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New answer. For any sequence $a=\langle a_\alpha\mid\alpha<\ell(a)\rangle$, define $\Box a$ to the sequence $\langle b_\alpha\mid\alpha<\ell(a)\rangle$, where $b_\alpha=\min\{a_\beta\mid\alpha<\beta\} …

The answer is no. In the Boolean algebra $P(\omega)/\text{Fin}$ every strictly descending sequence $A_0>A_1>A_2>\cdots$ has a nonzero lower bound, by the famous construction of Hausdorff. Namely, one …

The easy case occurs when $\kappa^{<\kappa}=\kappa$. Under GCH, this includes every regular cardinal. As Emil mentions in the comments, the result for this case follows from general model-theoretic co …

As Sam mentioned in the comments, the answer to question 1 is yes. One can map every condition $p$ in the partial order to the lower cone, the set $S_p=\{q\in P\mid q\leq p\}$ of conditions below $p$. …

Yes, $P(\omega)/\text{fin}$ is fractal. If $A\subseteq^* B$ but not equivalent, then the interval $[A,B]$ in $P(\omega)/\text{fin}$ consists of the sets that almost contain $A$ and are almost containe …

Every such cofinal family $\mathcal{A}'$ must have size continuum. The reason is that there is an almost disjoint family $\mathcal{D}$ of size continuum, a family of infinite co-infinite sets $A\subse …

Let me mention one easy observation, which is that the order $2^{<\kappa}$ is universal for all linear orders of size $\kappa$. View the order as $\{-1,1\}^{<\kappa}$, where strings are ordered lexica …

Consider the surreal line $\langle\newcommand\No{\text{No}}\No,\leq\rangle$, in its order structure only. This is a proper class linear order, which is universal for all set-sized linear orders, as we …

The usual definition of ideal also includes the requirement that $F$ is closed under unions. But in this case, I claim that Chvatal's property holds for all infinite ideals. First, if $a\in Y\in F$, t …

A partial order relation $\leq$ on a set $A$ is a binary relation that is reflexive, transitive, and antisymmetric. A preorder relation $\unlhd$ (also sometimes known as a quasi order or pseudo order) …

Regarding the dense embedding, perhaps this is helpful. Statement 1 can be taken as a definition of density, which makes the connection with topology by means of the lower-cone topology. Theorem. Supp …

Here is a counterexample showing that the quotient of $\text{Fct}(X,Y)$ is not necessarily a lattice. Let $X=\{0,1,2\}$ and let $Y=\{0,1\}$. Let $g(0)=g(1)=0$ and $g(2)=1$, while $f(0)=0$ and $f(1)=f …

This is a fantastic question! I spent the whole morning thinking about it, and I finally have a solution. The answer is no, not necessarily. To build a counterexample, I claim first that there is a …

If one takes the reflexive order relation as fundamental, then this is just the strict product order. It is good practice to take the reflexive order relation as the primary relation, since in the c …

Here is an example, which I think is similar to what you are asking for. A graded digraph is a structure $\langle A,\rightharpoonup,\leq\rangle$, where $\langle A,\rightharpoonup\rangle$ is a digraph …