Timeline for Model ${\sf ZF}$ that "spreads" members of ${\cal P}(X)$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2018 at 9:23 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 11, 2018 at 8:49 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 8, 2018 at 23:23 | answer | added | Amitayu Banerjee | timeline score: -1 | |
| Nov 29, 2017 at 15:50 | comment | added | Alex Kruckman | Ah, so it is easy. Thanks, Péter. I'll also alert @Gro-Tsen who asked the question as well. | |
| Nov 21, 2017 at 6:39 | comment | added | Péter Komjáth | Alex: Choose for each $a$ with $|f(a)|>k$, a $k+1$-element subset of $f(a)$. Under the axiom of choice, $|X|^{k+1}=|X|$, so there are at most $|X|$ such elements of $P(X)$. Similarly, the number of those elements of $P(X)$ for which $|f(a)|\leq k$ is also $\leq|X|$, so we get that $|P(X)|\leq|X|$, contradicting Cantor's theorem. | |
| Nov 13, 2017 at 15:21 | comment | added | Alex Kruckman | I'm also curious how to prove in ZFC there there is no infinite $X$ and injective $f\colon \mathcal{P}(X)\to \mathcal{P}(X)$ such that $|f(a)\cap f(b)|\leq k$ for all $a\neq b$, even for $k=1$. | |
| Oct 30, 2017 at 23:47 | comment | added | Asaf Karagila♦ | What @Gro-Tsen says is true. If you weaken this to a finite intersection then $\Bbb N$ produces an example without needing choice. You don't need choice to produce a continuum sized almost-disjoint family of subsets of $\Bbb N$ (and before you post this as a question, this was asked on MSE a few times, and probably here also, like in the second comment by Gro-Tsen). | |
| Oct 30, 2017 at 21:43 | comment | added | Gro-Tsen | Where can I find a proof of the fact that the statement with $|f(a)\cap f(b)|\leq 1$ is refutable in ZFC? How about with $|f(a)\cap f(b)|\leq k$ (with $k$ constant)? (Maybe these questions are stupid.) | |
| Oct 30, 2017 at 16:00 | history | edited | Dominic van der Zypen | CC BY-SA 3.0 |
added a comment by @Gro-Tsen as a note
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| Oct 30, 2017 at 15:55 | comment | added | Dominic van der Zypen | Thanks @Gro-Tsen - I am aware of this result; if you strengthen the finiteness condition you mention to "$f(a)\cap f(b)$ is empty or at most a singleton" (or contains at most $n$ elements for some fixed $n\in\mathbb{N})$ then the resulting statement is false in ZFC, but maybe not in ZF. But coming back to your statement, I will include it in the problem statement -- thank you again! | |
| Oct 30, 2017 at 15:37 | comment | added | Gro-Tsen | (In fact, the proof of the statement in my previous comment is so easy I might as well write it here: map an infinite binary sequence to the sequence of its finite prefixes, and use obvious bijections to get a map $f\colon\mathcal{P}(\mathbb{N})\to\mathcal{P}(\mathbb{N})$ with the same properties.) | |
| Oct 30, 2017 at 15:34 | comment | added | Gro-Tsen | It might be worth pointing out that (it is an easy theorem of ZF that) there exists $f\colon\mathcal{P}(\mathbb{N})\to\mathcal{P}(\mathbb{N})$ such that $f(a)\cap f(b)$ is finite for every $a\neq b$ in $\mathcal{P}(\mathbb{N})$ (see, e.g., Jech, Set Theory, third millennium ed., lemma 9.21). | |
| Oct 30, 2017 at 13:43 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |