Timeline for Model ${\sf ZF}$ that "spreads" members of ${\cal P}(X)$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2018 at 10:19 | comment | added | Amitayu Banerjee | @AsafKaragila Yes, we can't say they are comparable in ZF. I didn't assume $\vert \mathcal{P}(X)\vert$ and $\vert X \vert^{2}$ are comparable. I was trying to obtain $\vert \mathcal{P}(X)\vert \leq \vert X \vert^{2}$, which is not possible since there is no injection from $\mathcal{P}(X)$ to $X^{2}$. | |
| Feb 9, 2018 at 8:59 | comment | added | Asaf Karagila♦ | But you seem to assume that $|\mathcal P(X)|$ and $|X|^2$ are comparable. Why are they comparable? | |
| Feb 9, 2018 at 8:46 | comment | added | Amitayu Banerjee | @AndrésE.Caicedo What if we use the Theorem 4.2 from the Combinatorial Set Theory book by Lorenz J. Halbeisen ? I think then we can claim in ZF that $\vert\mathcal{P}(X)\vert \not\leq \vert X \vert^{2}$ for any infinite set X. I edited. | |
| Feb 9, 2018 at 8:40 | history | edited | Amitayu Banerjee | CC BY-SA 3.0 |
added 7 characters in body
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| Feb 9, 2018 at 1:05 | comment | added | Andrés E. Caicedo | You cannot prove in $\mathsf{ZF}$ that $|\mathcal P (X)|>|X|^2 $ for $X$ infinite. | |
| Feb 8, 2018 at 23:56 | comment | added | Nate Eldredge | "There are atmost $\vert X \vert^{2}$ many elements $a$ of $\mathcal{P}(X)$ such that $\vert f(a)\vert > 1$." How do you prove this in ZF? | |
| Feb 8, 2018 at 23:43 | review | Late answers | |||
| Feb 8, 2018 at 23:47 | |||||
| Feb 8, 2018 at 23:28 | review | First posts | |||
| Feb 9, 2018 at 1:31 | |||||
| Feb 8, 2018 at 23:23 | history | answered | Amitayu Banerjee | CC BY-SA 3.0 |