Timeline for Cantor's diagonal argument and ZF
Current License: CC BY-SA 3.0
28 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2013 at 2:01 | comment | added | Joel David Hamkins | Ah, and that case conforms with my conjecture, since $\aleph(X)=\omega$, which injects into $P(X)$ if it isn't Dedekind finite. (I'm sorry, I'm just catching up to you guys here...) | |
| Feb 21, 2013 at 1:32 | comment | added | François G. Dorais | But there is a $d_X$ when $X$ is Dedekind finite and $\mathcal{P}(X)$ isn't. Hopefully this is the only case. | |
| Feb 20, 2013 at 22:46 | comment | added | Joel David Hamkins | Perhaps we can simply refute the existence of $d_X$ when $X$ is Dedekind finite but not finite? And perhaps the truly finite case is subsumed by the well-ordered cardinal case? | |
| Feb 20, 2013 at 22:09 | comment | added | François G. Dorais | Looks like we're stuck in the Dedekind finite case. Can we do it in the really finite case and generalize? | |
| Feb 19, 2013 at 23:32 | comment | added | Joel David Hamkins | That's true. So I shall let hope for my conjecture stay alive! | |
| Feb 19, 2013 at 23:11 | comment | added | Asaf Karagila♦ | Hang on. I see a problem with my argument. The choice of $d_X(i)$ should be independent of $i$, so if $i_1(X)=i_2(X)$ we should have $d_X(i_1)=d_X(i_2)$. But my construction is very dependent of the choice of $i$. | |
| Feb 19, 2013 at 22:43 | comment | added | Asaf Karagila♦ | I am writing one right now. I went to check something in the vast bibles of choice principles. I'll finish my answer in a short time. (Thanks for the vote of confidence!) | |
| Feb 19, 2013 at 22:28 | comment | added | François G. Dorais | I vote for Asaf :) | |
| Feb 19, 2013 at 22:25 | comment | added | Joel David Hamkins | I should have said, Asaf or François! | |
| Feb 19, 2013 at 22:06 | comment | added | François G. Dorais | OK. Everything seems valid: If $X$ is Dedekind infinite, then there is such a map $d_X$ if and only if $\aleph(X) \preceq \mathcal{P}(X)$; if $X$ is Dedekind finite, then there is such a map $d_X$ regardless. | |
| Feb 19, 2013 at 21:57 | comment | added | Joel David Hamkins | Asaf, why don't you post an answer explaining it all? | |
| Feb 19, 2013 at 20:32 | comment | added | Asaf Karagila♦ | Francois, it seems so! We made quite the complements here! :-D | |
| Feb 19, 2013 at 20:31 | comment | added | François G. Dorais | And I think your argument, Asaf, works for all Dedekind finite $X$. I think this is it! | |
| Feb 19, 2013 at 20:28 | comment | added | Asaf Karagila♦ | Francois, it's quite easy to show that if $\omega\leq\alpha\lt\aleph(X)$ then $\alpha+X\sim X$. So in fact your argument works for all Dedekind-infinite sets. | |
| Feb 19, 2013 at 20:16 | comment | added | François G. Dorais | Actually, my argument only needs $\alpha+X \cong X$ for every $\alpha < \aleph(X)$. Note that $1+X \cong X$ iff $X$ is Dedekind infinite. | |
| Feb 19, 2013 at 20:04 | comment | added | Asaf Karagila♦ | Joel, so it seems we have established in the comments the following: (1) If $X+X\sim X$ then the existence of $d_X$ is equivalent to $\aleph(X)\leq\mathcal P(X)$; (2) If $X$ is amorphous then $d_X$ exists, but both the assumption and the consequence of (1) fail in this case. So how can we characterize cardinals such that $X+X\gt X$ and still $d_X$ exists? Is there any "nice" characterization for them at all? Maybe it is consistent that $d_X$ exists for all $X$ and choice fails, but maybe it's not... | |
| Feb 19, 2013 at 19:24 | comment | added | Asaf Karagila♦ | That's great! Amorphous sets are the best. :-) | |
| Feb 19, 2013 at 19:20 | comment | added | Joel David Hamkins | Yes! That shows that amorphous sets refute my suggestion on the Hartog number, since you've proved that they have a $d_X$, but meanwhile $\aleph(X)$ does not inject into $P(X)$, because you said $\aleph(X)=\aleph(P(X))$. | |
| Feb 19, 2013 at 19:17 | comment | added | Asaf Karagila♦ | [...] The induction has to stop at a finite point, otherwise we found a countable sequence. So we have a singleton $\lbrace x_n\rbrace$ which is not in the range of the injection; and we can define this as $d_X(i)$. $$\vphantom{space!}$$ Is this good enough? | |
| Feb 19, 2013 at 19:16 | comment | added | Asaf Karagila♦ | Hmmm. First thing we can observe is that all but finitely many elements of $X$ are sent to sets of the same cardinality (cardinals below $P(X)$ are linearly ordered of order type $\omega+\omega^*$, maps from amorphous to linearly ordered sets have finite range). So now I claim this, if no set was sent to the empty set then we are done; otherwise not all the singletons are in the range of $i$. Let $x_0$ be the element sent to $\varnothing$; if $\lbrace x_0\rbrace$ is not in the range we are done; otherwise pick $x_1$ to be the preimage of this singleton; and continue by induction [...] | |
| Feb 19, 2013 at 19:02 | comment | added | Joel David Hamkins | Asaf, that is a good suggestion for a critical case to check. If $X$ is amorphous, then you point out that $\aleph(X)$ does not inject into $P(X)$. But is there $d_X$ mapping every injective image of $X$ in $P(X)$ to a set that is missed? For amorphous $X$, it would seem that every injection of $X$ into $P(X)$ must be essentially trivial in some way. | |
| Feb 19, 2013 at 18:50 | comment | added | Asaf Karagila♦ | I have to admit that I still don't fully understand the question (but maybe it's this long day that's beating my mental faculties). In my experience counterexamples to things which work for $X+X\sim X$ would usually fail when we try to apply them for amorphous sets, in this case note that if $X$ is amorphous then $\aleph(X)=\aleph(\mathcal P(X))=\omega$. But beyond this I feel that I cannot be of much help in verifying if that which should hold, holds... | |
| Feb 19, 2013 at 18:29 | comment | added | Joel David Hamkins | Ah, yes, I think that would overcome the difficulty. I wonder if it is equivalent in general? | |
| Feb 19, 2013 at 18:24 | comment | added | François G. Dorais | I think it works when $X + X \cong X$. To make the recursion work, we need a function $g:\mathcal{P}_{\preceq X}(\mathcal{P}(X))\to\mathcal{P}(X)$ such that $g(A) \notin A$ for every $A \subseteq \mathcal{P}(X)$ with $A \preceq X$. What we have is a function $g':\mathcal{P}_{\cong X}(\mathcal{P}(X))\to\mathcal{P}(X)$. But if $X + X \cong X$ then defining $g(A) = g'(A \cup \lbrace\lbrace x\rbrace : x \in X\rbrace)$ works. | |
| Feb 19, 2013 at 11:27 | comment | added | Joel David Hamkins | I backed off my previous general claim about the Hartog number, since I don't see how to push the induction through. | |
| Feb 19, 2013 at 11:26 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Corrected remarks on Hartog
|
| Feb 19, 2013 at 4:42 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 552 characters in body
|
| Feb 19, 2013 at 4:13 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |