Timeline for Is this principle of internalization of external injections inconsistent?
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10 events
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| Oct 27, 2021 at 16:52 | comment | added | Zuhair Al-Johar | @Holo, how can you prove that $j_\beta$ is a set? We don't have the symbol $j$ as a primitive constant, so when we come to define the functional relation that defines $j_\beta$ [for a limit $\beta$] we must put that for each $\alpha \in \omega_1$ we have $j_{\alpha+1}= j_\alpha \cup \{(\alpha, J(\alpha))\}$, but once this enters the definition of $j_\beta$, then we cannot prove it being a set, since its definition contains an upper case variable, that is $J$. The problem is in the definition of limit stages of $j$,, they'll always use the symbol $J$, thereby precluding proving them being sets. | |
| Oct 26, 2021 at 17:43 | comment | added | Holo | @ZuhairAl-Johar Where did I use replacement on $J$? $j_0$ is a set because it is the empty set, if $α∈ω_1$ then $J(α)∈ω$ so it is also a set, hence $(α,J(α))$ (from normal ZF arguments) is a set, so $j_{α+1}$ is a set. And if $β∈ω_1$ is limit, we can define the functional relation $φ$ such that $φ(x,y,β)⇔(x∈β\text{ and }y=j_{x})\text{ or }(x∉β\text{ and }y=0)$, this is a formula only on sets, then there exists a set $b$ such that $a∈b$ if and only if there exists $c∈β$ such that $φ(a,c,β)$, that is, $a=j_{α}$ for some $α∈β$. | |
| Oct 26, 2021 at 4:34 | comment | added | Zuhair Al-Johar | @Holo, No! because set Replacement is fully written in set variables (lower case) , so you cannot use the upper case symbol $J$. | |
| Oct 25, 2021 at 22:20 | comment | added | Holo | Can't you take x,y to be aleph numbers, e.g. aleph1, aleph0, and then by induction on aleph1 show that $J$ is a set, which will lead to inconsistency? e.g. If $J:\omega_1↣\omega$, let $j_0=\emptyset$ and for $\alpha\in\omega_1$ let $j_{\alpha+1}=j_\alpha\cup\{(\alpha, J(\alpha))\}$, and at limit points take the union. Every stage is a set, as well as the union of all stages. That union is injective iff $J$ is injective, which is impossible | |
| Oct 25, 2021 at 12:05 | comment | added | Zuhair Al-Johar | @MarkSaving, it's Tarski infinite, I will update the post. Thanks! | |
| Oct 25, 2021 at 12:00 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Oct 24, 2021 at 21:52 | comment | added | Mark Saving | Since you didn't include the axiom of choice, how are you defining "infinite"? Also, what is $\rightarrowtail$? | |
| Oct 24, 2021 at 16:57 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Oct 24, 2021 at 12:32 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Oct 24, 2021 at 12:26 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |