Timeline for large cardinal tree properties as properties of sheaves
Current License: CC BY-SA 3.0
35 events
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| Oct 19, 2012 at 0:08 | comment | added | Nate Ackerman | If you are interested here is a link to the slides of the talk I gave at the joint meetings a couple of years ago Sheaf Induction As for the paper, I am in the process of putting the final touches on it and hope to post it soon. | |
| Oct 19, 2012 at 0:08 | comment | added | Nate Ackerman | @O a: Thanks for the clarification. With regards to the connection between sheaves and trees, part of my thesis actually is dedicated to that connection and how one can generalize transfinite recursion to "well-founded" sheaves which aren't trees. While I had to discover the connection by myself, and while I haven't seen it anywhere else, it wouldn't surprise me of someone else had noticed it before me. | |
| Oct 15, 2012 at 22:39 | history | edited | o a | CC BY-SA 3.0 |
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| Oct 15, 2012 at 22:37 | comment | added | o a | Nate: the question is whether there is a similar characterisation for supercompacts, as well as a request for references: probably someone noticed that a tree is a sheaf before, and maybe used some category theory... | |
| Oct 15, 2012 at 22:36 | comment | added | o a | Dear Joel, you are right, the definition $\kappa$-tree is correct only for $\kappa$ inaccessible, as Nate writes. (Although I must admit that I only meant that a $\kappa$-tree is a sheaf, perhaps with other properties). For an arbitrary cardinal $\kappa$, one has to require size-of-level condition explicitly. | |
| Oct 15, 2012 at 22:33 | history | edited | o a | CC BY-SA 3.0 |
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| Oct 15, 2012 at 21:01 | comment | added | Nate Ackerman | @O a: Also, could you please highlight exactly where your question is in the post. I read through it and couldn't identify it. Thanks. | |
| Oct 15, 2012 at 21:00 | comment | added | Nate Ackerman | But if $F^*$ is the corresponding tree then $|F^*(\omega)| = |2^{\omega}|$ | |
| Oct 15, 2012 at 21:00 | comment | added | Nate Ackerman | @O a: I agree with Joel that your definition of a sheaf on $\kappa$ is guaranteed to be a $\kappa$-tree only if $\kappa$ is inaccessible. Further, for non-inaccessible cardinals, if $F$ is the sheaf corresponding to $T$, then it may be the case that $T$ is a $\kappa$-tree but there is some $\alpha < \kappa$ with $|F(\alpha)| \geq \kappa$. For example consider the tree $T^*$ consisting of all maps from $n\rightarrow \{0,1\}$ for $n \in \omega$ along with all constant maps $f_\alpha:\alpha \rightarrow\{0\}$ for $\alpha \leq \omega_1$. $T^*$ is clearly an $\omega_1$ tree. | |
| Oct 15, 2012 at 19:47 | comment | added | Joel David Hamkins | Dear O a, isn't your definition of $\kappa$-tree still wrong in light of the size-of-level issue? I don't see how you've addressed this. | |
| Oct 15, 2012 at 18:59 | comment | added | o a | Joel and Nate: I added the assumption of inaccessibility into my definition of weakly compact cardinal. I am not entirely sure it is really necessary but it is best to avoid this point here... | |
| Oct 15, 2012 at 18:58 | comment | added | o a | Nate: thank you for your corrections! I corrected the definition of a sheaf corresponding to a tree; really the union is the wrong thing there, one should actually do the inverse limit...I take care of it now, and hope it is correct now. | |
| Oct 15, 2012 at 18:55 | history | edited | o a | CC BY-SA 3.0 |
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| Oct 15, 2012 at 18:52 | comment | added | Nate Ackerman | 3) Generally $\cup X$ means take the union over all element in the set $X$ where as $\cup_{P} X(x)$ means take the union over the set $\{X(x): P(x)$ holds $\}$. So I believe it would be clearer to write $\cup_{\beta < \alpha} F(\beta)$ and not $\cup \beta < \alpha F(\beta)$ And finally to anyone who isn't o a who is reading this. I apologize if it isn't correct etiquette to put so many corrections in the comments, but I don't have enough reputation points to edit the post myself (and I would welcome being told of any other way to pass along the info if there is one) | |
