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As follows from this talk Large Properties for Small Cardinals, p.7,p.4 http://www2.dm.unito.it/paginepersonali/viale/SEMINARS-TORINO/Fontanella-Torino-19.1.2012.pdf, the definitions of weakly compact and strongly compact large cardinals can be formulated stated in the language of sheaf theory. I am wondering whether sheaf theory can be of any help, and whether supercompactness can be reformulated in a similar way.

Let me describe these reformulations.

For an inaccesible cardinal $\kappa$, a $\kappa$-tree (in set theory) is a sheaf of 0-1 valued functions on ordinal $\kappa$ as a topological space with the initial interval topology: a subset of $\kappa$ is open iff it is an initial segment $\alpha\leq\kappa$. A $\kappa$ tree has a cofinal branch iff the sheaf has a global section, i.e. $T(\kappa)$ is non-empty. For an arbitrary cardinal $\kappa$, a $\kappa$-tree $T(\kappa)$ corresponds to a sheaf $F:\kappa^{op}\longrightarrow Sets$ of 0-1 valued functions such that for every successor ordinal $\beta+1<\kappa$, $|F(\beta+1)|<\kappa$. (Corrections thanks to Joel and Nate's comments).

(WC) A inaccessible cardinal $\kappa$ is weakly compact iff any such sheaf of functions of $\kappa$ such that there is a section for every proper open subset $\alpha<\kappa$, has a global section, i.e. eqv., any $\kappa$-tree has a cofinal branch.

Given a cardinal $\kappa$, define a Grothendieck topology on (the partial order viewed as) the category $(P(\lambda), \subseteq)$ (where $P(\lambda)=$ {$ X: X\subseteq \lambda $} as follows: a collection $S=${$U_i\subseteq U$}$_i$ forms a covering of $U$ iff for every $Y\subseteq U$ of size $|Y|<\kappa$, there exists $U_i\subseteq U$ in $S$ such that $U_i\supseteq Y$. (By definition, a Grothendieck topology is a collection of these coverings; we shall denote $P(\lambda), \subseteq)_\kappa$ this category equipped with this topology.)

(SC) An inaccessible cardinal $\kappa$ is strongly compact iff forevery sheaf $T:(P(\lambda), \subseteq)_\kappa \longrightarrow Sets$ of Sets (with that topology), the following implication holds: if $0<|T(X)|<\kappa$ for every $|X|<\kappa$, then $T(\lambda)\neq \emptyset$, i.e. the sheaf has a global section. (Being somewhat vague, one may drop the assumption of inaccessibility and require instead that every sheaf of functions has a global section). Set-theoretically, this is said as any $(\kappa,\lambda)$-tree has a cofinal branch.

If I understand correctly, an interesting question is whether for every(?) $\kappa$ every sheaf of sets with the condition $|X|<\kappa$ implies $|T(X)|<\kappa$ is consistent.

Let me say more about the definitions of sheaves and how to pass from a tree to a sheaf. Recall (as e.g. stated in the talk above) a $\kappa$-tree $T(\kappa)$ can be viewed as a subset of $2^{<\kappa}$. Define a sheaf $F$ on $\kappa$ as follows: (i) for limit ordinal $\alpha$ $F(\alpha)$ is the set of all functions $f:\alpha\longrightarrow 2$ such that for every $\beta<\alpha$, the restriction of $f$ to $f|\beta:\beta \longrightarrow 2$ is in $T(\kappa)$. (ii) for a successor ordinal, $F(\alpha+1)=Lev_{\alpha+1}(T)$ where $Lev_{\alpha+1}(T)$ is the set $2^{\alpha+1}\cap T(\kappa)$ of functions $f:\alpha+1\longrightarrow 2$ of tree $T(\kappa)$.

The condition (*) is the sheaf condition for this topology: $\alpha=\cup{\beta<\alpha}\beta$ is an open covering of open set $\alpha$, and (*) says that every function on the open set $\alpha$ is uniquely defined by its restrictions on the sets in an open covering (and vice versa, any compatible set of functions on $\beta$'s defines a functions on $\alpha$). Note that this condition is only non-trivial for limit ordinals: for successors there is no non-trivial covering.

Finally, let me say why (WC) implies that $\kappa$ is inaccessible. Let $\kappa=2^\alpha$ and $\alpha<\kappa$. Let $f_i:\alpha\longrightarrow 2, i<\kappa$ be an enumeration of $2^\alpha$. Define $F(j+1)=${$g: g|\alpha\neq f_i \forall i\leq j$} for every ordinal $j<\kappa$, and at limit stage define $F$ by sheaf property $(*)$ above. Then $F(j)$ is non-empty for every $j<\kappa$, yet $F(\kappa)$ is empty.

PS: I thank Joel and Nate for many corrections.

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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. –  Joel David Hamkins Oct 15 '12 at 10:57
    
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 $2^{<\kappa}$. –  o a Oct 15 '12 at 11:33
    
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 $\kappa$. –  Joel David Hamkins Oct 15 '12 at 12:22
    
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? –  Nate Ackerman Oct 15 '12 at 14:18
    
Thank you for your comments; oh indeed there are some misprints. I'll update the post with some explanations and corrections. –  o a Oct 15 '12 at 14:44
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