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8 votes
4 answers
630 views

closure of separative quotients

Does there exist a partial order, nontrivial for forcing, that is countably closed, but whose separative quotient is not countably closed? Supposing the answer is yes, then is there a partial order, ...
5 votes
1 answer
356 views

Calculate the $\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$ cofinalities of the poset of nontrivial finitary partitions of $\omega$

Let $(P,\le)$ be a poset. For a point $x\in P$ let $${\downarrow}x=\{p\in P:p\le x\}\quad\text{and}\quad{\uparrow}x=\{p\in P:x\le p\}$$be the lower and upper sets of the point $x$, and for a subset $...
3 votes
2 answers
432 views

When is a filter generated by a (countable) chain?

In any partial order $(P,\leq)$ it is easy to see that every chain generates (i.e., by taking the upwards closure) a filter, and any filter built as a result of the Rasiowa-Sikorski lemma in forcing ...
4 votes
3 answers
395 views

Problem understanding a passage of the proof of $\mathfrak{p}=\mathfrak{t}$ involving forcing

I've a problem with a passage of the proof of Claim 14.7 of the paper "Cofinality spectrum theorems in model theory, set theory, and general topolgy" by Malliaris and Shelah, or equivalently ...
11 votes
2 answers
682 views

On Applications of Forcing in Domain Theory

An interesting feature of domain theory is to use partial orders in order to provide a mathematical model for the computational approximation in a potentially infinite computational process (e.g. ...
3 votes
2 answers
689 views

A problem about posets similar to Suslin's problem

Suslin's problem is: Given a complete dense linear order without endpoints, if it has the ccc must it be isomorphic to $\mathbb{R}$? The answer is that it's independent of ZFC. The related ...