| Oct 15, 2012 at 18:44 | comment | added | Nate Ackerman | (although the correct answer can be inferred from the context we have to read the rest of the sentence to understand what the object is, or even that it is a single object and not two). As an example, writing $\{f_i:\alpha⟶2,i<\kappa\}$ takes care of this issue. 2) I would suggest never writing a quantifier after the thing it is quantified. Specifically I would write $F(j+1)=\{g:(\forall i \leq j)g|\alpha\neq f_i\}$ instead of $F(j+1)=\{g:g|\alpha\neq f_i \forall i \leq j\}$ | |
| Oct 15, 2012 at 18:35 | comment | added | Nate Ackerman | Lastly, there are a couple of presentation issues which I think will make the post much cleaner (although obviously these are my personal preference). 1) When defining a sequence it is helpful to have some form of delimiter joining the description of the sequence as well as the conditions on the indexes. Specifically: $f_i:\alpha⟶2,i<\kappa$ it is unclear if you are defining a sequence here, or are defining a unique function with an index $i$ that you are requiring to be less than $\kappa$ | |
| Oct 15, 2012 at 18:33 | comment | added | Nate Ackerman | The second reason this proof doesn't work (or is at least incomplete) is that you haven't shown that $\kappa$ is regular. So, for example this proof doesn't show that $\beth_\omega$ doesn't have (WC). | |
| Oct 15, 2012 at 18:33 | comment | added | Nate Ackerman | Third, I don't believe your argument that (WC) implies inaccessibility is correct for two reasons. First you are making fundamental use of the assumption that $\kappa = |2^\alpha|$. However it is possible to have $\kappa$ not be inaccessible and yet for this still not to hold. For example consider the case $|2^{\omega}| = \omega_2$. In this situation $\omega_1$ is not equal to $|2^{\alpha}|$ for any $\alpha$. | |
| Oct 15, 2012 at 18:32 | comment | added | Nate Ackerman | While I am not trying to do self-promoting here you might want to check out my slides Sheaf Recursion to see an explanation in the case of $\kappa=\omega$. | |
| Oct 15, 2012 at 18:32 | comment | added | Nate Ackerman | Second, according to your definition $F(1) \subseteq F(\omega \cdot \alpha)$ for any $\alpha$. If $y \in F(1)$ then what is $y|_{\omega, 2}$ (the restriction of $y$ as an element of $F(\omega)$ to an element of $F(2)$? I think what you want to do is use the fact that every $\kappa$-branching tree can be viewed as a subtree of $\kappa^{<\kappa}$ and then use the fact that $\kappa^{<\kappa}$ is a collection of functions to define your sheaf. | |
| Oct 15, 2012 at 18:32 | comment | added | Nate Ackerman | @o a: Thanks for the updated version. It is much clearer what is going on. However I still do have several points. First, your definition of a Grothendieck topology isn't quite right as the covers aren't sieves (see Grothendieck topology). What you have is a Grothendieck pretopology. The difference is, approximately, the difference between a topology and a basis for a topology. | |
| Oct 15, 2012 at 16:27 | comment | added | o a | But here it is important that I am talking about a sheaf of functions. I can also talk about an arbitrary sheaf of sets, and then I do need to require $Lev_\alpha<\kappa$. That is, $\kappa$ has the tree property iff for every sheaf $T:\kappa^{op}\longrightarrow Sets$ of sets on $\kappa$, if for every $\alpha<\kappa$ it holds $0<|T(\alpha)|<\kappa$, then also $T(\kappa)$ is non-empty. | |
| Oct 15, 2012 at 16:17 | comment | added | o a | namely, if $Lev_\alpha$ is of size $\kappa$, from level $\alpha$ onwards start "chopping off" a single branch at each level (and similarly for $|Lev_\alpha|>\kappa$. Does this make more sense? | |
| Oct 15, 2012 at 16:15 | comment | added | o a | Joel: thank you, right, not every sheaf of functions on $\kappa$ is necessarily a $\kappa$-tree. (I do not claim this explicitly but I do claim this implicitly..sorry.). In set theory terminology, my requirement is that $\alpha$-th level should have size at most $2^\alpha$. Now, my claim about inaccessibility is for such "trees": that is, if every such "tree" of height $\kappa$ necessarily has a $\kappa$-branch, then $\kappa$ is inaccessible. Essentially it is the argument saying that it is boring to consider trees without the condition that $\alpha$'s-level is of size $<\kappa$: | |
| Oct 15, 2012 at 15:39 | comment | added | Joel David Hamkins | Please correct me if I am wrong, but I don't see anything in your sheaf account that implements the requirement for a $\kappa$-tree that the $\alpha$-th level should have size strictly less than $\kappa$, which is part of what it means to be a $\kappa$-tree. Similarly, I think your argument on inaccessibility is wrong, since in general the tree property on $\kappa$, that is, the assertion that every $\kappa$-tree has a $\kappa$-branch, is known NOT to imply that $\kappa$ is inaccessible. It is possible, for example, that $\omega_2$ has the tree property. | |
| Oct 15, 2012 at 15:18 | comment | added | o a | Joel: yes, I misstated the condition on accessibility; see the corrected proof in the post. I also clarified the passage from trees to sheaves, in response to your second question: informally, every limit stage in a tree splits into two steps: first to you pass to all the limits, and then pass to the subset appearing in the tree. | |
| Oct 15, 2012 at 15:16 | comment | added | o a | Nate: I corrected the post. (i) $U_i$'s is the collection covering $U$; (ii) Yes, X=Y (a misprint) (iii) yes | |
| Oct 15, 2012 at 15:15 | history | edited | o a | CC BY-SA 3.0 |
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| Oct 15, 2012 at 15:09 | history | edited | o a | CC BY-SA 3.0 |
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| Oct 15, 2012 at 14:44 | comment | added | o a | Thank you for your comments; oh indeed there are some misprints. I'll update the post with some explanations and corrections. | |
| Oct 15, 2012 at 14:18 | comment | added | Nate Ackerman | I am confused by your notation. In your definition of the Grothendieck topology you say $P(\gamma) = X: X \subseteq \gamma$ do you mean $P(\gamma) = \{X: X \subseteq \gamma\}$? Also, you say "$U_i \subseteq U_i$ form a covering..." I am unsure what that means. What is the collection of sets here? What are they covering? Also on that same line you have both $X$ and $Y$. Do you want them all to be the same? If not what is $X$ here? Finally, am I correct that when you define $(P(\gamma), \subseteq)$ you want a $\kappa$-subscript? | |
| Oct 15, 2012 at 12:22 | comment | added | Joel David Hamkins | I don't follow you, but this is probably my fault because I don't usually think in terms of sheaves. It seems to me that one could have an $\omega_1$-tree in your sense, whose $\omega^{th}$ level had size continuum. Could you explain why not? (Set theorists define that an $\omega_1$-tree is a tree of height $\omega_1$ with all levels countable.) I don't understand your remark about inaccessibility, since $2^{\lt\kappa}\geq\kappa$ is true for every cardinal $\kappa$. | |
| Oct 15, 2012 at 11:33 | comment | added | o a | Thanks; Yes, I think it does imply that $\kappa$ is inaccessible: I am talking about a sheaf of functions, i.e. $T(alpha)$ is a set of functions $\alpha\longrightarrow 2$. If $2^{<\kappa}\geq\kappa$ then there is a sheaf of functions on $\kappa$ that has no global section i.e. no global branch: just make sure $f_{\alpha}\not\in T(\alpha+1)$ where $f_i$'s are some enumeration of functions in $2^{<\kappa}$. | |
| Oct 15, 2012 at 10:57 | comment | added | Joel David Hamkins | Does your sheaf account of $\kappa$-tree somehow include the requirement that every level of the tree has size less than $\kappa$? This is free if $\kappa$ is inaccessible, but one can consider the tree property on non-inaccessible cardinals. For this reason, should you add the phrase `and $\kappa$ is inaccessible' to your account of weak compactness? A cardinal is weakly compact if and only if it is inaccessible and has the tree property. | |
| Oct 15, 2012 at 8:25 | history | asked | o a | CC BY-SA 3.0 |