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clique concept – Ramsey ultrafilters on partial organize Answer

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clique concept – Ramsey ultrafilters on partial organize

$ newcommand { U} { mathcal {U}} $
$ newcommand { P} { mathbb {P}} $
$ newcommand { Q} { mathbb {Q}} $
$ newcommand { F} { mathcal {F}} $
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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