# clique concept – Ramsey ultrafilters on partial organize Answer

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## clique concept – Ramsey ultrafilters on partial organize

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Recall the next equal definitions of a Ramsey ultrafilter through $$omega$$:

Set (Ramsey ultrafilter). Permit $$U$$ breathe a non-major ultrafilter over $$omega$$. TFAE:

1. For every partition $$F: [omega]^ n to okay$$, there’s a homogeneous clique $$H in U$$.

2. For every partition $$F: [omega]^ 2 to 2$$, there’s a homogeneous clique $$H in U$$.

3. For every partition of $$omega$$, $${A_n: n in omega }$$, there’s a consequence $$langle {x_n: n in omega} rangle$$ in order that $$x_n in A_n$$ and $${x_n: n in omega } in U$$.

4. Given a reducing sequence of units $$A_0 supseteq A_1 supseteq cdots$$, there’s a lot $${x_i } _ {i in omega} in U$$ in order that for everybody $$n in omega$$, $$x_ {n + 1} in A_ {x_n}$$.

I might love to generalize this time period to countable posets and I’m extra all in favour of gauge (4) particularly. More exactly, let refer $$( P, leq)$$ breathe a countable poset, and let $$U$$ breathe an ultrafilter over $$P$$. Let’s say we are saying that $$U$$ is a Ramsey ultrafilter Above $$P$$ if for a partition $$F: [P]^ n to okay$$, there’s a homogeneous clique $$H in U$$.

I need to attain an equivalence that’s roughly as follows:

Supposition. Permit $$U$$ breathe a non-principal ultrafilter over a countable poset $$( P, leq)$$ (with presumably extra assumptions required on $$( P, leq)$$). TFAE:

1. $$U$$ is Ramsey. That is, for every partition $$F: [omega]^ n to okay$$, there’s a homogeneous clique $$H in U$$.

2. (?) Permit $$F = {A_p: p in P }$$ breathe a household of subsets of $$P$$ in order that $$p . Then there’s a lot Q subseteq P , Q in U so for everybody p, q in Q , if p then q in A_p .$$

If that is usually grievance, what further assumptions ought to we make? $$( P, leq)$$? Several assumptions I might love to make are:

1. Each train in $$P$$ is effectively sorted.

2. For all $$p, q in P$$, there are solely a finite variety of them $$r$$ in order that $$p .$$

But I need to remain away from the linear organize illustration.

TO EDIT: Allow me to elucidate the primary obstruction I countenance in generalizing equivalence: The most accessible proof of (3) $$implies$$ (4) appears to breathe Jech’s bespeak, Lemma 9.2. His proof is as follows:

[Let] $$X_0 supseteq X_1 supseteq cdots$$ breathe clique $$D$$ [where $$D$$ is a Ramsey ultrafilter]. Since $$D$$ is a p level, exists $$Y in D$$ so everybody $$Y – X_n$$ is finite. Let’s outline a sequence $$y_0 in and as follows: y_0 = the least y_0 in Y in order that {y in Y: y> y_0 } subseteq X_0 . y_1 = the least y_1 in Y in order that {y in Y: y> y_2 } subseteq X_ {y_0} . dots y_n = the least y_n in Y in order that {y in Y: y> y_ {n-1} } subseteq X_ {y_ {n-1}} . For every n , Permit A_n = {y in Y: y_n. Since D is Ramsey, there’s a lot {z_n } _ {n = 0} ^ infty in order that z_n in A_n for all n . We behold this in everybody n , z_ {n + 2} in X_ {z_n} : Since z_ {n + 2}> y_ {n + 2} , we’ve got z_ {n + 2} in X_ {y_ {n + 2}} , and since y_ {n + 1} geq z_n , we’ve got X_ {y_ {n + 1}} subseteq X_ {z_n} and due to this fact z_ {n + 2} in X_ {z_n} . So if we let a_n = z_ {2n} and b_n = z_ {2n + 1} , for all n , then both {a_n } _ {n = 0} ^ infty in D or {b_n } _ {n = 0} ^ infty in D ; and in each circumstances we get a sequence that [the property (4)].$$

For common posets, we won’t simply splinter it into two subsets $$a_n = z_ {2n}$$ and $$b_n = z_ {2n + 1}$$. If there are infinitely many branches, we can not assure that such a subset will breathe included $$D$$. There is too no level in fascinated by it $$sigma$$-complete Ramsey ultrafilter.

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