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A partial order $\mathbb{B}$ is universal for a class $\cal{P}$ of partial orders if every order in $\cal{P}$ embeds order-preservingly into $\mathbb{B}$. For example, every partial order $\langle\ma …

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 …

Suppose that I have a partial order P and an increasing sequence $a_0< a_1<a_2<\cdots$ of elements of $P$. A pair of elements (b,c) from P is said to be an exact pair for this sequence, if Both $b$ …

Question. How many (isomorphism types of) finite groups of size at most n are there? What is the asymptotic growth rate? And the same question for rings, fields, graphs, partial orders, etc. Motiva …

You are looking for the concept of saturated model. A model $M$ is $\kappa$-saturated if any type consisting of fewer than $\kappa$ many assertions that is consistent with the elementary diagram of $M …

Which are the rigid suborders of the real line? If A is any set of reals, then it can be viewed as an order structure itself under the induced order (A,<). The question is, when is this structure ri …

Yes. By Hartogs' theorem, there is an ordinal that has no injection into $R$. The minimal such ordinal is the smallest well-ordered cardinal not injecting into $R$. It is naturally well-ordered by the …

Yes. This is both studied by set theorists and interesting. I personally find some of the related questions below extremely interesting, connected with some very deep questions about the nature of mat …

Let's think about the countable case like this: think of the binary tree $2^{\lt\omega}$, which has size $\omega$, but has $2^\omega$ many branches. Each branch describes a cut in the natural lexical …

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) …

Very nice question! They are not isomorphic. What I claim is that when we take the quotient with respect to density, there is a countably infinite antichain above $0$ having a minimal upper bound, b …

The answer is yes in ZFC. We can construct a dense infinite set $A\subset\mathbb{R}$ such that the only order-preserving map $f:A\to A$ is the identity. In particular, $A$ is not order-isomorphic with …

There is no such recursive ordinal, because in fact every computable ordinal is the order type of a polynomial time computable relation on $\mathbb{N}$. In other words, the least ordinal not describab …

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 …

This is a partial answer, and I am unsure about part of it. I claim that these functions are well-ordered by eventual domination, and the order type is at most the ordinal $\epsilon_0$. First, your …