing the problem here. I don't want to add a primitive to the language, but could this be fixed if we add a quine atom, and a self membered set that is not equal to the quine atom that contains the quine atom as it's only non-self element, i.e. we have $a=\{a\}$ and $b=\{\mathcal S \, x: x=a \lor x=b \}$ where $b\neq a$; then we stipulate that only those can breach foundation. Now, those are definable! | |
| Jan 28 at 6:43 | history | edited | paste bee | CC BY-SA 4.0 |
the theory does actually have power set
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| Jan 28 at 6:41 | comment | added | paste bee | ...Oh, I see now. I had missed that the Replacement axiom was a biconditional. In the hereditarily countable sets, if $A$ is labelled then $B$ is labelled, but the converse isn't always true. | |
| Jan 28 at 6:38 | comment | added | Zuhair Al-Johar | the formula I mentioned would send every singleton of a subclass of $x$ to its element, and so it is one-to-one relation, now clearly $x$ is the sum of all atoms of its parts, by replacement since $x$ is labeled then, its power class must be labeled too. In set theoretic terms: we have one-to-one relation between the class of all singletons of subclasses of $x$ and the power class of $x$, since the set union of the later is $x$ itself, and $x$ do have a label, then the set union of the former must be labeled, this is saying the power class of $x$ must have a label. | |
| Jan 28 at 6:32 | comment | added | Zuhair Al-Johar | In atomic Mereology every object is a sum of its atoms, this is like saying every class is the union of all singleton subsets of it. When it is said a set is the sum of atoms of its parts, is like saying every class is the union of all singleton subsets of subsets of it; that is, $x= \bigcup \{ y \mid \exists z \subseteq x: y \subseteq z \land \exists k (y=\{k\}) \}$. Now, the sum of all labels of parts of $x$ is in set theoretic terms the class union of all singletons of subclasses of $x$, which is the power class of $x$, | |
| Jan 28 at 5:58 | comment | added | paste bee | I feel like at least one of us is making mistakes because of the mereological language, which at least for me is not something I'm used to. Is the set-theoretic statement "for any set $A$, $\{b : \exists a \in A \exists c (\varphi(a,c) \land b \in c)\}$ is a set" equivalent? (assuming $\varphi$ acts like a function and always "outputs" sets) | |
| Jan 28 at 5:50 | comment | added | paste bee | @ZuhairAl-Johar "clearly $x$ is the sum of all atoms of its parts" I don't see why this is true. If $x$ is $\mathbb{N}$, then the set of even numbers is one of its parts, but the set of even numbers is not a natural number, so the label of the set of even numbers is not part of $x$. If you mean that $x$ is the union of all of its parts that are atoms, that's true but doesn't seem helpful, it just means you'd end up with the set of all atoms that are part of $x$, which is not the power set. | |
| Jan 28 at 5:36 | comment | added | Zuhair Al-Johar | Power set axiom is provable from Replacement. Let $x$ be labeled (i.e. $\exists l: l=\{x\}$). Now, $\mathcal P(x)$ is the totality of all labels of parts of $x$, clearly $x$ is the sum of all atoms of its parts, and we have $l=\{y\} \land l \subseteq \mathcal P(x) \land y \subseteq x$ is 1-to-1, so by Replacement $\mathcal P(x)$ must be labeled. $\square$. In a similar way you can prove Set Unions. | |
| Jan 28 at 4:46 | comment | added | paste bee | @ZuhairAl-Johar The issue is that we can't just "fix a specific global choice function". None of them are definable, in particular we can't definably choose one of the Quine atoms. Also, my argument doesn't just require a particular global choice function; it requires the fact that the global choice function the interpretation is given (as $C$) is the one defined by the other interpretation, in order to get back out encoded information, and ensure that the interpretations are inverses of each other. | |
| Jan 28 at 4:08 | comment | added | paste bee | Adding a constant function for global choice would make it synonymous with "MK with a constant function for global choice", by applying my earlier argument. I don't know if that's synonymous with MK, but I think it probably isn't. | |
| Jan 28 at 4:03 | comment | added | paste bee | How does it prove power set? And which of the axioms is false in my "hereditarily countable sets" model? | |
| Jan 28 at 3:59 | comment | added | Zuhair Al-Johar | but why we cannot fix a specific global choice function, and work your older argument in reference to it? Also, another way if we add a constant function standing for global choice instead of adding a primitive constant denoting one of the Quine atoms, would that work to achieve synonymy. | |
| Jan 28 at 3:51 | comment | added | Zuhair Al-Johar | Thank you! As regards Infinity you are right, this I missed to add, I'll correct it. But this theory proves the power set axiom. So, after all it is not synonymous with MK because we cannot distinguish between the two atoms, that we need to add a constant to do that. Actually, that's what I wanted to really know. | |
| Jan 27 at 21:48 | history | answered | paste bee | CC BY-SA 4.0 |